expected standard deviation question

bodi · #1 07-11-2005, 12:11 PM

I'm trying to figure out how unlucky i've gotten over the past 20000 hands. I know i'm supposed to get about 20000/169 = 118 of each suited and offsuit starting hands. What's the standard deviation on this #. I think you should use the binomial sd, ie sqrt(n * p * (1-p)) but it doesn't look like it's working out. for a given hand, aks, over 20k hands i'm expected to get 118 with sd (20k * (1/169^2))^.5, which comes out to .8, which seems wrong. can someone correct this ??

BruceZ · #2 07-11-2005, 12:32 PM

[ QUOTE ]
I'm trying to figure out how unlucky i've gotten over the past 20000 hands. I know i'm supposed to get about 20000/169 = 118 of each suited and offsuit starting hands. What's the standard deviation on this #. I think you should use the binomial sd, ie sqrt(n * p * (1-p)) but it doesn't look like it's working out. for a given hand, aks, over 20k hands i'm expected to get 118 with sd (20k * (1/169^2))^.5, which comes out to .8, which seems wrong. can someone correct this ??

[/ QUOTE ]

For p = 1/169:

p*(1-p) = 1/169 * 168/169 = 168/169^2

(20k * 168/169^2)^0.5 =~ 10.8

But the 169 starting hands do not have equal probability. The ratio is offsuit:suited

airs = 12:4:1. So it doesn't make sense to set p = 1/169.

For AKs, p = 4/1326 = 1/331.5

(20k * 330.5/331.5^2)^0.5 =~ 7.8

PairTheBoard · #3 07-11-2005, 12:39 PM

[ QUOTE ]
I'm trying to figure out how unlucky i've gotten over the past 20000 hands. I know i'm supposed to get about 20000/169 = 118 of each suited and offsuit starting hands. What's the standard deviation on this #. I think you should use the binomial sd, ie sqrt(n * p * (1-p)) but it doesn't look like it's working out. for a given hand, aks, over 20k hands i'm expected to get 118 with sd (20k * (1/169^2))^.5, which comes out to .8, which seems wrong. can someone correct this ??

[/ QUOTE ]

The (1/169^2) term should be (1/169)(168/169) or (168/169^2) giving a sd of about 10. Even this seems small but I guess it's right.

PairTheBoard

LetYouDown · #4 07-11-2005, 12:48 PM

It's been forever since I've dealt with standard deviation, as I've rarely had a poker application for it. As a quick refresher.

Say I'm trying to calculate the standard deviation for pocket pairs in 10000 hands. Your expected amount is roughly 588. Standard deviation being:

(10000 * 1/17 * (1-1/17))^.5

(10000 * 16/17^2)^.5

~23.53

So this says that you can expect 588 pairs +-23.53, on average for every 10000 hand trial?

BruceZ · #5 07-11-2005, 01:24 PM

[ QUOTE ]
It's been forever since I've dealt with standard deviation, as I've rarely had a poker application for it. As a quick refresher.

Say I'm trying to calculate the standard deviation for pocket pairs in 10000 hands. Your expected amount is roughly 588. Standard deviation being:

(10000 * 1/17 * (1-1/17))^.5

(10000 * 16/17^2)^.5

~23.53

So this says that you can expect 588 pairs +-23.53, on average for every 10000 hand trial?

[/ QUOTE ]

Right, so using the normal distribution with this mean and standard deviation to approximate the binomial distribution, we would expect the number of pairs to fall in this range about 68% of the time.

LetYouDown · #6 07-11-2005, 02:08 PM

Thanks. What's the formula/reasoning behind the binomial distribution %age?

AaronBrown · #7 07-11-2005, 02:35 PM

In addition to the excellent answers you've had from BruceZ, LetYouDown and PairTheBoard, I'd add that you have to be careful intepreting these numbers.

Let's say we're only looking at pairs, and to keep the numbers round, that you played 22,100 hands instead of 20,000. You expect 100 of each pair, with a standard deviation just under 10 (9.977 to be precise). Therefore, you might think there is a 95% chance that you will have more than 80 and less than 120 of any pair (that's the mean plus or minus two standard deviations).

However, the chance of getting 80 or fewer of some pair is 1 in 4, the odds of getting 120 or more of some pair is 1 in 3. Those total to 7/12, more than 1/2, not 5%.

The reason is that the Normal approximation works best around the center of the distribution, extreme events are more likely than the Normal predicts. Also, even under a Normal you expect some pair to have a surprisingly low or high number of occurrences. So if you look at see that you got only 72 AA; don't decide that you are cursed.

BruceZ · #8 07-11-2005, 03:04 PM

[ QUOTE ]
Thanks. What's the formula/reasoning behind the binomial distribution %age?

[/ QUOTE ]

For a normal distribution, 68% lie within +/- 1 standard deviation of the mean. The normal distribution is a good approximation to the binomial distribution when N*p*(1-p) is large. This is due to the central limit theorem. Define a random variable Xi which is 1 when hand i is a pair, and 0 when hand i is a non-pair. Each Xi has a mean of 1/17, and a variance of 1/17 - (1/17)^2. The sum of the Xi for a large number of hands N gives the number of pairs in N hands, which is distributed as a binomial distribution. The central limit theorem tells us that the distribution of this sum of random variables Xi approaches a normal distribution with mean N*1/17 and variance N*[1/17 - (1/17)^2].

Siegmund · #9 07-11-2005, 03:35 PM

[ QUOTE ]
It's been forever since I've dealt with standard deviation, as I've rarely had a poker application for it.

[/ QUOTE ]

This surprises me.

At risk of both threadjacking and sounding preachy: most applications of expected values, including all sample means, also require accompanying measures of uncertainty. It's one of my big complaints with programs like PokerTracker: they spit forth all kinds of estimates and averages, unaccompanied by the rest of the information needed to know which of them are significant.

LetYouDown · #10 07-11-2005, 03:40 PM

I think your point is certainly valid. It's been a while since I've actually had to grunt any of it out. I think if anything was way off base (for instance in PokerTracker), I'd have enough "feel" to know that it didn't seem normal. It would be nice if PT did include this info.

I've never been an "online poker is rigged" theorist...so I've had enough faith to believe that those sorts of numbers fall within statistically reasonable bounds. Sure, it affects winrate, etc...but that's never really been my focus. I play to win, not for the money...so my winrate is purely emotional as opposed to numerical.