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Suppose you have a casino with n poker players. Each player has a win rate - the amount of money he wins or loses per hand. We assume that these win rates are normally distributed with a mean of 0. (We also assume that the players don't pay the casino any money.) Our goal is to estimate the variance V of the distribution.

For each player x, we have observed a number of hands; we know how much money x has won or lost on each of these hands.

How would you go about estimating V? Can we get a better estimate if we add some empirical assumptions (in the vein of "in the long run, no-one can sustain a win rate of more than 1$/hand")?

EDIT: Let me try to clarify what I mean by "win rate". If a player wins 500.000$ by playing a million hands then his observed win rate is 0.5$/hand. With a million hands it's also likely that his actual win rate is close to 0.5$/hand. The idea is that a player has an actual win rate which cannot be observed directly, but which is a function of the player's skill. For example, if all players are equally skilled they will all have an actual win rate of 0; in this case, we also have V=0. The question above is concerned with actual win rates.

EDIT: My motivation for asking this question was to estimate how many players have an actual win rate of, say, more than 0.3$/hand. If you disagree with the assumptions made above, feel free to base your estimate on other assumptions.

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    $\begingroup$ smalll point: it is a bit odd to describe "amount won per hand" as "win rate," since the amounts won per hand in poker vary depending on the size of the pot. Poor players win larger proportion of hands but for smaller amounts than good ones, who win fewer proportion of hands but take down bigger pots. You want "avg. hand value" or some such. $\endgroup$ – dmk38 Jun 12 '11 at 15:37
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    $\begingroup$ Well, I think my use of "win rate" reflects how it's used in the poker community (an example: pokerterms.com/winrate.html), though it may be confusing to a statistician. $\endgroup$ – user181813 Jun 12 '11 at 15:53
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    $\begingroup$ Poker would be a strange game indeed if the distribution of win rates were normal: this obviously is a contra-factual assumption. $\endgroup$ – whuber Jun 12 '11 at 17:30
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    $\begingroup$ @user If there are more than two players, it won't even be close. On average the wins should be several times as great as the losses. I would expect an actual dataset of wins to be heavily positively skewed, with a mode equal to the minimum stakes to enter each game (a negative value for the "win"). This suggests that it's a poor idea to use the variance to make inferences about sustainable play. $\endgroup$ – whuber Jun 12 '11 at 19:05
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    $\begingroup$ Even with your rewording, it is not obvious that the mean is zero: winners tend to play more games than losers as they do not run out of money, so a straight mean across all players of mean net winnings per game will probably be negative. An average weighted by numbers of games played by each player is probably more easily analysed as the average of the total net winnings of each player. $\endgroup$ – Henry Jun 12 '11 at 23:26
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Here is an approach to calculate how many players have an actual win rate of over 0.3 dollar per hand.

0: Check whether individual player results are normally distributed.
1: Calculate the standard deviation for each player.
2: Given his mean and standard deviation, calculate the probability that his win rate is over 0.3
3: Add all these probabilities of all players and this is a good estimate of how many players actually have a win rate of over 0.3.

Perhaps you may have to stop at step 0 in real life, but you can try to find a proper destribution or just use this as a guideline.

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  • $\begingroup$ I doubt anyone would get beyond step 0: please see my comments beneath the question itself. Note, too, that the OP asks about an amount of money (even though it is perversely termed a "rate"), not a count of players. $\endgroup$ – whuber Jan 2 '12 at 16:09

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