How much games will take to recover winrate?

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AtomEistee

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How to find the amount of games to play to get winrate equal 50%? Imagine if i played 27 games, i won 6 and lost 21, my winrate right now is 22%, im pretty sure what if i will continue play right now i have chance of win like 67% or even more, what is the formula to get the answer?

P.S Im doing it just for my small site and im really bad at math, so please explain it fully to make me understand :)
i already tried to do something by my self but its was weird what with 67 and 83 i got same ~50 games to recover :(
Im not sure do you need it but here is my code where i tried to find the amount of games, if it possible can you explain whats wrong and what should i do? I know its not the programming forum, but i need just the sense of what to do :)
 

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BigBeachBanana said:
Playing more games give you a chance to recover a 50% win rate. It does not provide you with certainty to recover a 50% win rate.
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What do you mean by that? If i know what i will win with 60% or even more, my winrate should grow with time, and i wand to find out how much games i need to play to get same amount of loses and wins. Sorry if im looking stupid :)
 
AtomEistee said:
If i know what i will win with 60% or even more, my winrate should grow with time, and i wand to find out how much games i need to play to get same amount of loses and wins.
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I don't understand what you mean by "If I know what I will win with 60% or even more,..." Do you mean that in each game, you have a probability of winning is 60% or more?
 
AtomEistee said:
Yes, exactly, sorry for bad explanation
Click to expand...
Although you have 60% of winning per game, you also have 40% of losing. That's why playing more games doesn't guarantee an improvement in win rate, in fact, you have can also decrease your win rate. If casinos operate on your theory, then they'd be out of business. The more you gamble doesn't guarantee that you will win more money, but you could also lose all the additional money you put in right? Similarly, playing more games doesn't guarantee a higher win rate, because you can still lose.

The best you can do is calculate the probability they will recover a 50% win rate within the n-number of games, given the player has >60% of winning.
 
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BigBeachBanana said:
Although you have 60% of winning per game, you also have 40% of losing. That's why playing more games doesn't guarantee an improvement in win rate, in fact, you have can also decrease your win rate. If casinos operate on your theory, then they'd be out of business. The more you gamble doesn't guarantee that you will win more money, but you could also lose all the additional money you put in right? Similarly, playing more games doesn't guarantee a higher win rate, because you can still lose.

The best you can do is calculate the probability they will recover a 50% win rate within the n-number of games, given the player has >60% of winning.
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Oh, so its impossible to find exact amount of games, okay, thank you for help
 
Here's the probability of winning 15 games (or more) if you play the number of games in the first column and the probability of winning each individual game is 60%...
Code: 
15  .000470184984576
16  .0032912948920320
17  .01231884659589120
18  .032781297124638720
19  .0696137080763842560
20  .12559897272303747072
21  .20024599225190842366
22  .28982241568655356713
23  .38835648146466322482
24  .48908019314895309661
25  .58577495636587137233
26  .67367928656306980232
27  .74986303940064177053
28  .81315415714262462910
29  .86378705133621090493
30  .90294315617925094242
31  .93231023481153094842
32  .95373092746095868673
33  .96896342001166281284
34  .97954599378373088091
35  .98674214394873707928
36  .99153957739206394836

So, if you played 25 games then you'd have a 59% chance of obtaining 15 or more wins. This was found using the Binomial_distribution
Code: 
losses=15 # 21-6
p=0.6     # 60% prob of individual win
for (g=losses; g<37; ++g) { # number of games to play
  #
  s=0 # sum the probabilities of obtaining "w" wins, where w>=losses
  for (w=losses; w<=g; ++w)
    s += choose(g,w) * p^w * (1-p)^(g-w)
  print g, "  ", s, "\n";
}

EDIT: I'd be VERY surprised if you've found a betting game where a player can become skilled enough to win 60% of the time (if the house and the player stand to loose the same amount of money). If your guess at 60% is wrong, then the whole table above will be wrong too!
 
Last edited:
Cubist said:
Here's the probability of winning 15 games (or more) if you play the number of games in the first column and the probability of winning each individual game is 60%...
Code: 
15  .000470184984576
16  .0032912948920320
17  .01231884659589120
18  .032781297124638720
19  .0696137080763842560
20  .12559897272303747072
21  .20024599225190842366
22  .28982241568655356713
23  .38835648146466322482
24  .48908019314895309661
25  .58577495636587137233
26  .67367928656306980232
27  .74986303940064177053
28  .81315415714262462910
29  .86378705133621090493
30  .90294315617925094242
31  .93231023481153094842
32  .95373092746095868673
33  .96896342001166281284
34  .97954599378373088091
35  .98674214394873707928
36  .99153957739206394836

So, if you played 25 games then you'd have a 59% chance of obtaining 15 or more wins. This was found using the Binomial_distribution
Code: 
losses=15 # 21-6
p=0.6     # 60% prob of individual win
for (g=losses; g<37; ++g) { # number of games to play
  #
  s=0 # sum the probabilities of obtaining "w" wins, where w>=losses
  for (w=losses; w<=g; ++w)
    s += choose(g,w) * p^w * (1-p)^(g-w)
  print g, "  ", s, "\n";
}

EDIT: I'd be VERY surprised if you've found a betting game where a player can become skilled enough to win 60% of the time (if the house and the player stand to loose the same amount of money). If your guess at 60% is wrong, then the whole table above will be wrong too!
Click to expand...
Oh wow, thank you!
 
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