Gambling win rate
 

Not sure if this question can be answered, but I was wondering if anyone can tell me what an acceptable amount of bets would be to calculate an accurate win rate. Or how this could be worked out to some degree of accuracy.

Thanks.

Re: Gambling win rate
 

You're not going to like the answer.

A 95% confidence interval (95% CI) for the win rate per hand (Xbar) is given by Xbar ± 1.96 * S / sqrt(n), where S is the standard deviation per hand and n is the number of hands. The half-width of this interval, that is, 1.96 * S / sqrt(n) is a conventional measure of the precision of X-bar, in this case, the estimate of the win rate per hand. So, to determine how big n has to be give us a desired level of precision for our win rate, we can set the above expression equal to the desired level of precision, and solve for n. Let's say we want to estimate our win rate per 100 hands to ± 0.5 BB/100. This is equivalent to measuring our win rate per hand to ± 0.005. So, our equation for n becomes

n = (1.96 * S / 0.005)^2.

A reasonable value for S is 1.5. Solving for n, we find we need about 346,000 hands to estimate our win rate per 100 hands to within 1/2 a BB. Like I said, you won't like the answer.

If you are satisfied with worse precision, the following table might be helpful. The first column is the desired precision (BB/100), the second column is the number of hands required.

0.5 ... 345744
1 ..... 86436
1.5 ... 38416
2 ..... 21609

Re: Gambling win rate
 

[ QUOTE ]
You're not going to like the answer.

[/ QUOTE ]

You're right, I don't like the answer. See the Bayesians vs frequensists thread for why the calculations that are given in that post may not be the correct way to go (they overestimate the true win rate of a player who has a positive return over a number of hands).

Re: Gambling win rate
 

[ QUOTE ]
[ QUOTE ]
You're not going to like the answer.

[/ QUOTE ]

You're right, I don't like the answer. See the Bayesians vs frequensists thread for why the calculations that are given in that post may not be the correct way to go (they overestimate the true win rate of a player who has a positive return over a number of hands).

[/ QUOTE ]

Well, if you don't like my answer, you're going to like the Bayesian answer even less, because the Bayesian estimate of win rate is a weighted average of the estimate under the prior distribution and the estimate under the posterior distribution. If you assume, as in the post you've linked, that your win rate under the posterior distribution is negative, but in fact, it is positive, then you will need a larger sample size than you would using a classical approach to overcome what is essentially a bias you've introduced by making a false (or at least a very conservative) assumption about your prior distribution.

My question is, Why would you treat yourself as a randomly selected member of a the general poker-playing population? For one thing, you're not randomly selected, you're yourself; and for another, you don't seem to fit the model, as, presumably, you've studied books, participatd in these forums, and have thought a lot more about correct poker playing than the general population.

To my way of thinking, using Bayesian statistics to estimate your own win rate is falling into Fancy Statistics Syndrome. When in doubt, follow the KISS principle: Keep It Simple, Stupid.

Jay

Re: Gambling win rate
 

I knew my stat class was good for something!

The equation given in a previous post works for ring games; for tourneys I use:

z=(p-pi)/sqrt(pi(1-pi)/n), where

z=# of SD away from the mean
p=observed "success" rate (here, ITM)
pi=theoretical success rate
n=number of attempts

so, if you participate in 100 tourneys (average 14% payout) and you ITM 27 times, your score is:

p=0.27
pi=0.14
n=100

z=3.71 SD, which on my chart coorolates to prob<0.0002 (actually my chart runs out at z=3.59)

Now what this means is that a typical donk, playing 100 tourneys, should acheive your results approximately once every 5000 times. It's up to you to decide if that means you're good or really lucky.

BTW, 1.96 SD means that said donk will get similar results once every 40 tries. Because of this, I consider 2 sigma merely "suggestive" of an outcome, and I wouldn't call myself a "proven" winner until 3.0 sigma at minimum.

Re: Gambling win rate
 

To further expand on the previous quote: if you want to get a feel for the lower end of your winrate, you can either solove the equation for pi with z=2 (very messy) or choose a value for pi and "plug & chug" until you wind up with your desired z value. For the example I gave, with p=0.27 and n=100, the highest value (rounded to 2 digits) for pi that allows z>=1.96 is pi=0.19, where z=2.04 and prob=0.0207.

Thus, using my criteria, the poker player in my example can say with (relative) certainty that he is doing better at ITM than the average donk, and data is suggestive that he ought to continue to beat 19% ITM. Note that this says nothing about how one fares once making the money. You could, if you wanted, change the equation to make it "how often I make the final table" or "how often I win it all" but since these events occurr less frequently, you'd need a much larger sample size.