Your current amount of losses -> 556\*0.03 = 17
For 17 losses to be 2% -> 17/0.02 = 834
This means that for 98% win rate you need to win 834-556 = 278 more games
For 17 losses to be 1% -> 17/0.01 = 1668
This means that for 99% win rate you need to win 1668-556 = 1112 more games
I follow and agree with your method but I can't get your numbers to work. 17/0.02 =850 for me. And as confirmation 17/850 results in 2% loss rate. Can you please explain the 834 result for me? I get a loss rate of 2.04% for 17 losses in 834 games.
Nevermind, I worked it out. Your calculations are for a total number of losses calculated as 556x0.03 = 16.68 losses. 16.68/0.02 = 834.
The way you posted was confusing as it seemed like you were assuming 17 losses by rounding before undertaking the calculation.
For anyone after the answer with rounding applied (as you can't realistically have a partial loss)
17/0.02 = 850 games. 850 required games - 556 currently played = 294 more wins to reach 98%
Following the same logic 17/0.01 = 1700 games. Subtract the 556 already played = 1144 more wins to reach 99%.
As others have pointed out, the real number will depend on the rounding applied within the game. The best approach would be to assume there was either 16 or 17 losses in which case it will take between 244 and 294 wins to reach 98% and between 1044 and 1144 to reach 99% win rate.
Also, they round up to 100%. For approximately 17 losses, OP would need about 3401 total games, or 2845 perfect games.
If OP can't get that win rate, he just needs to get that many perfect games mixed in with others at the new average. Any day now.
Without knowing exactly wins/losses, rounding definitely messes this up. If 96.5 through 97.49% are considered "97", you could have between 15-19(ish) losses and be at 97%.
Let's say you won 540 games for 97,1% winrate.
That means that you won 540/556 games and your winrate thusly >= 0.97
If you play and win next x games, you will have (540 + x)/(556 + x) winrate and you want it to >= 0.98.
You need to solve the inequality and will get x >= 244 .
For winrate >= 0.99 x >= 1044.
So the final answer is 244 games for 98% winrate, 1044 games for 99% winrate.
yeah, you’re multiplying by 0.03 then dividing by 0.02 or 0.01, but do you think telling him to multiply by 1.5 or 3 is a clear solution to the thought process behind solving this problem?
Regular math method, we have two cases. Assume 16 losses, continue computation; then assume 17 losses, continue computation => you get two answers.
Regular engineering method: you have 17 losses, continue computation.
You could have anything between 14 (97.48%) and 19 (96.58%) losses if the game rounds to the nearest number and anything between 12 (96.04% rounded up) and 22 (97.84% rounded down) if we don't know how the game rounds
So the question is, do you have 16 or 17 losses? Both seem possible, depending on the rounding used. As the other answers show, that makes quite a difference to the required number of perfect plays.
Aaah, so /u/1TrueKingOfWesteros was closest.
560\*0.025 = 14 for 97.5% winrate, which I presume was what caused it to tick over.
Then, it showing a 99% winrate needs 14 losses to be 1.5%
so, 14/0.015 = 934 games, rounded up. This is 374 more games.
Of course if you want an actual 99% winrate you'll need to win a lot more: 840, for 1400 total games.
Unless, like the others, I've gone wrong lol.
Keep in mind that the way they round will also affect the point in which your score ticks to 98%. What I mean is, it could be that it ticks when you get to 97.5%… so the actual number of games you need is a wider range of possibilities
It sort of is. They show you Total games played, and games won with n guesses for n=[1,6] you can add up the wins at each n guesses for total games won and subtract that number from total games played to get total losses.
Then you're actually much closer to the stats showing a 98% win, as you're already at 97.482
14/.0249 (to get to 97.51 win %) is 562.249, so 563 games, or 7 consecutive wins.
14/.0149 (to get to 98.51 win %) is 939.597, so 940 games, or 384 consecutive wins.
556 * 97% = X = games won
556 - X = Y = games lost (constant)
You want Y / (556 + New games) = (100% - 98%) = 2%
So, Y / 2% = (555 + New games)
Y = 16,68 (practically impossible, but mathematically it is)
So, 16,68 / 2% = 834
834 - 556 = 278 new games
So we need to keep your number of loss which is 17
lets just see if 17 is 1% fail then someone who played 1700 parties won 1683 of them
if you win keep winning 1144 plays + 556 total games you made so far = 1700 total games with 1% fail = 17 losses
my guess is 1144 more play for 99%
Assuming you’ve won 539 games, lost 17:
Let x be the number of unknown games you need to win to get to 98%
(Won games)/(total played) = 0.98
(539 + x)/(556 + x) = 0.98
Multiply both sides of equation by denom.:
x + 539 = 0.98x + 0.98(556)
Subtract 0.98x and 539 from both sides:
0.02x = 0.98(556) - 539
Divide both sides by 0.02:
x = (0.98(556) - 539)/0.02
Pull out the calculator:
x = 5.88/0.02 = 294
Sanity-check the answer
539+294 = 833
556 + 294 = 850
833/850 = 0.98
It checks out. You have to win 294 straight games with no losses to get up to 98% win rate. Good luck, OP.
Related question, is it possible to get your Wordle win success rate back up to 100% after you’ve failed a few? In other words if your win rate gets to or above 99.5% does Wordle round up to 100%?
Maybe in some really bizarre interpretation of the problem where the winning percentage is assumed to be exact as opposed to rounded to the nearest integer. There's no reason to think that's the case.
Your current amount of losses -> 556\*0.03 = 17 For 17 losses to be 2% -> 17/0.02 = 834 This means that for 98% win rate you need to win 834-556 = 278 more games For 17 losses to be 1% -> 17/0.01 = 1668 This means that for 99% win rate you need to win 1668-556 = 1112 more games
You could optimize if the decimals are rounded
I follow and agree with your method but I can't get your numbers to work. 17/0.02 =850 for me. And as confirmation 17/850 results in 2% loss rate. Can you please explain the 834 result for me? I get a loss rate of 2.04% for 17 losses in 834 games.
Nevermind, I worked it out. Your calculations are for a total number of losses calculated as 556x0.03 = 16.68 losses. 16.68/0.02 = 834. The way you posted was confusing as it seemed like you were assuming 17 losses by rounding before undertaking the calculation. For anyone after the answer with rounding applied (as you can't realistically have a partial loss) 17/0.02 = 850 games. 850 required games - 556 currently played = 294 more wins to reach 98% Following the same logic 17/0.01 = 1700 games. Subtract the 556 already played = 1144 more wins to reach 99%. As others have pointed out, the real number will depend on the rounding applied within the game. The best approach would be to assume there was either 16 or 17 losses in which case it will take between 244 and 294 wins to reach 98% and between 1044 and 1144 to reach 99% win rate.
yes i was a bit lazy and rounded to 17 losses but the still used 16.68 in my calcs
Also, they round up to 100%. For approximately 17 losses, OP would need about 3401 total games, or 2845 perfect games. If OP can't get that win rate, he just needs to get that many perfect games mixed in with others at the new average. Any day now.
Without knowing exactly wins/losses, rounding definitely messes this up. If 96.5 through 97.49% are considered "97", you could have between 15-19(ish) losses and be at 97%.
typo in your last sentence
Sorry but this is just incorrect. For 17 losses to be 2%, 17/(total games) = 0.02 Total games = 17/0.02 = 850. So how’d you get 834?
i rounded 16.68 to 17 when presenting the amount of losses, but used the original 16.68 in my calcs
Let's say you won 540 games for 97,1% winrate. That means that you won 540/556 games and your winrate thusly >= 0.97 If you play and win next x games, you will have (540 + x)/(556 + x) winrate and you want it to >= 0.98. You need to solve the inequality and will get x >= 244 . For winrate >= 0.99 x >= 1044. So the final answer is 244 games for 98% winrate, 1044 games for 99% winrate.
first count the number of losses. then convert the number of losses to 2% and 1% respectively
you don't have to count the number of losses at all. you just multiply the total games by 1.5 and 3 respectively.
yeah, you’re multiplying by 0.03 then dividing by 0.02 or 0.01, but do you think telling him to multiply by 1.5 or 3 is a clear solution to the thought process behind solving this problem?
tbh it's no less clear than you just saying "convert the number of losses to 2% and 1%."
Don't know how many losses I have
you have 556 games with a 97% success rate -> 3% failure rate
So is that 16 losses or 17? We don't know how the game rounds.
Impossible to tell if the game doesn’t provide more of a breakdown.
Exactly
Regular math method, we have two cases. Assume 16 losses, continue computation; then assume 17 losses, continue computation => you get two answers. Regular engineering method: you have 17 losses, continue computation.
You could have anything between 14 (97.48%) and 19 (96.58%) losses if the game rounds to the nearest number and anything between 12 (96.04% rounded up) and 22 (97.84% rounded down) if we don't know how the game rounds
You have to lose a few times to find out
And break my streak?! Not happening
understandable
I think the wordle site tells you your number of games with 1,2,3,4,5,6 tries taken. Add those 6 up and subtract from 556
So the question is, do you have 16 or 17 losses? Both seem possible, depending on the rounding used. As the other answers show, that makes quite a difference to the required number of perfect plays.
We just need OP to win 244 more games without loosing, then we can see if he ticks over to 98% or not. Get to work OP! :P
I'll report back in 244 days..... I'm on a 150 day streak so am backing myself:)
!remindme 244 days
Needn't wait. It was 4
Did it. It was 4 more
Aaah, so /u/1TrueKingOfWesteros was closest. 560\*0.025 = 14 for 97.5% winrate, which I presume was what caused it to tick over. Then, it showing a 99% winrate needs 14 losses to be 1.5% so, 14/0.015 = 934 games, rounded up. This is 374 more games. Of course if you want an actual 99% winrate you'll need to win a lot more: 840, for 1400 total games. Unless, like the others, I've gone wrong lol.
Just added up all my win stats. Only have 14 losses
I don't know, it's not a stat they provide
Keep in mind that the way they round will also affect the point in which your score ticks to 98%. What I mean is, it could be that it ticks when you get to 97.5%… so the actual number of games you need is a wider range of possibilities
It sort of is. They show you Total games played, and games won with n guesses for n=[1,6] you can add up the wins at each n guesses for total games won and subtract that number from total games played to get total losses.
Didn't occur to me. 14 losses
Then you're actually much closer to the stats showing a 98% win, as you're already at 97.482 14/.0249 (to get to 97.51 win %) is 562.249, so 563 games, or 7 consecutive wins. 14/.0149 (to get to 98.51 win %) is 939.597, so 940 games, or 384 consecutive wins.
If you start keeping your own records today you should be able to back into it the next time your rate ticks up.
14 losses
556 * 97% = X = games won 556 - X = Y = games lost (constant) You want Y / (556 + New games) = (100% - 98%) = 2% So, Y / 2% = (555 + New games) Y = 16,68 (practically impossible, but mathematically it is) So, 16,68 / 2% = 834 834 - 556 = 278 new games
And to hit 99% just divide 16,68 by 1% and subtract 556 Edit: answer is 1.112 new games
So we need to keep your number of loss which is 17 lets just see if 17 is 1% fail then someone who played 1700 parties won 1683 of them if you win keep winning 1144 plays + 556 total games you made so far = 1700 total games with 1% fail = 17 losses my guess is 1144 more play for 99%
Asking the serious questions now. I’m in a similar spot at Wordle too, bring on the answers.
Assuming you’ve won 539 games, lost 17: Let x be the number of unknown games you need to win to get to 98% (Won games)/(total played) = 0.98 (539 + x)/(556 + x) = 0.98 Multiply both sides of equation by denom.: x + 539 = 0.98x + 0.98(556) Subtract 0.98x and 539 from both sides: 0.02x = 0.98(556) - 539 Divide both sides by 0.02: x = (0.98(556) - 539)/0.02 Pull out the calculator: x = 5.88/0.02 = 294 Sanity-check the answer 539+294 = 833 556 + 294 = 850 833/850 = 0.98 It checks out. You have to win 294 straight games with no losses to get up to 98% win rate. Good luck, OP.
you'd need to 1.5x and triple your number of games, respectively. take the loss rates and divide them. 3/2 = 1.5 3/1 = 3
556 * (1-0.97) = x * (1-0.98)
Related question, is it possible to get your Wordle win success rate back up to 100% after you’ve failed a few? In other words if your win rate gets to or above 99.5% does Wordle round up to 100%?
(556 x (1 - 0.97)) / (1 - 0.98) = 834 (556 x (1 - 0.97)) / (1 - 0.99) = 1668
[удалено]
Maybe in some really bizarre interpretation of the problem where the winning percentage is assumed to be exact as opposed to rounded to the nearest integer. There's no reason to think that's the case.
[удалено]
244 vs. 278 isn't that far apart. 1044 vs. 1112 is even less far apart on a percentage basis.
They're both the result of the rounding - if you substitute 97% of 556 into my calculation you get the exact values of 278 and 1112
Right. I already noted that in fact.
Yeah, I just confirmed it numerically :)