normal distribution

[jason1990]

I started a thread in "Poker Theory", but it was suggested it better belongs here. The original post was this:

[ QUOTE ]
Does anyone know, roughly, how many poker hands a person needs to play before their winnings become approximately normally distributed? For example, suppose someone has a winrate of 2BB/100 and a standard deviation of 15BB/100. They might think that if they play 100 hands, then they have a 95% chance of winning somewhere between -28BB and 32BB (a spread of 2 SD's). But this is only true if the result of playing 100 hands is (roughly) normally distributed. Is it?

[/ QUOTE ]

I think it was then misinterpreted, so I posted this:

[ QUOTE ]
Let me rephrase: suppose you play 1000 sessions of 100 hands each. You plot the result of each session on a graph. Will this graph have a bell-shaped curve?

For example, if you plot the result of each hand on a graph, that graph will not have a bell-shaped curve. It will have a huge spike at 0, because of all the folding, and will not be symmetrical about the mean, since (on any particular hand) it is very likely that you will lose a small amount and much less likely you will win a large amount. What this shows is that the result of a single hand is not normally distributed.

But what about the result of a 100 hand session?

[Edit: By the way, I should clarify for non-mathematicians: By "normally distributed", I mean that it has a Gaussian distribution which is characterized by the bell-shaped curve. I do not mean, in any way, that the result is "normal" (or "typical") in the ordinary sense of the word, i.e. I am not asking whether the result of 100 hands can be used as a reliable indicator of long-term results. I know that this is very far from the truth.]

[/ QUOTE ]

Then, finally, I decided to be very specific and posted this:

"It occured to me that I should probably be more specific. I am assuming that the winrate and SD are completely accurate. So...

Suppose a player plays a 100 million hands and, based on this, determines that his winrate is 2BB/100 and his SD is 15BB/100. Converting this to various different units, we get

winrate: 0.2BB/10 = 2BB/100 = 20BB/1000 = 200BB/10000 = 2000BB/100K
SD: 4.7BB/10 = 15BB/100 = 47BB/1000 = 150BB/10000 = 474BB/100K.

Now, after this player has already played these 100 million hands, he sits down at the poker table and asks himself 5 questions:

1. 'Do I have about a 95% chance of winning between -9.2BB and 9.6BB on the next 10 hands I play?'

2. 'Do I have about a 95% chance of winning between -28BB and 32BB on the next 100 hands I play?'

3. 'Do I have about a 95% chance of winning between -74BB and 114BB on the next 1000 hands I play?'

4. 'Do I have about a 95% chance of winning between -100BB and 500BB on the next 10000 hands I play?'

5. 'Do I have about a 95% chance of winning between 1052BB and 2948BB on the next 100K hands I play?'

The point of these questions is this: you have a 95% chance of falling within 2 SD's of your mean provided the thing you're asking about has the bell-shaped property. People always say you need 100K hands to determine if you're a winning player. However, the bell-shaped property is going to emerge long before 100K hands. So the answer to questions 4 and 5 is a definite "yes". The answer to question 1 is an obvious no, as experience alone can tell us. This is because the bell-shaped property does not emerge after only 10 hands. But how about questions 2 and 3? Mathematical experience indicates to me that the answer to question 3 is very likely to be yes. So question 2 is the one that interests me most."

Hopefully, people here will have some insight into this question.

[alThor]

Answer: it depends.

First, if you want to be anally technical, it never is "exactly" normal. The distribution only tends towards normal as n (# hands) goes to infinity. You probably knew that, so...

Ok, less anally, it depends on the game. In looser games, you will win fewer but bigger pots, creating more skew. That would take more hands than in tigher games. So there is no single answer to your question, covering all situations.

My guess is that in the 100-200 range, it's looking pretty normal in most games. If I had some data, I could set up a quick sim and give a better answer. I would need to know how often (%) you win/lose 1BB, 2BBs, 3BBS, ... etc. If anyone wants to post that data...

alThor

[Izverg04]

Without running any simulations, I can say that the distribution of wins over 100 hands will be very close to normally distributed. This wouldn't be the case for example in video poker, however in limit poker the underlying distribution is fairly normal already, so that the win over 100 hands for all practical purposes is normally distributed.

[pzhon]

You want an effective version of the Central Limit Theorem. Since the amount you can win in a hand is bounded, the 3rd moment is finite, and you can use the Berry-Esseen theorem (http://www.wordiq.com/definition/Berry-Esséen_theorem).

[jason1990]

Frankly, I didn't expect anyone to reply with this comment. I guess there are more mathematicians here than I realized. Anyway, I already tried this. Using the naive bound on the 3rd moment, which we get using the fact that the amount you can win in a hand is bounded, Berry-Esseen says we're not even close to a normal until 100K hands. In practice, I'm sure the 3rd moment is much smaller. However, it cannot be any smaller than the SD, and using the SD in place of the 3rd moment in Berry-Esseen doesn't improve the result very much. The problem, I think, is that Berry-Esseen is a generic bound which is meant to work no matter what the underlying distribution is. In our case, the underlying distribution is probably nice enough to give a much faster rate of convergence than that given by Berry-Esseen.

[pzhon]

[ QUOTE ]
Using the naive bound on the 3rd moment, which we get using the fact that the amount you can win in a hand is bounded, Berry-Esseen says we're not even close to a normal until 100K hands.

[/ QUOTE ]
Is the problem the Berry-Esséen theorem, or the bad estimate you are using for the 3rd moment? I would not be surprised if the Berry-Esséen estimate could be improved in a more restricted context, but I think you may get a significant improvement by estimating the 3rd moment from actual data.

[jason1990]

[ QUOTE ]
[ QUOTE ]
Using the naive bound on the 3rd moment, which we get using the fact that the amount you can win in a hand is bounded, Berry-Esseen says we're not even close to a normal until 100K hands.

[/ QUOTE ]
Is the problem the Berry-Esséen theorem, or the bad estimate you are using for the 3rd moment? I would not be surprised if the Berry-Esséen estimate could be improved in a more restricted context, but I think you may get a significant improvement by estimating the 3rd moment from actual data.

[/ QUOTE ]

The problem is definitely with Berry-Esseen, because the problem remains if you use the SD as the value of the 3rd moment. And by Jensen's inequality, the true 3rd moment (i.e. the cube root of rho, in the notation of the link you provided) is always greater than or equal to the SD.

[alThor]

[ QUOTE ]
The problem is definitely with Berry-Esseen

[/ QUOTE ]

The problem is with the expectation that such a theorem would be of value here. This theorem provides an upper bound for all distributions (satisfying the assumptions); in particular, it applies to "worst case" distributions. While I wouldn't call the distribution from one poker hand "very close" to normal (close is relative anyway...), it is certainly closer than some "worst case" examples we could dream up. So clearly this case would converge quicker.

alThor

[jason1990]

[ QUOTE ]
The problem is with the expectation that such a theorem would be of value here. This theorem provides an upper bound for all distributions (satisfying the assumptions); in particular, it applies to "worst case" distributions. While I wouldn't call the distribution from one poker hand "very close" to normal (close is relative anyway...), it is certainly closer than some "worst case" examples we could dream up. So clearly this case would converge quicker.

[/ QUOTE ]

Absolutely. However, one thing to notice is that the "worst case" among distributions that share the same first 3 moments with the distribution of one poker hand is "nicer" than the "worst case" among those that share only the mean and SD. And, under mild assumptions, the more moments we know, the closer we get to describing the true shape of the distribution.

So, does anyone know, is there a Berry-Esseen-type theorem which utilizes the 4th moment? How about one which uses the first n moments? Something like this might be of use, provided we had raw data on these higher moments. In a case like that, the "worst case" would be much closer to the actual case, since we are specifying more moments and, hence, specifying more of the shape of the distribution.

(I feel I've been rambling and unclear. Hopefully, this makes sense.)

[irchans]

Jason1990 asked, "1. Do I have about a 95% chance of winning between -9.2BB and 9.6BB on the next 10 hands I play?" assuming an edge of 0.02 BB per hand and a standard deviation of 1.5 BB per hand.


To answer your question I did some numerical calculations.

First I had to guess at the distribution for one hand. My guess was

70.00% 0 win
20.67% lose 1 big bet
7.06% win 2 to 7 Big Bets
2.27% lose 2 to 6 Big Bets

That distribution has a standard deviation of 1.5 BB and an expectation of 0.02 BB. I also assumed that you always lost or gained an integral number of BB.

I then computed the probabilities for the possible results after playing 10 hands. My results are printed at the bottom of this post.

I found that 94.7261% of the time, we get a result between -9.2 and +9.6 BB. So the answer to question 1 is YES. The Gaussian estimate is accurate even after only 10 hands.

This seems surprising considering that the distribution does not look like a bell shaped curve as noted by others. The Berry-Esseen theorem indicates that the cumulative distribution for 10 hands and the cumulative normal distribution must differ by less than

C rho /(sigma^3 * sqrt(10))
= 0.8* 11.4/(3.37 * 3.16) = 0.85.

When we measure the actual difference, we get differences as large as 0.13. But at the tails of the distribution, the differences are much smaller. So, Berry-Esseen is way too pessimistic for the purposes of these questions.


<font class="small">Code:</font><hr /><pre>
win
or
loss probability
-19 0.000019357419535699934
-18 0.00003600421040582269
-17 0.00006456969332220574
-16 0.00012481787038983035
-15 0.0002612687168103279
-14 0.0005247804087985602
-13 0.0009241500966841833
-12 0.0014654484488216134
-11 0.002395737665248195
-10 0.004454606841964618
-9 0.008541200959866983
-8 0.014395733990861418
-7 0.020100017620170867
-6 0.02570073403965464
-5 0.037722712056824786
-4 0.06518731850759737
-3 0.10496371582704672
-2 0.12977770479990375
-1 0.1114191806112588
0 0.07075359035811012
1 0.05433695876840763
2 0.05865557747163329
3 0.058014682745855146
4 0.05476807202964498
5 0.04626930435645343
6 0.03333247094900252
7 0.021922502380769975
8 0.016717751477048966
9 0.014681501169795198
10 0.01198293096170944
11 0.009280628047952062
12 0.006662885943047578
13 0.004428813219035153
14 0.002913464124472338
15 0.002112229124996278
16 0.0016192430382377574
17 0.0011682355314015158
18 0.0008031858302970064
19 0.0005240124719644412
20 0.0003300763164111946
21 0.0002113263017840343
22 0.00014380247684095702
23 0.00009954016566822779
24 0.00006540745765697863
25 0.00004122060143457475
26 0.000025060547021882077
27 0.00001503408835227676
</pre><hr />

[jason1990]

Wow, you're probably right -- even if your guess at the one-hand distribution is off. I hadn't thought about this so carefully. Winning 9.6BB means winning a fairly large pot, considering what you've put into it as well. And losing 9.2BB in a single hand is probably a pretty bad beat. So this is not that surprising after all. Nice post -- thanks.

[jason1990]

Wow, please ignore me. I think I need a break. Nothing I just said made any sense. I was confusing one hand with ten hands. I will have to look at this again later after I sleep.