Variance of a poker tournament

Printable View

April 8th 2011, 09:11 AM
DannyOcean

Variance of a poker tournament

This is just an indepedent project I'm working on because I think it's fun.

Suppose we have a 9-man poker tournament. Buy in is 6.50, payouts go to 1st, 2nd and 3rd place like such

1st - 27
2nd - 16.20
3rd - 10.80

We have a player who plays n of these tournaments, and has some distribution of finishes

1st - x%
2nd - y%
3rd - z%
out of the money = 1-x-y-z = oom

To calculate profit per tournament (PPT), we would just take 27x +16.2y+10.8z-6.5

Return on Investment (ROI) is another statistic of importance, and is equal to PPT/(Buy-in), so in this case PPT/6.5

How can we find the variance of PPT and ROI? This step I feel like I know how to do, but just want to double check. Once we do find the variance of PPT and ROI, can we build confidence intervals for them?
April 8th 2011, 05:25 PM
theodds

I typed up a response but decided not to post it because it isn't clear to me what exactly you are trying to get at. Which of the following are you trying to do?

(a) Given some tournament results, calculate an estimate of PPT and ROI and then put a confidence interval on it.

(b) Given some assumptions about x, y, and z, calculate the variance of PPT/ROI in order to get an understanding of how variable your winnings are.

(c) Something else.

I have played a lot of SNGs in my day, and have done all of these things at one point or another.

If you don't care at all about the underlying mathematics and just want to learn about your profitability, I know that there are spreadsheets available on 2+2 forums and elsewhere that are designed to handle (a) and other related things. I think the usual software packages (e.g. Poker Tracker, Holdem Manager) will do this as well, but maybe not confidence intervals. I don't think I've come across (b) publicly, although I'm sure it has been done many, many times.
April 8th 2011, 05:40 PM
theodds

I typed up a response but decided not to post it because it isn't clear to me what exactly you are trying to get at. Which of the following are you trying to do?

(a) Given some tournament results, calculate an estimate of PPT and ROI and then put a confidence interval on it.

(b) Given some assumptions about x, y, and z, calculate the variance of PPT/ROI in order to get an understanding of how variable your winnings are.

(c) Something else.

I have played a lot of SNGs in my day, and did both of these things at one point or another..

If you don't care at all about the underlying mathematics and just want to learn about your profitability, I know that there are spreadsheets available on 2+2 forums and elsewhere that are designed to handle (a) and other related things. I think the usual software packages will do this as well, but maybe not confidence intervals. I don't think I've come across (b) publicly, although I'm sure it has been done many, many times.
April 9th 2011, 08:06 AM
DannyOcean

I'm looking for (B)

I am a member of the 2p2 forums as well, and an SNG grinder. I'm more interested in the underlying math.

My thought about the variance of the PPT of such an SNG is that it would be something like "

(SUM[([prize-buyin]*probability)-PPT])/n

and then, because ROI = PPT/Buy-in, V(ROI)= (1/(buy-in)^2)*V(PPT)

But I'm not sure if these are correct.
April 9th 2011, 08:09 AM
DannyOcean

Then, assuming we can calculate a variance for ROI, I'm interesting in creating a confidence interval for ROI.

I know this isn't a typical distribution like a normal, so we might not have any predetermined chart to look at. My thought was that there could be an application of chebyshev's rule to try to create an interval (although chebyshev is super conservative of course).
April 9th 2011, 08:33 AM
theodds

No need to invoke Chebychev, you can use the central limit theorem.

Let $X_1, X_2, ..., X_n$ be the results from $n$ tournaments, which we'll assume are independent and identically distributed. The operator $\mbox E$ is the expected value and $\mbox{Var}$ is the variance. Then obviously:

$\mbox E X_1 = 20.50 x + 9.7 y + 4.3 z - 6.50 (1 - x - y - z)$

and

$\mbox E X_1 ^ 2 = 420.25 x + 94.09 y + 18.49 z + 42.25 (1 - x - y - z)$

so $\mbox{Var}X_1 = \mbox E X_1 ^ 2 - \mbox E X_1$ , which we'll denote $\sigma^2$ . Your (empirical) PPT is given by $\bar X = n^{-1} \sum X_i$ and your (empirical) ROI is PPT / 6.5. It follows that

$ \mbox{Var PPT} = \frac {\sigma^2}{n} $

and

$ \mbox{Var ROI} = \frac{\sigma^2}{(6.5^2) n}. $

From the central limit theorem, for suitably large $n$ we have that both PPT and ROI are approximately normal, with the correct mean and variance as given above. So, you now have the asymptotic distribution of both of these statistics.

With this you can answer a few somewhat interesting questions, like "what is the probability that I will be a losing player over my next 1000 games" for various values of x, y, and z.
April 9th 2011, 08:49 AM
theodds

Also, you wouldn't typically construct a confidence interval, per se, in problem (b). What you would do, rather, is make a prediction interval for PPT. If you let $\theta$ be your true PPT then we would predict that 95% of the time we will observe

$\theta - 1.96 \frac{\sigma}{\sqrt{n}} \le \bar X \le \theta + 1.96 \frac{\sigma}{\sqrt{n}}$

In problem (a) you are more interested in estimating $\theta$ , so you would make a confidence interval for $\theta$ whereas in problem (b) you know $\theta$ and are trying to learn about what you should expect to observe in terms of your results.
April 10th 2011, 08:57 AM
DannyOcean

Quote:

Originally Posted by theodds

No need to invoke Chebychev, you can use the central limit theorem.

Let $X_1, X_2, ..., X_n$ be the results from $n$ tournaments, which we'll assume are independent and identically distributed. The operator $\mbox E$ is the expected value and $\mbox{Var}$ is the variance. Then obviously:

$\mbox E X_1 = 20.50 x + 9.7 y + 4.3 z - 6.50 (1 - x - y - z)$

and

$\mbox E X_1 ^ 2 = 420.25 x + 94.09 y + 18.49 z + 42.25 (1 - x - y - z)$

so $\mbox{Var}X_1 = \mbox E X_1 ^ 2 - \mbox E X_1$ , which we'll denote $\sigma^2$ . Your (empirical) PPT is given by $\bar X = n^{-1} \sum X_i$ and your (empirical) ROI is PPT / 6.5. It follows that

$ \mbox{Var PPT} = \frac {\sigma^2}{n} $

and

$ \mbox{Var ROI} = \frac{\sigma^2}{(6.5^2) n}. $

From the central limit theorem, for suitably large $n$ we have that both PPT and ROI are approximately normal, with the correct mean and variance as given above. So, you now have the asymptotic distribution of both of these statistics.

With this you can answer a few somewhat interesting questions, like "what is the probability that I will be a losing player over my next 1000 games" for various values of x, y, and z.

isn't variance E(X^2)-((E(X))^2, not what you have above? I think you may have left out a ^2.
April 10th 2011, 09:14 AM
theodds

Yeah, that's a typo. Unfortunately I can't edit it to fix it (Angry)