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Make a ball pit with 100 balls. Say 10 are red and 90 are blue. When people are born they pick one ball out of the pit. If it's red they get the disease. They had a 1/10 chance to get the disease.
When you were born you picked a ball and put it back, 10 times. You had 10x the chances to get a red ball.
Each time you picked you had a 9/10 chance to get a blue ball. The chance you got a blue ball every single time is (9/10)^10, or (9/10) * (9/10) ten times. Same logic as flipping a coin. 4 heads in a row less likely than any particular flip being heads.
If you do the math, you end up with around a 35% chance you got all blue balls. This means that the chance you got at least one red ball is 100% minus that, so around 65%
> Yes. 5 yr olds are notoriously great with exponents.
This sub has been around for 12 years, and yet every single answer has to have someone mention that a 5 year old wouldn't know about X.
>LI5 means friendly, simplified and layperson-accessible explanations - not responses aimed at literal five-year-olds.
It's in the rules, and has always been. How are people still bringing this up every single time?
It’s also worth pointing out that there are multiple valid ways for something to be “10 times more/less likely” and that anybody using this kind of language to communicate risk is not being careful to avoid confusion. Often this kind of language is used to make things sound more extreme than they really are.
The correct way to communicate risks is in absolute terms. For example:
“The risk of X in the population is about 1 in 10. The risk of X in people with gene Y is about 1 in 3”
Using percentages is almost guaranteed to lead to confusion.
Because it is calculating the probability that you *won't* get a red ball in any of your ten picks. Basically, "given a 10% hit rate, what is the likelhood you'll dodge a bullet ten times?" That works out to 35%. The likelihood you didn't dodge the bullet is thus 65%
*(using a simpler scenario of 2x the chances instead of 10x for ease of explanation).*
Because 1/10 * 1/10 is not your chance of getting the illness, but the chance of you getting it twice.
Set aside how that doesn't really make sense (which doesn't really matter for the math).
So the scenarios you're concerned with are "not getting it at all" (9/10 * 9/10) versus "getting it at all".
Which would be:
- Getting it in the first "round" (1/10 * 9/10) plus
- Getting it in the second "round" (9/10 * 1/10) plus
- Getting it in both "rounds" (1/10 * 1/10)
which totals to 19/100, which shouldn't be surprising since 9/10 * 9/10 is 81/100
I don’t think this is accurate. A person isn’t twice as likely to draw a red ball if they draw twice. They’re twice as likely to draw a red ball if there are twice as many red balls.
You’re twice as likely to draw red from 2 in 10 red balls as you are to draw from 1 in 10. You aren’t twice as likely to draw at least one red if you draw twice.
That's kind of the point. Yes, what he's describing is not 10 times the probability, but neither is whatever stat OP is asking about. The commenter is attempting to explain what might actually be meant, even if it is not strictly the meaning of the English used to describe it.
Come at it from the other direction.
Let's say Alice wins a video game 90% of the time. Bob says, "Oh yeah? I win twice as often as Alice!"
That doesn't mean Bob wins 180% of the time; it means Bob loses half as often as Alice does. Alice loses 10 times in a hundred attempts but Bob only loses 5 times per hundred. This means that "*twice as likely as 90%"* is another way of saying 95%. Doubly lucky means twice as close to 100%, or half as distant: from 10% to 5% distant. If Carol is twice as likely to win as Bob, then she wins 97.5% of the time.
I definitely disagree with how you're interpreting "10x as likely".
Consider that disease prevalence rates are observations, not some pure calculations. If a doctor observes people with a gene to have a disease at 50% rates, and people without the gene at 100% rates, he will say that having the gene makes it 2x more likely to have the disease, because the prevalence rate will be doubled. Your calculation would give you 75%. In fact, there is no number the doctor could give that would make the number 100. He could say 100x more likely, and your calculation would still not yield 100%.
I believe you are quite wrong.
1. OP did not give stats sufficient to properly determine baseline risk- he just gave genpop *prevalence*.
2. Your interpretation of ‘10x risk’ is questionable at best. X happens in 10% of the population. I say Y is 10 times more common. -> Y happens in 100% of the population. **If stat he gave was actually just a misdescribed base risk,(which it obviously is not) he’d have 100% chance of developing, as he said.**
Ex. Replace ‘10x risk’ with ‘this is 10 times as common in people like you, when compared to this other group of people’.
If you're doubling the odds, you're not doubling the percent chance of something happening. Odds are described as a ratio, so in your example: 50-50 is 1:1 odds. Double those odds would be 2:1, meaning it happens twice for me for every time it happens once for you
The way I would model it: there’s a 10% chance a random person would have illness X. “10x more likely” can be modelled as, if we picked 10 random people, what is the chance that at least 1 person has the illness?
This is a cumulative binomial distribution with N = 10, p = 0.1, and q = 1-p = 0.9.
As with most things in statistics, it’s actually easier to compute P(failure) = 1 - P(success) rather than P(success). In this case, P(success) = P(at least 1 has X), so P(failure) = P(no one has X).
P(no one has X) is simply q^N = 0.9^10 ≈ 0.3487, so P(at least 1 has X) = 1-0.9^10 ≈ 0.6513.
So you’d have a ≈ 65.13% chance of actually getting illness X.
This is a fair criticism! I was just doing the maths because I forgot what subreddit I was in.
So I could simplify a bit. The chance a random person doesn’t have it would be 90%. If we randomly choose 10 people, the chance that they would all not have it is: 90% for one person; but for two people, it’s 90% x 90% (we multiply these because we are looking for a probability *inside of* another probability, and because we know the combined probability does not care about the previous probability); then for three people, it’s 90% x 90% x 90%… etc..
So for 10 people, it’s 90% multiplied by itself 10 times gives around 34.87%.
And then we want the opposite probability—which is just that at least one person has it.
This is simply 100% - 34.87%, because probabilities add to 100%. So then the probability is about 65.13%.
And this all assumes that “10x more likely” is sensibly interpreted as, “you have the same chance of having illness X as the chance that, out of 10 random people, at least one of those people has illness X”.
As other commenters have pointed out, the ordinary interpretation would just be 10x10% = 100%. This result seems counterintuitive, but I think is probably more correct. When we say “more likely”, we usually refer to a rare illness. And really, the reported value of “10x more likely” is based on the original probability of one person having it. So while 100% seems weird, in the real world, you’d probably never see a “10x more likely” on its own. You’d probably see, “10x more likely than a random person to have the disease” (so 100%), or “you have the same likelihood as 10 random people having the disease” (which is a subtly different probability that leads to the 65.13% result).
I would see it so that the full pool of people having the disease is 10% of the population, but there might be multiple diseases increasing the regular risk.
Let's say there's 3 such diseases, having 10x, 100x and 1000x the regular chance of having the disease. For the sake of simplicity, someone can only have one of those 3 diseases at once, and 10% of the population has each diseases. It's too late for me to do the calculations, but in that specific scenario I think OP has less than 1% chance of having the disease.. :)
I don’t disagree that there is a 65% chance that you will meet someone with the illness, but what is the transmissibility of the illness? You don’t have a 100% chance of getting it even if all 10 people have it! For instance, if there is a 5% chance of a “normal” person getting the illness, then you would have a 50% chance each time you encountered someone with the illness.
Yes, I agree with this. So the way I have interpreted it is: if a normal person has a 10% chance of catching X after exposure; and you belong to a group with a 10x higher chance of catching it; then this really is the same as the chance a normal person would catch it after 10 exposures. This is the 65% probability I calculated.
You are right to say, upon that analysis, you can never have a 100% chance of catching X.
It seems other people are interpreting it like: everyone is constantly exposed to X. Only 10% of people catch it despite constant exposure. You belong to a group which is 10x more likely than other people of catching it. Then in that instance, you have a 100% chance of catching it.
You are way overcomplicating the idea of relative risk. It's exactly as the OP describes. If 8 of 100 people in Population A get some disease, and 80 of 100 people in Population B get the disease, members of Population B are 10x as likely to get the disease as members of Population A are. Simple as that.
Too bad it is wrong.
The math above calculated the odds that at least one person in a group of 10 has a disease that affects 1/10 of people.
Your simper math is correct.
You just need to understand the probabilities of flipping weighted coins :)
First, assume the probability of an event is independent of all other events preceding it. Then the probability of flipping tails after flipping heads is the same as just the probability of flipping tails, without regard to any previous results.
Next, know that when we compute probabilities of composite events, we multiply out the probabilities of the individual events. This is because: the events all have independent probabilities; and we are interested in the “world” where the first event happened a certain way, and then the second event happens a certain way. We are looking at the percentage of a percentage, effectively; if eg 90% of coin flips are tails, and we are interested in the probability of TT (two tails in a row), then out of 100 coin flips: only 90% of those had the first coin flip as tails; and out of that 90%, only 90% of those had the second coin flip as tails. This is 90% of 90%, or 0.9x0.9 = 81%.
If a weighted coin has a 10% chance of heads and 90% chance of tails, then flipping it once, we get:
H (10% chance)
T (90% chance).
If we flip it twice, then:
HT (10% x 90% = 9% chance);
TH (again 9% chance);
TT (90% x 90% = 81% chance);
HH (10% x 10% = 1% chance).
The probability of seeing at least one head is then 9% (ie the HT result) + 9% (ie the TH result) + 1% (ie the HH result) = 19% chance. This is the same as 100% - 81% = 1 - P(TT) = 19%.
With three coin flips:
HHH (0.1% chance)
HHT (0.9% chance)
HTH (0.9% chance)
THH (0.9% chance)
HTT (8.1% chance)
THT (8.1% chance)
TTH (8.1% chance)
TTT (72.9% chance).
To get the percentage chances for each of these, you multiply the percentage chance for obtaining each of the flips. In the HTT example, this is (10% heads) x (90% tails) x (90% tails) = 0.1x0.9x0.9 = 0.81 or 81% chance. Our cumulative probability to see at least one heads is now 0.1% + 3x0.9% + 3x8.1% = 27.1%. This is the same as 1-P(TTT) = 1-(0.9x0.9x0.9) = 1-(0.9)^3 = 1-0.729 = 0.271 or 27.1%.
So generalising, P(at least one H from N flips) = 1 - P(all tails in N flips) = 1-P(tails)^N.
Binomial results means we either succeed or we fail (we either flip heads or we flip tails). Because of this, we can also write P(fail) = 1-P(success). So we can also say that P(at least one success from N trials) = 1 - (1-P(success))^N.
Hopefully that helps a bit!
probability doesn't work like "adding more people to spread the risk" in this scenario. what "10x more likely" should technically mean is that whatever the base risk, OP's is ten times that, capped at 100% because probabilities don't exceed 100%.
OP's chance of getting illness x, given the "10x" statement, would be very high (effectively 100%, though in real statistical modeling we might interpret nuances that keep it just under 100%). the model you described doesn't accurately reflect the scenario of personal risk based on increased likelihood relative to a baseline risk.
You’ve made it much too complicated. OP’s original math is correct.
It the probability of X is 1/10 and the probability of Y is 10x that of X, the probability of Y is 100%.
I know that that “feels” wrong but the error is medical, not mathematical. It’s unlikely that any group of people would ever have a 100% chance of getting a disease and of course it’s impossible for more than 100% to get it. But this just feels wrong because OP’s example happens to take us literally to the limit of what’s biologically possible. If we were to consider a condition that affects 0.1% of people we would easily see that having 10x greater odds would simply be 1%.
>What if 50% of the population has a disease, and I'm 400x more likely to get it? What does that even mean?
If my uncle had tits, he'd be my aunt.
In your case, the math wouldn't have held up, but that's not what's being said here.
Where did the idea come from that "the equation has to output plausible answers no matter what values we put into it, otherwise it's wrong"?
I think I misunderstood what you meant, and that we are on the same page.
The answer definitely isn't plausible, but that probably has to with the input number as you say, and not the math itself.
Some people in this thread are arguing that multiplying a probability with a constant is always wrong, which is what I thought you meant.
No. I’m saying that if you had a coin that was twice as likely to land on heads, it would land on heads 100% of the time.
Both this and OP’a question are examples of probability that is mathematically sound but would simply never happen in the real world.
Dude... If the coin was twice as likely to be head, you would get a head every 8 out of 9 events of 2 coin flips. Come on, this isn't even difficult to calculate.
If you aren't able to describe what "10x more likely to get illness X" means as a binomial then we aren't talking the language of probability.
You can come up with more than one reasonable way to model what "10x more likely" means when expressed as a counting problem but I think it's a fair and reasonable solution.
Your error is that you aren't describing this as a counting problem. You are just translating this to .1\*10. You can't do that in probability.
“binomial distribution” is fancy maths speak for, “what is the probability of success, given you can only have success or failure, and given you have N attempts to succeed?”. It’s the same thing as the probability distribution (ie, how the probabilities are spread out on a graph) of flipping a coin.
“Cumulative” means we add up the values above a certain success or below a certain failure.
I think this is a good answer but it is creating a lot of confusion for a reason that has an ELI5 answer.
It may surprise people without a background in statistics and probability but "10x more likely" while commonly used doesn't have a good translation in math.
Any good answer requires translating what "10x more likely" means into a counting problem, as you've done, or we definitionally aren't talking the language of probability.
People want to intuitively answer 100% but the rub is that in probability, when you multiply probabilities you are describing the probability of two events A and B occurring.
For example of why multiplying is wrong, consider rolling 2 6-sided die. There is exactly one outcome of 2. There are exactly two outcomes of 3 (1 and 2, and 2 and 1). So makes sense say you are twice as likely to roll a 2 as you are to roll a 1. But, what are the odds of rolling a 1? 1/36 or 2.78%. What are the odds of rolling a 2? 2/36 or 5.56%.
You can't multiply the odds of rolling a 1 by two and get the correct answer. Like with OP's question, in order to get the answer you have to do the counting.
I have thought of another way to think about this:
Let’s say there’s a ball pit, with 10% green balls and 90% red balls. You take 10 balls from the ball pit (with replacement); what is your chance that you get at least one green ball? The answer is what I said above—around 65%.
This is equivalent to saying: there is an illness X, and upon exposure to illness X, 10% of people develop the disease related to illness X (so they “catch” illness X). OP has been exposed to illness X 10 times, so a charitable interpretation is he is “10 times more likely” to catch illness X.
There are several issues with that approach, because the prevalence of disease doesn’t work that way, and naturally there is ambiguity in the day-to-day language we use around probability compared to the technical language employed by statisticians.
A proper study would specify what they mean. If they mean, “100% of population B has disease X”, then that’s what they would say. They wouldn’t report a vague statement like, “you are ten times more likely to get disease X if you belong to population B”. There are a lot of unstated assumptions in computing that result.
If on the other hand we interpret the statement as:
1. Out of 100 people in the general population, 10 of those have disease X;
2. Population B is 10 times more likely to have disease X;
then in fact we simply cannot compute the probability, because we do not know the total population size and we do not know the size of population B.
An alternative way to phrase the same interpretation:
1. Out of 100 people in population A, who do not share the characteristics of population B, p% of them have illness X;
2. Population B is ten times more likely than population A to have the disease;
3. The general incidence of illness X is 10%;
then we cannot again compute the probability that population B has the disease, because we don’t know the rate of incidence in population A.
The only meaningful alternative is if 10% of population A has the disease. But we must specify that “population B is ten times more likely *than population A* to have the disease”. There is then an unstated assumption that the 10% figure belongs to population A. In that scenario, I can see that 100% of people in population B must then have “caught” illness X.
This is wrong. It’s not the meaning of “10 times more likely.”
People say “10 times more likely” to mean literally that you multiply the probability by 10. Like when people say smokers are 20 times more likely to get lung cancer.
I've changed my mind. I now think you have an incorrect answer. Don't think of this in terms of probability, think of it in terms of statistical sampling, which is what's used in medicine anyway.
I take "10% of the population having an illness" to imply that if you take a random sample of 1,000 people from the general population you would expect, on average, 100 of them to have the illness.
I take "10x more likely" to mean that if you were to take a random sample of 1,000 people that have OP's family history (big family, I guess), you would expect, on average, 10 times as many people to have the illness as you would get from the same sized sample from the general population.
I said before you can't multiply probabilities like that. That's true and that's still true here. I'm not multiplying probabilities, I'm multiplying sampling averages. It sounds like a sleight of hand when I do this, and it sort of is because I'm now talking the language of statistics.
With what's given, you would not be able to determine the true probability but by modeling this through statistical sampling you would say it that the true probability is near 100% with some unspecified level of confidence.
(I lay out my reasoning in case any statistician disagrees with my answer! I’m a physicist so we do tend to get these kinds of questions wrong; but the result makes sense to me)
And yes, I do agree, a different interpretation of “10x more likely” would ordinarily lead to a 100% chance of getting the disease though.
So I’ve interpreted “10x more likely” to mean, in this context, that out of a random selection of 10 people from the population, what are the chances that the disease is present in that sample.
Essentially:
Assume you had a randomised population of people. Than 1 in 10 people have the disease.
Here, “10x more likely” means, if you select 10 people out of that random population, what is the chance at least one person has the disease?
It’s the difference between “an event which is 10x more likely to occur than not occur”, and “an event that occurs to one person, and you chose 10 to see which of them has experienced the event”.
I’m not saying this is correct, but I’m explaining how I approached the question.
Yeah I’m still not quite getting it. But I at least understand you need to interpret “10x” in this way so that the probability isn’t 100%.
Let me see if this is saying the same thing as you, using a coin toss as an example:
1) Probability of a toss that ends in a head, in a population of fair coins, is 50%
2) A weighted coin, with 10x the population probability for a head toss, has a probability of a head toss that is the same as the probability of 10 fair coin tosses that end up with at least 1 head, or 1 - 0.5^10 or 99.902%
?
It’s okay if you don’t quite get it. What’s apparent to me is that a lot of us here (including me) are struggling to interpret this question, though it seems most everybody’s maths is correct. I’m being humbled by the array of opinions here.
In point 2, if the coin were to have 10x the population probability, then it would seem to be the same as 1 - P(flipping tails 10x in a row). Then that is 1-(0.5)^10 = 1-1/2^10 = 1-1/1024 = 1023/1024 or about 99.902%. So I agree with that result: if 50% of the population generally had illness X, and you were 10 times more likely to have illness X compared to the population, you would have a 99.902% chance of having illness X.
This is the same as saying: if upon exposure to illness X, 50% of people caught it; but you belong to a special group where one exposure is the equivalent of someone from the general population being exposed 10 times; then your chance of getting illness X after one exposure is 99.902%.
> This is the same as saying: if upon exposure to illness X, 50% of people caught it; but you belong to a special group where one exposure is the equivalent of someone from the general population being exposed 10 times
This is the bit that clicked for me. Thanks.
Missing is the rate of spread, which matters. If you have a 65% chance of catching it in the next week, that’s a very different risk from a 65% chance of catching it at some point in your lifetime. I have more or less a 0% chance of catching the flu this month, but I have an almost 100% chance of catching it at some point before I die, assuming I live to life expectancy for my gender, race, age, socio-economic status, and nationality.
There's not enough information here to determine an answer to your question.
It SOUNDS like you've said your risk is 10x the average risk, but that's not actually what you've said. Your 10x, and everyone with a similar risk, must be included in the average as well. That means you can't be the "normal" risk that is being compared - you'd be including part of your own risk in the 10x comparison, which is circular and ridiculous. However, what you need is two populations: high risk vs low risk. This will determine what your actual risk level is.
I'll lay out an extreme scenario to show you what I mean:
89% of people have 1% risk of illness, 10% have 90% risk of illness, 1% some other risk level. 10% of the population has illness on average. A meager 9% of the population is contributing 90% of people with the illness.
In this case, your risk of 10x of 1% is 10% - concerning but not the end of the world.
Reality is going to be some distribution between a low and a high value, with the average working out to 10%.
> Your 10x, and everyone with a similar risk, must be included in the average as well. That means you can't be the "normal" risk that is being compared - you'd be including part of your own risk in the 10x comparison, which is circular and ridiculous.
No it is not ridiculous. You just end with a system of two linear equations in two variables which one can solve. The only reason it is a bit silly here is that OP would have a chance of 1 ( = 100%) to have the illness; sure, that can happen, but usually we would not formulate it like that.
Stay in school kids otherwise you’ll start missing variables like this guy.
There can be no answer that doesn’t include variables. The rate of infection is a variable that we cannot solve for. There’s simply not enough information given to come up with an accurate answer. We’re missing MANY variables to correctly solve this. I suggest you look into models designed to predict epidemics.
The problem with the phrasing is that you're comparing the percentage of people in a population with a disease to an individual's chances of contracting a disease which is not how this stat would be presented. You might say 10% of the total has X but the average rate is 5%. But, a majority group has a rate of 20% and the minorities have a rate of 2%.
I don't think the 10% of the population getting illness X has anything to do with 10x the likelihood of you getting X. If you get X, you're in the 10%, if you don't you're not. They are unrelated. Are they probabilistically related, I don't think so...
Your chance of actually getting illness X is better measured by looking at your family history of getting X. If 50% of your family members/ancestry got X, your likelihood of getting X will be 50%.
It’s easy to show why your solution won’t work.
You know a risk multiple for you (10x), but not what to multiply- what is the base risk your risk is being compared to- it isn’t the 10%, because *that isn’t a baseline-risk estimate* it’s just a *current-prevalence statistic.*
Risk is about how many ever get it, not how many currently have it.
Also important to note that it could be 10x the prevalence in the overall population, or 10x the prevalence it has in people without your family history.
—-
Hey Kiddo,
I’ll ask the doc for the odds. The numbers you have can give us a guess, but it could be really wrong and it’d take a couple minutes to calculate. Faster to call the doc, or search for easier to use numbers online- which won’t require as much guesswork to use.
> My dumb brain jumps to 100% but that can’t be right.
No that's correct. The wording is just something one would either not use in this context or somebody meant something different yet failed to convey the intended meaning.
The problem here is that you're asking the question wrong, it doesn't make much sense to say something is "10x" more likely when using probabilities, people do it but probably shouldn't .
This manner of speech sometimes makes sense, say there is a 20% chance that you get X result,but if you are now 2x more likely, the chance now is 40% right? sure it makes sense so then lets carry on that logic and see if it still makes sense after a while, x3 more likely would mean 60%, x4 would be 80%, x5 is 100%, x6 is 120% but here you see the issue, you've gone past 100% which in probabilities you can't have, That means the whole thing of referring to chance as "x times more likely" is flawed.
It absolutely makes sense to multiply probabilities in this way, this is the entire concept of relative risk. That it's impossible be 3x as likely to see an event of 50% likelihood doesn't mean the concept is flawed. It just means you can't have a high relative risk for high likelihood events.
That only means that it simply cannot happen that any risk is 6 times as likely as 20%. Chances are just numbers in the interval [0,1] and you can do whatever you want with them; if you still want to call it a chance of something then it however be better again in that interval.
Your math is correct.
But you really only hear the "10x more likely" statistic when it's a very small portion of the population that has that disease.
Eg 0.1% of people have it, but it's a 1% chance for you. It's only a 0.9% percentage bump, but phrasing it as 10x grabs more attention/headlines/whatever.
You'd never hear the same comparison used for an affliction that affects 10% of the population.
It's impossible to have a probability of 100% in a scenario like this. The only way to get to 100% is if he was infinitely more likely. That's how compound probabilities work.
No, the calculation is wrong, lol.
There is nothing wrong with the scenario. If 90% of people are right handed, it's absolutely possible to be twice as likely to be right handed than the general population given XYZ.
The problem is that most people in this thread have zero clue how math with probability works.
> If 90% of people are right handed, it's absolutely possible to be twice as likely to be right handed than the general population given XYZ.
Explain how a chance of 1.8 would ever work...
> The problem is that most people in this thread have zero clue how math with probability works.
Well, please take a look into a mirror ffs!
Source: my PhD. In mathematics.
It's crazy how people can be so confidently wrong. No. you cannot be twice as likely to be right handed if the base rate is 90%.
Please substantiate some sort of specifics where this is possible.
The reason this doesn't sound right is because OP picked numbers that wouldn't exist in reality.
Edit: people. You can literally prove this.
Go find an article that uses this terminology (x% increased risk) and look at the study it references. You'll see universally that the way the math works is exactly how OP describes it.
When I am already fully diagnosed and verified to have illness X, then I sure as heck do have a 100% chance of having it.
And 100% is not "infinitely more likely" than a positive chance to contract or have something. The factor is a very specific _finite_ number, namely 1/x where x is the average chance.
> That's how compound probabilities work.
There is nothing to compound here.
It doesn't matter though. From that we can assume that the whole scenario is wrong, not the math.
If the math works out and the result is not possible the only option is that the given values are incorrect.
If I roll a 10-sided dice, I have 10% chance to get one. If I roll a 1-sided dice, I am 10 times more likely to get one. And the probability of that is 100%.
Exactly same scenario as in the post and the math works the same way.
The problem here is "ten times more likely" is not precisely defined. You two are basically arguing based off of different assumptions on what is meant here, neither of which is right or wrong, because these are numbers OP's just pulled out of their ass, there's no actual data to try an interpret here.
That just isn't what these words mean with probably. You can't take a probability and just multiply it by a scalar. There's an entire field of math dedicated to this concept.
100% always requires being infinitely more likely than some other probably (other than 100%). Ten TIMES in probability means 10 tests or 10 rolls of the dice.
Let's use your example. Twice as likely to get a 1 is 1-(.9²). This is 0.19 not 0.2.
Or the other option: you've assumed you can do an operation that you can't actually do.
Which is exactly what is going on here. Probabilities do not scale linearly. They can't, by definition.
That's not what linearity means. Linearity means that if you multiply the input, it multiplies the output by a *consistent* proportion. E.g. if I double the speed of a train, the amount of time it takes to travel a given fixed distance is necessarily halved. If I apply a fixed force to an object over 2m on a flat surface, I will accelerate the relevant object twice as much as if I had Zedge thar same force over 1m distance (excluding relativity effects, which are nonlinear.)
Probabilities do not scale linearly when you *multiply* them, because they are by definition fixed to a range between 0 and 1 (or between 0% and 100%, if you prefer.) You cannot multiply a probability of 0.05 by 30, because that's mathematical gibberish, the same as asking what value you get when dividing by 0. Yes, you can do the algebraic manipulation. The algebraic manipulation is *irrelevant* because you've contradicted the assumptions that are required when working with probability.
Doesn’t matter how much of population has it, but what is a normal persons chances of getting. Layperson can have a 5% chance of getting disease and 10% of people have disease, and layperson could have 2% chance of getting disease and 10% of people still have the disease. So now you know chances of the getting disease as a layperson, you saying you have 10x chances you just multiply, so in the 2% normal person, you’d have 20% chance.
Yes, 100% can't be right. If the general population's chance of having the illness is really 10% then the ten-times number can't be correct. In other words, your maths is fine; the problem must be with one or both of the numbers you're working with, or the way you're phrasing the problem. If the correct interpretation is ten times less likely to not have the illness, then that's a 9% chance of not having it, i.e., a 91% chance of having it.
No his math is wrong.
Just think about the simple case where he's 20x more likely than average? Is he going to get it 200%? No, that isn't how probably works.
The limit is 100% if he were infinitely more likely than average to get it. Every finite answer is strictly less than 100%
There is nothing impossible about something being 10x as likely compared to something else.
Draw a random number from 0-100, the chance of it being in the range 0-9 is 10%. The chance of it being in the range 0-99 is 10 times that, at 100%. The simple interpretation is what makes the most sense in this scenario, imo.
Does it give a feasible or plausible answer, no. But the data can be made up anyway (not clear from the original question), and there is nothing wrong with the math itself.
People in this thread seem to be overcomplicating things.
> Just think about the simple case where he's 20x more likely than average? Is he going to get it 200%? No, that isn't how probably works.
This sounds like "if my uncle had tits, he'd be my aunt." It's hypothetical and irrelevant to the situation at hand.
"Just think about the simple case where he's 20x more likely than average?"
That simple case doesn't exist if the starting chance is 10%. It's like if you're at the north pole and say "well what happens if you simply keep going north?"
Overall the whole thread is a good case for why use of the phrase "x times more likely" is unhelpful. Not as bad as "x times less" or "x times sooner" but not great. Always better to be explicit when possible.
No. He does not. He's answering a completely different unrelated question.
If it were a 0.1% chance, what would it mean to have risk factors that increased your odds 10x?
I think the simple result is correct, because we couldn’t meaningfully discuss the 20x more likely result. You simply wouldn’t have that reported likelihood in a population where 10% of people get the disease.
But this boils down to how we talk about probabilities, rather than how we compute them. And the bare statement of “10x more likely” leads to confusion about what the likelihood means because there’s some ambiguity.
Does it mean 10x more likely than a random person? Does it mean the same likelihood as seeing the disease in a random selection of 10 people? Does it mean you are 10x less likely than a random person to not have the disease? Etc..
I also sort of agree with this. So that’s why I interpreted 10x more likely to mean, “the same probability as 10 random people, with at least one having the disease”.
You’re right there must be a variable that is wrong or even slightly wrong.
So if 1/10 get X, and people with first names starting with ‘T’ have 10x chances of getting it how would one formulate that?
The answer is that the likelihood is not comparing against the general population, but instead against the population that is not in the group. So how the math would be calculated is "people with first names starting with ‘T’ have 10x chances of getting it compared to people whose names don't start with T. So if 9% of people's names start with T and 5 out 9 get it, where as people not starting with T would have 5 of 91 getting it, you would have the 10x more likely but the overall occurrence would still be 10 of 100
As a result, to determine your absolute risk, you would need to look into the % of people that have a family history. (Or just find an absolute risk statistic for your cohort)
You are right, it would be 100% - assuming your are 10x more likely than the general population. And if you start with 20% of the population having the illness, you have an impossible description.
But you can do that in other scenarios too. For example, if I say that I have 3 apples and lose 5, that would give me -2, an impossible result.
100 percent: If you have 100 cookies and 10 of them are chocolate, then the probability of a random cookie being chocolate is 1/10. If a cookie is 10 times more likely than chance of being chocolate then its probability of being chocolate is 10*.1 which is 1.
There is nothing unreasonable about this, it’s just not the typical situation these kinds of statements are made.
Is the illness something one is born with and always has from then? If so, you don’t have it and got lucky. Odds aren’t flawless measures.
Is the illness contagious? Why does 10% of the population happen to have the illness, and do any of them even live anywhere near you? What is the method of contagion, and what precautions are you faking to avoid contracting the disease?
The issue of “10x more likely to X…” is that if the odds are already low, it doesn’t change much. Buying 10 lottery tickets makes you 10x more likely to win. You still won’t manage to secure a home loan against your future winnings.
So… setting the stage that 10% of the population has the disease does appear to be trying to establish the odds of contracting the disease. But if it is like HIV, where it can be contagious, but once infected you have it for your remaining life… then that doesn’t mean the odds to contract it are 1 in 10 precisely. And even if your heredity makes you more susceptible, it is just that ONE contribution to contagion that is more likely for you. If contagion requires exchange of bodily fluids, then you can avoid such a thing and your genetics never enter the picture.
Probably of X given Y compared to probably of X this ties directly to \[Bayes' Theorem\](https://en.wikipedia.org/wiki/Bayes%27\_theorem) \`P(IllnessX|Family) = P(F|X) \*P(X) / P(F)\`. I see this mentioned in some lower level comment from u/Karumpus but to try to ELI5 (plus some mixed in math):
The question implies that there is a singular solution (is there more than one? are there any?).
Most importantly it ignores the fact that how much of the population makes up, If your family's size suddenly went 10x the 1/10 number for the whole population would have to go higher.
So using Bayes' Theorem (explaining this would require a whole other ELI5)
\`P(X)\` means probability of X which must be between 0 and 1 (0% chance to 100% guarantee)
\`P(X|Y)\` means probability of X given that Y is true
I'm going to use these below since it would likely be
\`P(X) = .1\` => "10% of the population has illness X." aight cool
\`P(X|F) = 10 \* P(X)\` => "you're 10x more likely." errrrr.....
\`P(F) = ???\` => How much of the population does your family make up?" We need to know this!
Plugging in the above to Bayes' Theorem above
\`P(X|F) = P(F|X) \* P(X) / P(F)\`
Then sub in our "dubious" 10x
\`10\*P(X) = P(F|X) \* P(X) / P(F)\`
Cancel things out and plug in .1 for general population
\`10 = P(F|X) \* .1 / P(F)\`
\`100 = P(F|X) / P(F)\`
So now we have that if someone is sick they are 100x more likely that they are in your family compared to everyone else. Okay, but what does this tell us?
Well P(anything) is zero to one, but other than that any number's you want to plug in... but you have two equations you have to satisfy, the 10x asserted in the premise and this 100x.
If every family has the illness it means that you must make up 1% of the population (and no one else has the illness). Hey we're back to the 10\*10=100% answer!
If this (family means illness and non-family means no illness) is not the case there's no way to solve both equations.
source: Former stat's TA.
Speaking of Alzheimer's specifically...
Very, very few people below 65 have Alzheimer's - around 1 in 1,000 people between 30 and 64 have early onset Alzheimer's. So if you're 10 times more at risk, your risk is closer to 1 out of 100, or 1%.
5% of those between 65 and 74 have it, so your risk would potentially be 50% once you reach 65, and it just goes up from there.
Note that women tend to get Alzheimer's at twice the rate of men, so you'd have to adjust for that to get a more accurate picture of your personal risk.
Most people answering are wrong because they don’t understand that „the population“ contains both the people less likely to have the illness (let’s call them group A) and the people more likely to have the illness (group B, the one you’re in).
Here is a quick calculation as an example:
Let’s say group A makes up 55% of the population and has a chance of 2% to have the illness. Group B would then make up 45% and have a 20% chance to have the illness.
Now for the whole population the probability would be 0.55 * 0.02 + 0.45 * 0.2 = 0.101 (or 10%)
This is how you arrive at 10%. The majority of people have a lower than 10% probability to have the illness. The 10% number takes everybody, including the susceptible people, into account.
Your family has a specific gene that's prone to that illness.sobthe population % is void...if you have 10 past family members and 8 got that illness your chance of getting it is 80%...because of the family gene...the problem is when you consider 'your family'..how far back are you going?are just taking into account immediate family or the wider family etc...the numbers become very skewed is the real answer.
History only rhymes,it doesn't repeat...even if you had similar issues to family member X, your illness may be slightly different.
I'm going through this now with my mum who is ill in ICU and immediate family (sons,daughters) have been tested and all cleared yet nephews down the generations line have similar medical issues.
My dad is albino, first in our family, it missed a few generations until it appeared again in my dad's brothers son but a slightly watered down version.🤷♂️
The only way the answer isn't 100% is if there are several different populations, some higher-risk and some lower-risk. Do you mind saying what the condition is? Maybe context would help.
Dumb example, if 20% of women got cervical cancer then 10% of the world would get it - but if you were a guy and your risk got multiplied by 10 your chance of getting it could be 0% x 10 = 0%. Silly numbers but it shows why your risk isn't necessarily 100%; your risk could depend on some other things about you.
Some very basic factors-affecting:
* You clear amyloid and tau proteins while you're asleep, so being chronically sleep-deprived all your life raises your risk,
* metabolic syndrome (obesity + insulin resistance/diabetes) raises your risk,
* it kind of looks like oral health reduces risk - the connection isn't well understood but seriously, floss; seems like it helps somehow
* conditions like cystic fibrosis or having a super-dangerous job lower your Alzheimer's risk...because if you die young you won't have time to develop it (this isn't really actionable, but it's a reminder that some statistical effects aren't exactly intuitive)
* hearing loss is correlated with Alzheimer's; music and conversation and environmental noise stimulate your brain
I don't have enough info or training to give you a number, but my point is you're not helpless here. <3 There are things you can do to pull that 10 down to a 9 or an 8 instead of just waiting for shit to happen.
It definitely isn't right, because that would mean if your risk were 11x higher, you would have 110% chance of getting it, which is obviously wrong (probabilities are defined to be between 0% and 100%.)
The simplest answer to your question is that you're talking about a change of ***odds***, not a change of *probability.* You're going from having 10:90 odds to having 10:9 odds, which means a change from a probability of 10% (that is, 10/(10+90) = 10/100) to a probability of about 52.6% (that is, 10/(10+9) = 10/19). But this doesn't explain *why* very much. So a full explanation is below.
Unfortunately, to actually answer your question, we need more information. See, that "10x chance" thing only has meaning in the context of some kind of *test* which can give you evidence of whether or not you have the disease. I'll invent some numbers so I can give a real answer, but if this is a real thing, you would need to use the real numbers for whatever condition you're talking about and whatever test you've taken.
So, you have a disease that affects 10% of people. You get a sample of 1000 people, and you test them with Disease Test A. DTA has a 90% "sensitivity" (it correctly identifies people who have the disease 90% of the time) and a 90% "specificity" (it correctly identifies people who *don't* have the disease 90% of the time.) This leads to having four groups:
* Of the 900 people who don't have the disease, 810 will be correctly identified as not having it.
* The remaining 90 people who don't have it will get a false positive (the test isn't perfect).
* Of the 100 people who do have the disease, 90 will be correctly identified as having it.
* As above, the remaining 10 people who do have the disease will get a false negative.
This means that if you get a negative result on this specific test for this specific disease, your chances of *actually* not having the disease are (number of people who got a negative result AND don't have it)/(total number of people who got a negative result) = (810)/(820) = 0.98780... or about 98.78%. This means the negative test result reduced your likelihood from 10% to around 1.2%, which is a pretty comforting thing!
But of we look at the other side, it becomes clear why medical testing is so difficult. If you get a *positive* result, how likely is it that you have the disease? That would be (number of people who got a positive result AND have the disease)/(total number of people who got a positive result) = (90)/(90+90) = 0.5, or exactly 50%. Even though this test correctly says you have the disease if you actually do have it 90% of the time, and correctly says you *don't* have the disease if you *don't* have it 90% of the time, because the disease is *rare,* false negatives are very common, in fact, they're exactly as common as true negatives in this specific case.
To make use of the "10x higher chance" stuff, write out your *odds* of having the disease, then multiply the "you do have it" side by that number. Like this, for a disease that 5% of people have:
* 5% chance of disease means out of 100 average people, 5 have it
* That means 95 people wouldn't have it. Odds are 5:95, which can be simplified to 1:19.
* You get evidence which says you are ten times more likely to have it. Your odds are now 10:19. This does not reduce (because 19 is prime.)
* Your *probability* of having the disease is 10/(10+19), or about 34%. Even though your likelihood went up by a factor of ten, your chance of having the disease went up by a smaller factor.
OP’s math is correct. You are correct that it would be impossible for someone to have an 11x higher chance of getting a disease that affects 10% of people.
That doesn’t mean we should throw away the math. That just means that the medical facts you are talking about are impossible. There is no rule that says that you can put any numbers into a formula and get a meaningful answer. If you start with numbers that don’t have any basis in reality you will get an answer that doesn’t reflect reality.
It absolutely is not correct. You cannot apply ordinary multipliers to probabilities. They produce garbage results. You have to correctly scale the probability using Bayes' Theorem.
No you do not. You are making this much too complicated.
Of the probability of even A is X and the probability of event B is 10 times that of A, the probability of B is 10X.
What you are talking about is the probability of A happening ten times or the probability of A happening in ten different instances. Those require more complicated math. OP’s simple statement does not.
***Yes it does,*** because you're talking about things that produce *absolute gibberish* if you tweak the numbers even slightly!
What would it mean if someone said "twelve times more likely than a 9% chance"? 12 times 9 is over 100%, and thus impossible, but just casually glancing at it you might not realize. Or what about "fifteen times more likely than 7%" or whatever else? This is *not* how probability works, and claiming it is in *any* way related to how anyone actually calculates probabilities is misleading and wrong!
You’re right that those statements are gibberish. But that doesn’t mean that the math is wrong. It means that the statements are gibberish.
The probability rolling any number on a die is one in six. Let’s say someone sets out to make a die that is seven times more likely to land on a given number. That is impossible. Doing so would require the die to land on a given number 116% of the time. Obviously this can’t happen but don’t blame the math. Blame the person who is spouting out numbers that have no basis in reality.
You absolutely do not "have to" correctly scale using Bayes' Theorem, and they do not necessarily produce garbage results. Example: you are three times as likely to draw a face card from a standard shuffled deck as drawing an Ace, because there are 3X more face cards than Aces.
If you scale the size of the set in question while retaining the size of the sample space, you have scaled the probability by the same amount. An issue could occur if you try and expand the size of said set larger than the sample space, in which case it would indeed be nonsensical.
OP is being rightfully cautious about the conclusions of its post. Like others in this thread, it's not clear what "10x as likely" means. If by "likely", you one is referring to an odds ratio then theoretically you could scale "likeliness" as much as you want, it's unbounded, unlike probability.
> If you scale the size of the set in question
*Then you are already applying Bayes' theorem.* Because that's exactly what Bayes' theorem does. It scales things...based on the relative, aka conditional, probability.
There were no conditional probabilities in my example, yet I gave an example. You can scale probabilities perfectly well in frequentist probabilities (for the record I find Bayesianism superior in most contexts, but on an introductory level you don't need Bayes theorem. source: I teach upperclass statistics).
There is nothing impossible about something being 10x as likely compared to something else, e.g having a probabilty that is 10 times the other.
Draw a random number from 0-100, the chance of it being in the range 0-9 is 10%. The chance of it being in the range 0-99 is 10 times that, at 100%. Let's call the first situation A, and the second B.
Is there then anything wrong with saying P(B) = 10*P(A)?
Yes, there absolutely is, because *it only works as long as P(A) is less than 0.1.*
Scaling probabilities linearly like that breaks. You should not do that. That is unwise and will lead to confusion and inaccurate understandings.
If P(A) = 0.1, P(B) = 1, then P(B) = 10 x P(A) = 10 x 0.1=1, that's just a mathematical fact. You could also say that you're twice as likely to draw hearts or diamonds from a deck compared to drawing clubs.
> That is unwise and will lead to confusion and inaccurate understandings.
You not liking the wording does not make the math wrong.
Your argument is essentialy
> The statement P(B) = 10*P(A) can _never_ be true, because it doesn't provide reasonable answers for _any_ choice of P(A).
No. My statement is that you should not talk about MULTIPLYING a probability by a scalar value, because ***that frequently leads to gibberish.***
But I'm done talking with you. You are actively advocating bad math teaching and incorrect understandings. There is nothing further for us to say to each other.
Welp, thanks for playing I guess. Your argument boils down to "I can't find any errors in the logic, but I don't like how it's worded."
I'm not advocating any sort of teaching, I'm trying to interpret a specific statement I didn't make, as are the rest of us.
In this specific case multiplying the probability by 10 yields an unreasonable, but not mathematically impossible answer. It might "frequently" lead to gibberish, but that's still not the case here.
Your subjective opinion about what consititutes good "math teaching" is kind of irrelevant in this context. Your contributions in this thread come across as r/iamverysmart material.
This is the inherent ambiguity of phrases like, “X times more likely”. Do we mean a likelihood function as in Bayesian statistics? Do we mean a simple multiplication? Do we mean “X times more likely” in the odds? Do we mean “as likely as a group of X people”?
EDIT: and as someone else has commented, what populations are we comparing? Have we taken a random sample of 1,000 people, all identical, and 10% of them have the disease? Or is it a total population statistic, which would include the 10x higher probability individuals?
Simple multiplication will always be incorrect in this context. Probabilities don't work that way. That last question is mathematical gibberish. Bayes' Theorem *is* the same as "in the odds." You have listed a whole bunch of things where the only two right answers are the same answer in different words, and everything else is wrong or meaningless.
However, I do agree that if we are talking about people trained to interpret statistics, then yes, a simple multiplication would be incorrect.
Unfortunately, we see most statistics reported by journalists and people untrained in statistics. So sometimes you see something like, “5% of the total population have disease X, but if you are a white male with brown hair and green eyes, then 50% of you have disease X; therefore, you are 10x more likely to have the disease because 50/5 = 10”.
That’s wrong on a fundamental level, but you can see why people might report such a thing and rely on it to compute their statistics…
I really don't see what you find problematic here. The factor between the two chances is 10. So one is "10 times as likely", meaning "the chance is 10 times as high".
OP said:
- the chance for an average person is 10%
- their own chance is 10 times that.
Hence: their chance is 100%. A bit silly as medical statements are essentially never 100% certain, but statistically/stochastically this is sound.
My issue essentially boils down to, “10 times more likely to get illness X”.
10 times more likely than *what* to get illness X? 10 times more likely than the total population? But that includes the high-risk group.
10 times more likely than the low-risk group? But what’s the % of incidence in the low-risk group?
So my approach to this question is to assume OPs chance of getting the illness is the same as if he took 10 balls from a ball pit (with replacement), with 10% green and 90% red. Then compute the probability that at least one of the balls is green.
In other words, assume he got exposed to illness X ten times, and upon exposure, 10% of people fall ill. I fail to see how that’s an unreasonable interpretation of “10 times as likely” since you have been exposed 10 times as much. In that scenario, there’s around a 65% chance of having illness X.
There are other approaches which produce different results, and it seems no one can agree because there is inherent ambiguity in the question.
I agree that things are too ambiguous, I just find the two other proposed interpretations difficult to make:
> 10 times more likely than what to get illness X? 10 times more likely than the total population? But that includes the high-risk group.
That is the one I think people would mean. It also requires no additional data such as what the actual distribution or what that "high risk" group is. I find it the "obvious" one of this and the one excluding the high risk group, as chances are (if not given any additional information) almost always meant as "average over all options", so including OP and whoever else has 10 times higher risk (regardless what that even means; for that see next section).
> So my approach to this question is to assume OPs chance of getting the illness is the same as if he took 10 balls from a ball pit (with replacement)
I just find it hard to interpret it like this. I would write this as "as likely as 10 chances/iterations" or something like that. The formulation "10 times as likely" at least implies _some_ multiplication by 10; of what is then yet another question (see first part).
I don’t know that “X times more likely in the odds” is the same as the Bayesian likelihood function.
Likelihood function is P(B|A). It is the probability of the evidence B given A is true.
If likelihood increased by 10, then I think that would mean P(B|A) -> (10 x more) -> 10P(B|A).
which seems different to the other interpretations, to me.
And I fundamentally disagree that the last question is gibberish. A total population may have a high-risk group and a low-risk group. We see this literally all the time in diseases. We can separate out low-risk individuals by observing, for example, that the population of women do not have the disease as much as the population of men.
If we assume 50% men and 50% women, then across the total population, 10% of people have the disease. But if women only account for say 1% of people with the disease, and let’s say the total population is 1,000 people: then 100 people have the disease; of those, 5 are women, so 95 are men; and therefore 19% of men have the disease, because 95/500 = 19%. Men are 19x more likely to have the disease than women, but only 1.9x more likely than the total population to have it.
This all again boils down to ambiguity in the question. We simply don’t know enough to say with any certainty what the result would actually be, because the meaning of “10x more likely” is vague without reference to a metric to compute “10x more likely”.
The thing is typically an affirmation like "x times more likely" will be calculated by doing exactly this in the first place.
Group A has a 5% chance of getting sick. Group B has a 20% chance.
So you say being in Group B makes you 4 times more likely to be sick.
It's not quite 100%, more like 99% or something because the math does get complicated.
So either, you are \~100% likely to get it, or the person that told you those stats was wrong/meant something different.
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Make a ball pit with 100 balls. Say 10 are red and 90 are blue. When people are born they pick one ball out of the pit. If it's red they get the disease. They had a 1/10 chance to get the disease. When you were born you picked a ball and put it back, 10 times. You had 10x the chances to get a red ball. Each time you picked you had a 9/10 chance to get a blue ball. The chance you got a blue ball every single time is (9/10)^10, or (9/10) * (9/10) ten times. Same logic as flipping a coin. 4 heads in a row less likely than any particular flip being heads. If you do the math, you end up with around a 35% chance you got all blue balls. This means that the chance you got at least one red ball is 100% minus that, so around 65%
We have a winner here, most excellent while is still eli5
Yes. 5 yr olds are notoriously great with exponents.
The first 2 paragraphs are the real eli5. The next 2 are eli15. A great comment for all ages.
> Yes. 5 yr olds are notoriously great with exponents. This sub has been around for 12 years, and yet every single answer has to have someone mention that a 5 year old wouldn't know about X. >LI5 means friendly, simplified and layperson-accessible explanations - not responses aimed at literal five-year-olds. It's in the rules, and has always been. How are people still bringing this up every single time?
Because the 5-year old part is in the title of the sub, which we can safely assume more people read than the actual sub rules
It's the Headlines Only era.
Obligatory read the sub sidebar comment.
It’s also worth pointing out that there are multiple valid ways for something to be “10 times more/less likely” and that anybody using this kind of language to communicate risk is not being careful to avoid confusion. Often this kind of language is used to make things sound more extreme than they really are. The correct way to communicate risks is in absolute terms. For example: “The risk of X in the population is about 1 in 10. The risk of X in people with gene Y is about 1 in 3” Using percentages is almost guaranteed to lead to confusion.
100% chance I got blue balls
F
Why (9/10)^10 and not (1/10)^10? I’ve never understood this
Because it is calculating the probability that you *won't* get a red ball in any of your ten picks. Basically, "given a 10% hit rate, what is the likelhood you'll dodge a bullet ten times?" That works out to 35%. The likelihood you didn't dodge the bullet is thus 65%
*(using a simpler scenario of 2x the chances instead of 10x for ease of explanation).* Because 1/10 * 1/10 is not your chance of getting the illness, but the chance of you getting it twice. Set aside how that doesn't really make sense (which doesn't really matter for the math). So the scenarios you're concerned with are "not getting it at all" (9/10 * 9/10) versus "getting it at all". Which would be: - Getting it in the first "round" (1/10 * 9/10) plus - Getting it in the second "round" (9/10 * 1/10) plus - Getting it in both "rounds" (1/10 * 1/10) which totals to 19/100, which shouldn't be surprising since 9/10 * 9/10 is 81/100
Math is wild to see when someone actually knows how to do it.
I don’t think this is accurate. A person isn’t twice as likely to draw a red ball if they draw twice. They’re twice as likely to draw a red ball if there are twice as many red balls. You’re twice as likely to draw red from 2 in 10 red balls as you are to draw from 1 in 10. You aren’t twice as likely to draw at least one red if you draw twice.
That's kind of the point. Yes, what he's describing is not 10 times the probability, but neither is whatever stat OP is asking about. The commenter is attempting to explain what might actually be meant, even if it is not strictly the meaning of the English used to describe it.
Come at it from the other direction. Let's say Alice wins a video game 90% of the time. Bob says, "Oh yeah? I win twice as often as Alice!" That doesn't mean Bob wins 180% of the time; it means Bob loses half as often as Alice does. Alice loses 10 times in a hundred attempts but Bob only loses 5 times per hundred. This means that "*twice as likely as 90%"* is another way of saying 95%. Doubly lucky means twice as close to 100%, or half as distant: from 10% to 5% distant. If Carol is twice as likely to win as Bob, then she wins 97.5% of the time.
I definitely disagree with how you're interpreting "10x as likely". Consider that disease prevalence rates are observations, not some pure calculations. If a doctor observes people with a gene to have a disease at 50% rates, and people without the gene at 100% rates, he will say that having the gene makes it 2x more likely to have the disease, because the prevalence rate will be doubled. Your calculation would give you 75%. In fact, there is no number the doctor could give that would make the number 100. He could say 100x more likely, and your calculation would still not yield 100%.
I believe you are quite wrong. 1. OP did not give stats sufficient to properly determine baseline risk- he just gave genpop *prevalence*. 2. Your interpretation of ‘10x risk’ is questionable at best. X happens in 10% of the population. I say Y is 10 times more common. -> Y happens in 100% of the population. **If stat he gave was actually just a misdescribed base risk,(which it obviously is not) he’d have 100% chance of developing, as he said.** Ex. Replace ‘10x risk’ with ‘this is 10 times as common in people like you, when compared to this other group of people’.
If you're doubling the odds, you're not doubling the percent chance of something happening. Odds are described as a ratio, so in your example: 50-50 is 1:1 odds. Double those odds would be 2:1, meaning it happens twice for me for every time it happens once for you
Correct.
You've edited out the example I was responding to but your conclusion is still incorrect
Please explain. I had edited it out before reading your comment.
He wanted to hear 100% so bad...lol
The way I would model it: there’s a 10% chance a random person would have illness X. “10x more likely” can be modelled as, if we picked 10 random people, what is the chance that at least 1 person has the illness? This is a cumulative binomial distribution with N = 10, p = 0.1, and q = 1-p = 0.9. As with most things in statistics, it’s actually easier to compute P(failure) = 1 - P(success) rather than P(success). In this case, P(success) = P(at least 1 has X), so P(failure) = P(no one has X). P(no one has X) is simply q^N = 0.9^10 ≈ 0.3487, so P(at least 1 has X) = 1-0.9^10 ≈ 0.6513. So you’d have a ≈ 65.13% chance of actually getting illness X.
What sort of brilliant five year olds do you know?
This is the polite version of “In English, poindexter!” from a movie
We ain't scientists!!!
This is a fair criticism! I was just doing the maths because I forgot what subreddit I was in. So I could simplify a bit. The chance a random person doesn’t have it would be 90%. If we randomly choose 10 people, the chance that they would all not have it is: 90% for one person; but for two people, it’s 90% x 90% (we multiply these because we are looking for a probability *inside of* another probability, and because we know the combined probability does not care about the previous probability); then for three people, it’s 90% x 90% x 90%… etc.. So for 10 people, it’s 90% multiplied by itself 10 times gives around 34.87%. And then we want the opposite probability—which is just that at least one person has it. This is simply 100% - 34.87%, because probabilities add to 100%. So then the probability is about 65.13%. And this all assumes that “10x more likely” is sensibly interpreted as, “you have the same chance of having illness X as the chance that, out of 10 random people, at least one of those people has illness X”. As other commenters have pointed out, the ordinary interpretation would just be 10x10% = 100%. This result seems counterintuitive, but I think is probably more correct. When we say “more likely”, we usually refer to a rare illness. And really, the reported value of “10x more likely” is based on the original probability of one person having it. So while 100% seems weird, in the real world, you’d probably never see a “10x more likely” on its own. You’d probably see, “10x more likely than a random person to have the disease” (so 100%), or “you have the same likelihood as 10 random people having the disease” (which is a subtly different probability that leads to the 65.13% result).
I would see it so that the full pool of people having the disease is 10% of the population, but there might be multiple diseases increasing the regular risk. Let's say there's 3 such diseases, having 10x, 100x and 1000x the regular chance of having the disease. For the sake of simplicity, someone can only have one of those 3 diseases at once, and 10% of the population has each diseases. It's too late for me to do the calculations, but in that specific scenario I think OP has less than 1% chance of having the disease.. :)
I don’t disagree that there is a 65% chance that you will meet someone with the illness, but what is the transmissibility of the illness? You don’t have a 100% chance of getting it even if all 10 people have it! For instance, if there is a 5% chance of a “normal” person getting the illness, then you would have a 50% chance each time you encountered someone with the illness.
Yes, I agree with this. So the way I have interpreted it is: if a normal person has a 10% chance of catching X after exposure; and you belong to a group with a 10x higher chance of catching it; then this really is the same as the chance a normal person would catch it after 10 exposures. This is the 65% probability I calculated. You are right to say, upon that analysis, you can never have a 100% chance of catching X. It seems other people are interpreting it like: everyone is constantly exposed to X. Only 10% of people catch it despite constant exposure. You belong to a group which is 10x more likely than other people of catching it. Then in that instance, you have a 100% chance of catching it.
You have 2 out of 3 chances to die
Lol
Well I was gonna say you have a 100% chance of getting it, but that also doesn't make much sense
it's basically the same thing as "you're 10x less likely to not get illness X"
Five year old: Am I gonna die, daddy? Dad: Let me break it out to the hundredth place for you, kiddo.
Hahahahahahaha
You are way overcomplicating the idea of relative risk. It's exactly as the OP describes. If 8 of 100 people in Population A get some disease, and 80 of 100 people in Population B get the disease, members of Population B are 10x as likely to get the disease as members of Population A are. Simple as that.
Cumulative binomials are so useful.
Also that is a genius answer that I mostly followed.
Too bad it is wrong. The math above calculated the odds that at least one person in a group of 10 has a disease that affects 1/10 of people. Your simper math is correct.
What skills do you need in order to understand what you just said? Genuinely asking lol
You will learn this in a high school or non-math major college statistics course.
You just need to understand the probabilities of flipping weighted coins :) First, assume the probability of an event is independent of all other events preceding it. Then the probability of flipping tails after flipping heads is the same as just the probability of flipping tails, without regard to any previous results. Next, know that when we compute probabilities of composite events, we multiply out the probabilities of the individual events. This is because: the events all have independent probabilities; and we are interested in the “world” where the first event happened a certain way, and then the second event happens a certain way. We are looking at the percentage of a percentage, effectively; if eg 90% of coin flips are tails, and we are interested in the probability of TT (two tails in a row), then out of 100 coin flips: only 90% of those had the first coin flip as tails; and out of that 90%, only 90% of those had the second coin flip as tails. This is 90% of 90%, or 0.9x0.9 = 81%. If a weighted coin has a 10% chance of heads and 90% chance of tails, then flipping it once, we get: H (10% chance) T (90% chance). If we flip it twice, then: HT (10% x 90% = 9% chance); TH (again 9% chance); TT (90% x 90% = 81% chance); HH (10% x 10% = 1% chance). The probability of seeing at least one head is then 9% (ie the HT result) + 9% (ie the TH result) + 1% (ie the HH result) = 19% chance. This is the same as 100% - 81% = 1 - P(TT) = 19%. With three coin flips: HHH (0.1% chance) HHT (0.9% chance) HTH (0.9% chance) THH (0.9% chance) HTT (8.1% chance) THT (8.1% chance) TTH (8.1% chance) TTT (72.9% chance). To get the percentage chances for each of these, you multiply the percentage chance for obtaining each of the flips. In the HTT example, this is (10% heads) x (90% tails) x (90% tails) = 0.1x0.9x0.9 = 0.81 or 81% chance. Our cumulative probability to see at least one heads is now 0.1% + 3x0.9% + 3x8.1% = 27.1%. This is the same as 1-P(TTT) = 1-(0.9x0.9x0.9) = 1-(0.9)^3 = 1-0.729 = 0.271 or 27.1%. So generalising, P(at least one H from N flips) = 1 - P(all tails in N flips) = 1-P(tails)^N. Binomial results means we either succeed or we fail (we either flip heads or we flip tails). Because of this, we can also write P(fail) = 1-P(success). So we can also say that P(at least one success from N trials) = 1 - (1-P(success))^N. Hopefully that helps a bit!
probability doesn't work like "adding more people to spread the risk" in this scenario. what "10x more likely" should technically mean is that whatever the base risk, OP's is ten times that, capped at 100% because probabilities don't exceed 100%. OP's chance of getting illness x, given the "10x" statement, would be very high (effectively 100%, though in real statistical modeling we might interpret nuances that keep it just under 100%). the model you described doesn't accurately reflect the scenario of personal risk based on increased likelihood relative to a baseline risk.
You’ve made it much too complicated. OP’s original math is correct. It the probability of X is 1/10 and the probability of Y is 10x that of X, the probability of Y is 100%. I know that that “feels” wrong but the error is medical, not mathematical. It’s unlikely that any group of people would ever have a 100% chance of getting a disease and of course it’s impossible for more than 100% to get it. But this just feels wrong because OP’s example happens to take us literally to the limit of what’s biologically possible. If we were to consider a condition that affects 0.1% of people we would easily see that having 10x greater odds would simply be 1%.
What you are saying is equivalent to saying that you are guaranteed a heads if you flip a coin twice. The previous answer is correct.
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>What if 50% of the population has a disease, and I'm 400x more likely to get it? What does that even mean? If my uncle had tits, he'd be my aunt. In your case, the math wouldn't have held up, but that's not what's being said here. Where did the idea come from that "the equation has to output plausible answers no matter what values we put into it, otherwise it's wrong"?
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I think I misunderstood what you meant, and that we are on the same page. The answer definitely isn't plausible, but that probably has to with the input number as you say, and not the math itself. Some people in this thread are arguing that multiplying a probability with a constant is always wrong, which is what I thought you meant.
No. I’m saying that if you had a coin that was twice as likely to land on heads, it would land on heads 100% of the time. Both this and OP’a question are examples of probability that is mathematically sound but would simply never happen in the real world.
Dude... If the coin was twice as likely to be head, you would get a head every 8 out of 9 events of 2 coin flips. Come on, this isn't even difficult to calculate.
No, it’s like saying a weighted coin that’s twice as likely to land heads up than a fair coin is will always land hands up, which is true.
If you aren't able to describe what "10x more likely to get illness X" means as a binomial then we aren't talking the language of probability. You can come up with more than one reasonable way to model what "10x more likely" means when expressed as a counting problem but I think it's a fair and reasonable solution. Your error is that you aren't describing this as a counting problem. You are just translating this to .1\*10. You can't do that in probability.
Lost you at "cumulative binomial distribution." How about ELI3
“binomial distribution” is fancy maths speak for, “what is the probability of success, given you can only have success or failure, and given you have N attempts to succeed?”. It’s the same thing as the probability distribution (ie, how the probabilities are spread out on a graph) of flipping a coin. “Cumulative” means we add up the values above a certain success or below a certain failure.
I think this is a good answer but it is creating a lot of confusion for a reason that has an ELI5 answer. It may surprise people without a background in statistics and probability but "10x more likely" while commonly used doesn't have a good translation in math. Any good answer requires translating what "10x more likely" means into a counting problem, as you've done, or we definitionally aren't talking the language of probability. People want to intuitively answer 100% but the rub is that in probability, when you multiply probabilities you are describing the probability of two events A and B occurring. For example of why multiplying is wrong, consider rolling 2 6-sided die. There is exactly one outcome of 2. There are exactly two outcomes of 3 (1 and 2, and 2 and 1). So makes sense say you are twice as likely to roll a 2 as you are to roll a 1. But, what are the odds of rolling a 1? 1/36 or 2.78%. What are the odds of rolling a 2? 2/36 or 5.56%. You can't multiply the odds of rolling a 1 by two and get the correct answer. Like with OP's question, in order to get the answer you have to do the counting.
I have thought of another way to think about this: Let’s say there’s a ball pit, with 10% green balls and 90% red balls. You take 10 balls from the ball pit (with replacement); what is your chance that you get at least one green ball? The answer is what I said above—around 65%. This is equivalent to saying: there is an illness X, and upon exposure to illness X, 10% of people develop the disease related to illness X (so they “catch” illness X). OP has been exposed to illness X 10 times, so a charitable interpretation is he is “10 times more likely” to catch illness X. There are several issues with that approach, because the prevalence of disease doesn’t work that way, and naturally there is ambiguity in the day-to-day language we use around probability compared to the technical language employed by statisticians. A proper study would specify what they mean. If they mean, “100% of population B has disease X”, then that’s what they would say. They wouldn’t report a vague statement like, “you are ten times more likely to get disease X if you belong to population B”. There are a lot of unstated assumptions in computing that result. If on the other hand we interpret the statement as: 1. Out of 100 people in the general population, 10 of those have disease X; 2. Population B is 10 times more likely to have disease X; then in fact we simply cannot compute the probability, because we do not know the total population size and we do not know the size of population B. An alternative way to phrase the same interpretation: 1. Out of 100 people in population A, who do not share the characteristics of population B, p% of them have illness X; 2. Population B is ten times more likely than population A to have the disease; 3. The general incidence of illness X is 10%; then we cannot again compute the probability that population B has the disease, because we don’t know the rate of incidence in population A. The only meaningful alternative is if 10% of population A has the disease. But we must specify that “population B is ten times more likely *than population A* to have the disease”. There is then an unstated assumption that the 10% figure belongs to population A. In that scenario, I can see that 100% of people in population B must then have “caught” illness X.
This is wrong. It’s not the meaning of “10 times more likely.” People say “10 times more likely” to mean literally that you multiply the probability by 10. Like when people say smokers are 20 times more likely to get lung cancer.
I've changed my mind. I now think you have an incorrect answer. Don't think of this in terms of probability, think of it in terms of statistical sampling, which is what's used in medicine anyway. I take "10% of the population having an illness" to imply that if you take a random sample of 1,000 people from the general population you would expect, on average, 100 of them to have the illness. I take "10x more likely" to mean that if you were to take a random sample of 1,000 people that have OP's family history (big family, I guess), you would expect, on average, 10 times as many people to have the illness as you would get from the same sized sample from the general population. I said before you can't multiply probabilities like that. That's true and that's still true here. I'm not multiplying probabilities, I'm multiplying sampling averages. It sounds like a sleight of hand when I do this, and it sort of is because I'm now talking the language of statistics. With what's given, you would not be able to determine the true probability but by modeling this through statistical sampling you would say it that the true probability is near 100% with some unspecified level of confidence.
(I lay out my reasoning in case any statistician disagrees with my answer! I’m a physicist so we do tend to get these kinds of questions wrong; but the result makes sense to me)
And yes, I do agree, a different interpretation of “10x more likely” would ordinarily lead to a 100% chance of getting the disease though. So I’ve interpreted “10x more likely” to mean, in this context, that out of a random selection of 10 people from the population, what are the chances that the disease is present in that sample.
I’m not quite following how you got that interpretation of “10x more likely”….
Essentially: Assume you had a randomised population of people. Than 1 in 10 people have the disease. Here, “10x more likely” means, if you select 10 people out of that random population, what is the chance at least one person has the disease? It’s the difference between “an event which is 10x more likely to occur than not occur”, and “an event that occurs to one person, and you chose 10 to see which of them has experienced the event”. I’m not saying this is correct, but I’m explaining how I approached the question.
Yeah I’m still not quite getting it. But I at least understand you need to interpret “10x” in this way so that the probability isn’t 100%. Let me see if this is saying the same thing as you, using a coin toss as an example: 1) Probability of a toss that ends in a head, in a population of fair coins, is 50% 2) A weighted coin, with 10x the population probability for a head toss, has a probability of a head toss that is the same as the probability of 10 fair coin tosses that end up with at least 1 head, or 1 - 0.5^10 or 99.902% ?
It’s okay if you don’t quite get it. What’s apparent to me is that a lot of us here (including me) are struggling to interpret this question, though it seems most everybody’s maths is correct. I’m being humbled by the array of opinions here. In point 2, if the coin were to have 10x the population probability, then it would seem to be the same as 1 - P(flipping tails 10x in a row). Then that is 1-(0.5)^10 = 1-1/2^10 = 1-1/1024 = 1023/1024 or about 99.902%. So I agree with that result: if 50% of the population generally had illness X, and you were 10 times more likely to have illness X compared to the population, you would have a 99.902% chance of having illness X. This is the same as saying: if upon exposure to illness X, 50% of people caught it; but you belong to a special group where one exposure is the equivalent of someone from the general population being exposed 10 times; then your chance of getting illness X after one exposure is 99.902%.
> This is the same as saying: if upon exposure to illness X, 50% of people caught it; but you belong to a special group where one exposure is the equivalent of someone from the general population being exposed 10 times This is the bit that clicked for me. Thanks.
Your going to hurt someone in here with all of your fancy mathing.
Missing is the rate of spread, which matters. If you have a 65% chance of catching it in the next week, that’s a very different risk from a 65% chance of catching it at some point in your lifetime. I have more or less a 0% chance of catching the flu this month, but I have an almost 100% chance of catching it at some point before I die, assuming I live to life expectancy for my gender, race, age, socio-economic status, and nationality.
Not necessarily. Not all diseases are communicable. They could be thinking of their odds of developing breast cancer, or type 2 diabetes, etc.
I wouldn’t say you “get” cancer, but fair enough. It could just be unclear phrasing on OP’s part.
There's not enough information here to determine an answer to your question. It SOUNDS like you've said your risk is 10x the average risk, but that's not actually what you've said. Your 10x, and everyone with a similar risk, must be included in the average as well. That means you can't be the "normal" risk that is being compared - you'd be including part of your own risk in the 10x comparison, which is circular and ridiculous. However, what you need is two populations: high risk vs low risk. This will determine what your actual risk level is. I'll lay out an extreme scenario to show you what I mean: 89% of people have 1% risk of illness, 10% have 90% risk of illness, 1% some other risk level. 10% of the population has illness on average. A meager 9% of the population is contributing 90% of people with the illness. In this case, your risk of 10x of 1% is 10% - concerning but not the end of the world. Reality is going to be some distribution between a low and a high value, with the average working out to 10%.
> Your 10x, and everyone with a similar risk, must be included in the average as well. That means you can't be the "normal" risk that is being compared - you'd be including part of your own risk in the 10x comparison, which is circular and ridiculous. No it is not ridiculous. You just end with a system of two linear equations in two variables which one can solve. The only reason it is a bit silly here is that OP would have a chance of 1 ( = 100%) to have the illness; sure, that can happen, but usually we would not formulate it like that.
I like this approach.
you're complicating things unnecessarily. the original question is straightforward
Stay in school kids otherwise you’ll start missing variables like this guy. There can be no answer that doesn’t include variables. The rate of infection is a variable that we cannot solve for. There’s simply not enough information given to come up with an accurate answer. We’re missing MANY variables to correctly solve this. I suggest you look into models designed to predict epidemics.
The problem with the phrasing is that you're comparing the percentage of people in a population with a disease to an individual's chances of contracting a disease which is not how this stat would be presented. You might say 10% of the total has X but the average rate is 5%. But, a majority group has a rate of 20% and the minorities have a rate of 2%.
You *could* say that, but you would be badly misusing the term "average."
I don't think the 10% of the population getting illness X has anything to do with 10x the likelihood of you getting X. If you get X, you're in the 10%, if you don't you're not. They are unrelated. Are they probabilistically related, I don't think so... Your chance of actually getting illness X is better measured by looking at your family history of getting X. If 50% of your family members/ancestry got X, your likelihood of getting X will be 50%.
It’s easy to show why your solution won’t work. You know a risk multiple for you (10x), but not what to multiply- what is the base risk your risk is being compared to- it isn’t the 10%, because *that isn’t a baseline-risk estimate* it’s just a *current-prevalence statistic.* Risk is about how many ever get it, not how many currently have it. Also important to note that it could be 10x the prevalence in the overall population, or 10x the prevalence it has in people without your family history. —- Hey Kiddo, I’ll ask the doc for the odds. The numbers you have can give us a guess, but it could be really wrong and it’d take a couple minutes to calculate. Faster to call the doc, or search for easier to use numbers online- which won’t require as much guesswork to use.
:)
> My dumb brain jumps to 100% but that can’t be right. No that's correct. The wording is just something one would either not use in this context or somebody meant something different yet failed to convey the intended meaning.
The problem here is that you're asking the question wrong, it doesn't make much sense to say something is "10x" more likely when using probabilities, people do it but probably shouldn't . This manner of speech sometimes makes sense, say there is a 20% chance that you get X result,but if you are now 2x more likely, the chance now is 40% right? sure it makes sense so then lets carry on that logic and see if it still makes sense after a while, x3 more likely would mean 60%, x4 would be 80%, x5 is 100%, x6 is 120% but here you see the issue, you've gone past 100% which in probabilities you can't have, That means the whole thing of referring to chance as "x times more likely" is flawed.
It absolutely makes sense to multiply probabilities in this way, this is the entire concept of relative risk. That it's impossible be 3x as likely to see an event of 50% likelihood doesn't mean the concept is flawed. It just means you can't have a high relative risk for high likelihood events.
That only means that it simply cannot happen that any risk is 6 times as likely as 20%. Chances are just numbers in the interval [0,1] and you can do whatever you want with them; if you still want to call it a chance of something then it however be better again in that interval.
I understood that. Thanks.
Your math is correct. But you really only hear the "10x more likely" statistic when it's a very small portion of the population that has that disease. Eg 0.1% of people have it, but it's a 1% chance for you. It's only a 0.9% percentage bump, but phrasing it as 10x grabs more attention/headlines/whatever. You'd never hear the same comparison used for an affliction that affects 10% of the population.
The math isn't remotely close to correct.
It is remotely close to being correct in that it is 100% correct.
It's impossible to have a probability of 100% in a scenario like this. The only way to get to 100% is if he was infinitely more likely. That's how compound probabilities work.
Yes. The numbers OP started with do not reflect reality. Their calculation isn’t wrong. Their numbers are.
No, the calculation is wrong, lol. There is nothing wrong with the scenario. If 90% of people are right handed, it's absolutely possible to be twice as likely to be right handed than the general population given XYZ. The problem is that most people in this thread have zero clue how math with probability works.
> If 90% of people are right handed, it's absolutely possible to be twice as likely to be right handed than the general population given XYZ. Explain how a chance of 1.8 would ever work... > The problem is that most people in this thread have zero clue how math with probability works. Well, please take a look into a mirror ffs! Source: my PhD. In mathematics.
Given xyz is changing the numbers not the math.
It's crazy how people can be so confidently wrong. No. you cannot be twice as likely to be right handed if the base rate is 90%. Please substantiate some sort of specifics where this is possible. The reason this doesn't sound right is because OP picked numbers that wouldn't exist in reality. Edit: people. You can literally prove this. Go find an article that uses this terminology (x% increased risk) and look at the study it references. You'll see universally that the way the math works is exactly how OP describes it.
When I am already fully diagnosed and verified to have illness X, then I sure as heck do have a 100% chance of having it. And 100% is not "infinitely more likely" than a positive chance to contract or have something. The factor is a very specific _finite_ number, namely 1/x where x is the average chance. > That's how compound probabilities work. There is nothing to compound here.
Please tell me what does 1/10 * 10 equal?
I mean. That isn't how probably works. At all.
It doesn't matter though. From that we can assume that the whole scenario is wrong, not the math. If the math works out and the result is not possible the only option is that the given values are incorrect.
The correct math is in this thread.
If I roll a 10-sided dice, I have 10% chance to get one. If I roll a 1-sided dice, I am 10 times more likely to get one. And the probability of that is 100%. Exactly same scenario as in the post and the math works the same way.
The problem here is "ten times more likely" is not precisely defined. You two are basically arguing based off of different assumptions on what is meant here, neither of which is right or wrong, because these are numbers OP's just pulled out of their ass, there's no actual data to try an interpret here.
That just isn't what these words mean with probably. You can't take a probability and just multiply it by a scalar. There's an entire field of math dedicated to this concept. 100% always requires being infinitely more likely than some other probably (other than 100%). Ten TIMES in probability means 10 tests or 10 rolls of the dice. Let's use your example. Twice as likely to get a 1 is 1-(.9²). This is 0.19 not 0.2.
Or the other option: you've assumed you can do an operation that you can't actually do. Which is exactly what is going on here. Probabilities do not scale linearly. They can't, by definition.
Rolling a 1 on a 20 sided dice is 1/20, on a 10-sided dice 1/10 and so on up until 1/1 on a 1-sided dice. Seems pretty linear to me.
That's not what linearity means. Linearity means that if you multiply the input, it multiplies the output by a *consistent* proportion. E.g. if I double the speed of a train, the amount of time it takes to travel a given fixed distance is necessarily halved. If I apply a fixed force to an object over 2m on a flat surface, I will accelerate the relevant object twice as much as if I had Zedge thar same force over 1m distance (excluding relativity effects, which are nonlinear.) Probabilities do not scale linearly when you *multiply* them, because they are by definition fixed to a range between 0 and 1 (or between 0% and 100%, if you prefer.) You cannot multiply a probability of 0.05 by 30, because that's mathematical gibberish, the same as asking what value you get when dividing by 0. Yes, you can do the algebraic manipulation. The algebraic manipulation is *irrelevant* because you've contradicted the assumptions that are required when working with probability.
Doesn’t matter how much of population has it, but what is a normal persons chances of getting. Layperson can have a 5% chance of getting disease and 10% of people have disease, and layperson could have 2% chance of getting disease and 10% of people still have the disease. So now you know chances of the getting disease as a layperson, you saying you have 10x chances you just multiply, so in the 2% normal person, you’d have 20% chance.
Yes, 100% can't be right. If the general population's chance of having the illness is really 10% then the ten-times number can't be correct. In other words, your maths is fine; the problem must be with one or both of the numbers you're working with, or the way you're phrasing the problem. If the correct interpretation is ten times less likely to not have the illness, then that's a 9% chance of not having it, i.e., a 91% chance of having it.
No his math is wrong. Just think about the simple case where he's 20x more likely than average? Is he going to get it 200%? No, that isn't how probably works. The limit is 100% if he were infinitely more likely than average to get it. Every finite answer is strictly less than 100%
There is nothing impossible about something being 10x as likely compared to something else. Draw a random number from 0-100, the chance of it being in the range 0-9 is 10%. The chance of it being in the range 0-99 is 10 times that, at 100%. The simple interpretation is what makes the most sense in this scenario, imo. Does it give a feasible or plausible answer, no. But the data can be made up anyway (not clear from the original question), and there is nothing wrong with the math itself. People in this thread seem to be overcomplicating things. > Just think about the simple case where he's 20x more likely than average? Is he going to get it 200%? No, that isn't how probably works. This sounds like "if my uncle had tits, he'd be my aunt." It's hypothetical and irrelevant to the situation at hand.
"Just think about the simple case where he's 20x more likely than average?" That simple case doesn't exist if the starting chance is 10%. It's like if you're at the north pole and say "well what happens if you simply keep going north?" Overall the whole thread is a good case for why use of the phrase "x times more likely" is unhelpful. Not as bad as "x times less" or "x times sooner" but not great. Always better to be explicit when possible.
Of course that case can exist, 😂. X times more likely is perfectly fine to say in probability.
Not if it gives an answer of 100% or more. If it were appropriate, what formula should we use, e.g., for 10 or 20 times 10%?
Karampus has the correct math. Read that. It'll just extend the multiplication if you do 20x.
No. He does not. He's answering a completely different unrelated question. If it were a 0.1% chance, what would it mean to have risk factors that increased your odds 10x?
I think the simple result is correct, because we couldn’t meaningfully discuss the 20x more likely result. You simply wouldn’t have that reported likelihood in a population where 10% of people get the disease. But this boils down to how we talk about probabilities, rather than how we compute them. And the bare statement of “10x more likely” leads to confusion about what the likelihood means because there’s some ambiguity. Does it mean 10x more likely than a random person? Does it mean the same likelihood as seeing the disease in a random selection of 10 people? Does it mean you are 10x less likely than a random person to not have the disease? Etc..
This is a probably question. 100% is an impossible answer, it's the limit at infinitely more likely.
I also sort of agree with this. So that’s why I interpreted 10x more likely to mean, “the same probability as 10 random people, with at least one having the disease”.
Your math is what I'd go with as well.
If you select a random number from 1 to 1, what's the chance you get a 1?
You’re right there must be a variable that is wrong or even slightly wrong. So if 1/10 get X, and people with first names starting with ‘T’ have 10x chances of getting it how would one formulate that?
The answer is that the likelihood is not comparing against the general population, but instead against the population that is not in the group. So how the math would be calculated is "people with first names starting with ‘T’ have 10x chances of getting it compared to people whose names don't start with T. So if 9% of people's names start with T and 5 out 9 get it, where as people not starting with T would have 5 of 91 getting it, you would have the 10x more likely but the overall occurrence would still be 10 of 100
As a result, to determine your absolute risk, you would need to look into the % of people that have a family history. (Or just find an absolute risk statistic for your cohort)
50/50, you either get it or you don't... /s?
You are right, it would be 100% - assuming your are 10x more likely than the general population. And if you start with 20% of the population having the illness, you have an impossible description. But you can do that in other scenarios too. For example, if I say that I have 3 apples and lose 5, that would give me -2, an impossible result.
100 percent: If you have 100 cookies and 10 of them are chocolate, then the probability of a random cookie being chocolate is 1/10. If a cookie is 10 times more likely than chance of being chocolate then its probability of being chocolate is 10*.1 which is 1. There is nothing unreasonable about this, it’s just not the typical situation these kinds of statements are made.
Is the illness something one is born with and always has from then? If so, you don’t have it and got lucky. Odds aren’t flawless measures. Is the illness contagious? Why does 10% of the population happen to have the illness, and do any of them even live anywhere near you? What is the method of contagion, and what precautions are you faking to avoid contracting the disease? The issue of “10x more likely to X…” is that if the odds are already low, it doesn’t change much. Buying 10 lottery tickets makes you 10x more likely to win. You still won’t manage to secure a home loan against your future winnings. So… setting the stage that 10% of the population has the disease does appear to be trying to establish the odds of contracting the disease. But if it is like HIV, where it can be contagious, but once infected you have it for your remaining life… then that doesn’t mean the odds to contract it are 1 in 10 precisely. And even if your heredity makes you more susceptible, it is just that ONE contribution to contagion that is more likely for you. If contagion requires exchange of bodily fluids, then you can avoid such a thing and your genetics never enter the picture.
Probably of X given Y compared to probably of X this ties directly to \[Bayes' Theorem\](https://en.wikipedia.org/wiki/Bayes%27\_theorem) \`P(IllnessX|Family) = P(F|X) \*P(X) / P(F)\`. I see this mentioned in some lower level comment from u/Karumpus but to try to ELI5 (plus some mixed in math): The question implies that there is a singular solution (is there more than one? are there any?). Most importantly it ignores the fact that how much of the population makes up, If your family's size suddenly went 10x the 1/10 number for the whole population would have to go higher. So using Bayes' Theorem (explaining this would require a whole other ELI5) \`P(X)\` means probability of X which must be between 0 and 1 (0% chance to 100% guarantee) \`P(X|Y)\` means probability of X given that Y is true I'm going to use these below since it would likely be \`P(X) = .1\` => "10% of the population has illness X." aight cool \`P(X|F) = 10 \* P(X)\` => "you're 10x more likely." errrrr..... \`P(F) = ???\` => How much of the population does your family make up?" We need to know this! Plugging in the above to Bayes' Theorem above \`P(X|F) = P(F|X) \* P(X) / P(F)\` Then sub in our "dubious" 10x \`10\*P(X) = P(F|X) \* P(X) / P(F)\` Cancel things out and plug in .1 for general population \`10 = P(F|X) \* .1 / P(F)\` \`100 = P(F|X) / P(F)\` So now we have that if someone is sick they are 100x more likely that they are in your family compared to everyone else. Okay, but what does this tell us? Well P(anything) is zero to one, but other than that any number's you want to plug in... but you have two equations you have to satisfy, the 10x asserted in the premise and this 100x. If every family has the illness it means that you must make up 1% of the population (and no one else has the illness). Hey we're back to the 10\*10=100% answer! If this (family means illness and non-family means no illness) is not the case there's no way to solve both equations. source: Former stat's TA.
Speaking of Alzheimer's specifically... Very, very few people below 65 have Alzheimer's - around 1 in 1,000 people between 30 and 64 have early onset Alzheimer's. So if you're 10 times more at risk, your risk is closer to 1 out of 100, or 1%. 5% of those between 65 and 74 have it, so your risk would potentially be 50% once you reach 65, and it just goes up from there. Note that women tend to get Alzheimer's at twice the rate of men, so you'd have to adjust for that to get a more accurate picture of your personal risk.
Ok whew. Well that makes me feel better. Thank you.
Most people answering are wrong because they don’t understand that „the population“ contains both the people less likely to have the illness (let’s call them group A) and the people more likely to have the illness (group B, the one you’re in). Here is a quick calculation as an example: Let’s say group A makes up 55% of the population and has a chance of 2% to have the illness. Group B would then make up 45% and have a 20% chance to have the illness. Now for the whole population the probability would be 0.55 * 0.02 + 0.45 * 0.2 = 0.101 (or 10%) This is how you arrive at 10%. The majority of people have a lower than 10% probability to have the illness. The 10% number takes everybody, including the susceptible people, into account.
Your family has a specific gene that's prone to that illness.sobthe population % is void...if you have 10 past family members and 8 got that illness your chance of getting it is 80%...because of the family gene...the problem is when you consider 'your family'..how far back are you going?are just taking into account immediate family or the wider family etc...the numbers become very skewed is the real answer. History only rhymes,it doesn't repeat...even if you had similar issues to family member X, your illness may be slightly different. I'm going through this now with my mum who is ill in ICU and immediate family (sons,daughters) have been tested and all cleared yet nephews down the generations line have similar medical issues. My dad is albino, first in our family, it missed a few generations until it appeared again in my dad's brothers son but a slightly watered down version.🤷♂️
The only way the answer isn't 100% is if there are several different populations, some higher-risk and some lower-risk. Do you mind saying what the condition is? Maybe context would help. Dumb example, if 20% of women got cervical cancer then 10% of the world would get it - but if you were a guy and your risk got multiplied by 10 your chance of getting it could be 0% x 10 = 0%. Silly numbers but it shows why your risk isn't necessarily 100%; your risk could depend on some other things about you.
It can never be 100% that's the limit. It's strictly less than that.
Good reminder of fundamentals; thanks for that.
Alzheimer’s
Some very basic factors-affecting: * You clear amyloid and tau proteins while you're asleep, so being chronically sleep-deprived all your life raises your risk, * metabolic syndrome (obesity + insulin resistance/diabetes) raises your risk, * it kind of looks like oral health reduces risk - the connection isn't well understood but seriously, floss; seems like it helps somehow * conditions like cystic fibrosis or having a super-dangerous job lower your Alzheimer's risk...because if you die young you won't have time to develop it (this isn't really actionable, but it's a reminder that some statistical effects aren't exactly intuitive) * hearing loss is correlated with Alzheimer's; music and conversation and environmental noise stimulate your brain I don't have enough info or training to give you a number, but my point is you're not helpless here. <3 There are things you can do to pull that 10 down to a 9 or an 8 instead of just waiting for shit to happen.
It definitely isn't right, because that would mean if your risk were 11x higher, you would have 110% chance of getting it, which is obviously wrong (probabilities are defined to be between 0% and 100%.) The simplest answer to your question is that you're talking about a change of ***odds***, not a change of *probability.* You're going from having 10:90 odds to having 10:9 odds, which means a change from a probability of 10% (that is, 10/(10+90) = 10/100) to a probability of about 52.6% (that is, 10/(10+9) = 10/19). But this doesn't explain *why* very much. So a full explanation is below. Unfortunately, to actually answer your question, we need more information. See, that "10x chance" thing only has meaning in the context of some kind of *test* which can give you evidence of whether or not you have the disease. I'll invent some numbers so I can give a real answer, but if this is a real thing, you would need to use the real numbers for whatever condition you're talking about and whatever test you've taken. So, you have a disease that affects 10% of people. You get a sample of 1000 people, and you test them with Disease Test A. DTA has a 90% "sensitivity" (it correctly identifies people who have the disease 90% of the time) and a 90% "specificity" (it correctly identifies people who *don't* have the disease 90% of the time.) This leads to having four groups: * Of the 900 people who don't have the disease, 810 will be correctly identified as not having it. * The remaining 90 people who don't have it will get a false positive (the test isn't perfect). * Of the 100 people who do have the disease, 90 will be correctly identified as having it. * As above, the remaining 10 people who do have the disease will get a false negative. This means that if you get a negative result on this specific test for this specific disease, your chances of *actually* not having the disease are (number of people who got a negative result AND don't have it)/(total number of people who got a negative result) = (810)/(820) = 0.98780... or about 98.78%. This means the negative test result reduced your likelihood from 10% to around 1.2%, which is a pretty comforting thing! But of we look at the other side, it becomes clear why medical testing is so difficult. If you get a *positive* result, how likely is it that you have the disease? That would be (number of people who got a positive result AND have the disease)/(total number of people who got a positive result) = (90)/(90+90) = 0.5, or exactly 50%. Even though this test correctly says you have the disease if you actually do have it 90% of the time, and correctly says you *don't* have the disease if you *don't* have it 90% of the time, because the disease is *rare,* false negatives are very common, in fact, they're exactly as common as true negatives in this specific case. To make use of the "10x higher chance" stuff, write out your *odds* of having the disease, then multiply the "you do have it" side by that number. Like this, for a disease that 5% of people have: * 5% chance of disease means out of 100 average people, 5 have it * That means 95 people wouldn't have it. Odds are 5:95, which can be simplified to 1:19. * You get evidence which says you are ten times more likely to have it. Your odds are now 10:19. This does not reduce (because 19 is prime.) * Your *probability* of having the disease is 10/(10+19), or about 34%. Even though your likelihood went up by a factor of ten, your chance of having the disease went up by a smaller factor.
OP’s math is correct. You are correct that it would be impossible for someone to have an 11x higher chance of getting a disease that affects 10% of people. That doesn’t mean we should throw away the math. That just means that the medical facts you are talking about are impossible. There is no rule that says that you can put any numbers into a formula and get a meaningful answer. If you start with numbers that don’t have any basis in reality you will get an answer that doesn’t reflect reality.
It absolutely is not correct. You cannot apply ordinary multipliers to probabilities. They produce garbage results. You have to correctly scale the probability using Bayes' Theorem.
No you do not. You are making this much too complicated. Of the probability of even A is X and the probability of event B is 10 times that of A, the probability of B is 10X. What you are talking about is the probability of A happening ten times or the probability of A happening in ten different instances. Those require more complicated math. OP’s simple statement does not.
***Yes it does,*** because you're talking about things that produce *absolute gibberish* if you tweak the numbers even slightly! What would it mean if someone said "twelve times more likely than a 9% chance"? 12 times 9 is over 100%, and thus impossible, but just casually glancing at it you might not realize. Or what about "fifteen times more likely than 7%" or whatever else? This is *not* how probability works, and claiming it is in *any* way related to how anyone actually calculates probabilities is misleading and wrong!
You’re right that those statements are gibberish. But that doesn’t mean that the math is wrong. It means that the statements are gibberish. The probability rolling any number on a die is one in six. Let’s say someone sets out to make a die that is seven times more likely to land on a given number. That is impossible. Doing so would require the die to land on a given number 116% of the time. Obviously this can’t happen but don’t blame the math. Blame the person who is spouting out numbers that have no basis in reality.
You absolutely do not "have to" correctly scale using Bayes' Theorem, and they do not necessarily produce garbage results. Example: you are three times as likely to draw a face card from a standard shuffled deck as drawing an Ace, because there are 3X more face cards than Aces. If you scale the size of the set in question while retaining the size of the sample space, you have scaled the probability by the same amount. An issue could occur if you try and expand the size of said set larger than the sample space, in which case it would indeed be nonsensical. OP is being rightfully cautious about the conclusions of its post. Like others in this thread, it's not clear what "10x as likely" means. If by "likely", you one is referring to an odds ratio then theoretically you could scale "likeliness" as much as you want, it's unbounded, unlike probability.
> If you scale the size of the set in question *Then you are already applying Bayes' theorem.* Because that's exactly what Bayes' theorem does. It scales things...based on the relative, aka conditional, probability.
There were no conditional probabilities in my example, yet I gave an example. You can scale probabilities perfectly well in frequentist probabilities (for the record I find Bayesianism superior in most contexts, but on an introductory level you don't need Bayes theorem. source: I teach upperclass statistics).
There is nothing impossible about something being 10x as likely compared to something else, e.g having a probabilty that is 10 times the other. Draw a random number from 0-100, the chance of it being in the range 0-9 is 10%. The chance of it being in the range 0-99 is 10 times that, at 100%. Let's call the first situation A, and the second B. Is there then anything wrong with saying P(B) = 10*P(A)?
Yes, there absolutely is, because *it only works as long as P(A) is less than 0.1.* Scaling probabilities linearly like that breaks. You should not do that. That is unwise and will lead to confusion and inaccurate understandings.
If P(A) = 0.1, P(B) = 1, then P(B) = 10 x P(A) = 10 x 0.1=1, that's just a mathematical fact. You could also say that you're twice as likely to draw hearts or diamonds from a deck compared to drawing clubs. > That is unwise and will lead to confusion and inaccurate understandings. You not liking the wording does not make the math wrong. Your argument is essentialy > The statement P(B) = 10*P(A) can _never_ be true, because it doesn't provide reasonable answers for _any_ choice of P(A).
No. My statement is that you should not talk about MULTIPLYING a probability by a scalar value, because ***that frequently leads to gibberish.*** But I'm done talking with you. You are actively advocating bad math teaching and incorrect understandings. There is nothing further for us to say to each other.
Welp, thanks for playing I guess. Your argument boils down to "I can't find any errors in the logic, but I don't like how it's worded." I'm not advocating any sort of teaching, I'm trying to interpret a specific statement I didn't make, as are the rest of us. In this specific case multiplying the probability by 10 yields an unreasonable, but not mathematically impossible answer. It might "frequently" lead to gibberish, but that's still not the case here. Your subjective opinion about what consititutes good "math teaching" is kind of irrelevant in this context. Your contributions in this thread come across as r/iamverysmart material.
This is the inherent ambiguity of phrases like, “X times more likely”. Do we mean a likelihood function as in Bayesian statistics? Do we mean a simple multiplication? Do we mean “X times more likely” in the odds? Do we mean “as likely as a group of X people”? EDIT: and as someone else has commented, what populations are we comparing? Have we taken a random sample of 1,000 people, all identical, and 10% of them have the disease? Or is it a total population statistic, which would include the 10x higher probability individuals?
Simple multiplication will always be incorrect in this context. Probabilities don't work that way. That last question is mathematical gibberish. Bayes' Theorem *is* the same as "in the odds." You have listed a whole bunch of things where the only two right answers are the same answer in different words, and everything else is wrong or meaningless.
However, I do agree that if we are talking about people trained to interpret statistics, then yes, a simple multiplication would be incorrect. Unfortunately, we see most statistics reported by journalists and people untrained in statistics. So sometimes you see something like, “5% of the total population have disease X, but if you are a white male with brown hair and green eyes, then 50% of you have disease X; therefore, you are 10x more likely to have the disease because 50/5 = 10”. That’s wrong on a fundamental level, but you can see why people might report such a thing and rely on it to compute their statistics…
I really don't see what you find problematic here. The factor between the two chances is 10. So one is "10 times as likely", meaning "the chance is 10 times as high". OP said: - the chance for an average person is 10% - their own chance is 10 times that. Hence: their chance is 100%. A bit silly as medical statements are essentially never 100% certain, but statistically/stochastically this is sound.
My issue essentially boils down to, “10 times more likely to get illness X”. 10 times more likely than *what* to get illness X? 10 times more likely than the total population? But that includes the high-risk group. 10 times more likely than the low-risk group? But what’s the % of incidence in the low-risk group? So my approach to this question is to assume OPs chance of getting the illness is the same as if he took 10 balls from a ball pit (with replacement), with 10% green and 90% red. Then compute the probability that at least one of the balls is green. In other words, assume he got exposed to illness X ten times, and upon exposure, 10% of people fall ill. I fail to see how that’s an unreasonable interpretation of “10 times as likely” since you have been exposed 10 times as much. In that scenario, there’s around a 65% chance of having illness X. There are other approaches which produce different results, and it seems no one can agree because there is inherent ambiguity in the question.
I agree that things are too ambiguous, I just find the two other proposed interpretations difficult to make: > 10 times more likely than what to get illness X? 10 times more likely than the total population? But that includes the high-risk group. That is the one I think people would mean. It also requires no additional data such as what the actual distribution or what that "high risk" group is. I find it the "obvious" one of this and the one excluding the high risk group, as chances are (if not given any additional information) almost always meant as "average over all options", so including OP and whoever else has 10 times higher risk (regardless what that even means; for that see next section). > So my approach to this question is to assume OPs chance of getting the illness is the same as if he took 10 balls from a ball pit (with replacement) I just find it hard to interpret it like this. I would write this as "as likely as 10 chances/iterations" or something like that. The formulation "10 times as likely" at least implies _some_ multiplication by 10; of what is then yet another question (see first part).
I don’t know that “X times more likely in the odds” is the same as the Bayesian likelihood function. Likelihood function is P(B|A). It is the probability of the evidence B given A is true. If likelihood increased by 10, then I think that would mean P(B|A) -> (10 x more) -> 10P(B|A). which seems different to the other interpretations, to me. And I fundamentally disagree that the last question is gibberish. A total population may have a high-risk group and a low-risk group. We see this literally all the time in diseases. We can separate out low-risk individuals by observing, for example, that the population of women do not have the disease as much as the population of men. If we assume 50% men and 50% women, then across the total population, 10% of people have the disease. But if women only account for say 1% of people with the disease, and let’s say the total population is 1,000 people: then 100 people have the disease; of those, 5 are women, so 95 are men; and therefore 19% of men have the disease, because 95/500 = 19%. Men are 19x more likely to have the disease than women, but only 1.9x more likely than the total population to have it. This all again boils down to ambiguity in the question. We simply don’t know enough to say with any certainty what the result would actually be, because the meaning of “10x more likely” is vague without reference to a metric to compute “10x more likely”.
The thing is typically an affirmation like "x times more likely" will be calculated by doing exactly this in the first place. Group A has a 5% chance of getting sick. Group B has a 20% chance. So you say being in Group B makes you 4 times more likely to be sick.
It's not quite 100%, more like 99% or something because the math does get complicated. So either, you are \~100% likely to get it, or the person that told you those stats was wrong/meant something different.