expected standard deviation question

[bodi]

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]

[ 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:pairs = 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]

[ 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]

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]

[ 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]

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

[AaronBrown]

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]

[ 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]

[ 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]

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.

[BruceZ]

[ QUOTE ]
It would be nice if PT did include this info.

[/ QUOTE ]

PT gives your standard deviation. It is buried under session notes if you click on "more detail".

[LetYouDown]

Interesting. I'll have to look at that when I get home. Thanks for pointing it out.

I typically use PT to categorize my opponents. I rarely even look at the financial information. I realize I'm certainly in the minority in that regard.

[PairTheBoard]

[ QUOTE ]
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.

[/ QUOTE ]

Especially: "The reason is that the Normal approximation works best around the center of the distribution, extreme events are more likely than the Normal predicts."

However correct this statement may be, it seems a bit misleading when coupled with the way you've switched from probabilties for a specific pair to probabilties for some pair out of the 13 kinds of pairs.

To make it clear, you're not saying that for some specific pair like Aces, there is 1 chance in 3 that the number you'll see in 22,100 trials will exceed 120? But I assume that there is 1 chance in 3 that there will be some pair out of the 13 kinds of pairs that will be an outlier and exceed 120. I don't think this has to do with problems with the Normal Aproximation to the Binomial for small p. If the chance for a specific pair is about 2.5% it's not supprising the chance for some pair out of 13 would be about 1 in 3 or somewhat less. However, the 1 chance in 4 for such an outlier on the low side below 80 does show the somewhat nonsymetric, skewed, nonnormal looking shape of the binomial for Np(1-p) this small.

I wonder what a measure of skewness for this distribution would be? Also, just how much more likely is it for a specific pair like Aces to have extreme results in this case, compared to those given by the Normal aproximation?

PairTheBoard

[Siegmund]

[ QUOTE ]

PT gives your standard deviation. It is buried under session notes if you click on "more detail".

[/ QUOTE ]

PT gives the standard deviation of your winrate.

It does not give the standard deviations of all the VPIP, PFR, AF, WSD etc estimates, or of your results by seat, or of your return on a particular set of hole cards. Nor does it give a meaasure of its confidence in assigning players to classes when it autorates them.

I am sure casual users don't care, and maybe are even glad not to see the screen cluttered with these. But it annoys me to no end that people blindly trust these numbers just because they are numbers - and when warned not to, they all post asking "how many hands does it take before I can trust them," when it would have been so easy to have an indication next to each one of how trustworthy it is.

[AaronBrown]

You're right, of course, I worded my answer misleadingly.

There are two things going on. First is that the Normal approximation begins to break down. There is 22% more chance of getting 120 or more of a specific pair than the Normal approximation suggests. There's 179% chance of getting 140 or more.

But as you point out, the much bigger factor is that some pair is very likely to have a deviant outcome. There may be only a 3% chance of a specific pair showing up 120 or more time, but there is (roughly) 13 times that chance of SOME pair showing up 120 or more times.

[PairTheBoard]

AaronBrown --
"There are two things going on. First is that the Normal approximation begins to break down. There is 22% more chance of getting 120 or more of a specific pair than the Normal approximation suggests. There's 179% chance of getting 140 or more."

That's interesting. So if the Normal Aproximation was a 2.5% chance of over 120, the correct probabilty would be more like 3%. The chance of being over 140 is so small it's off my Normal Chart. But if the probablity according to the Normal was say, .00000001 the correct chance would be more like .00000003.

I guess that kind of improvement in accuracy can mean the difference between a missle hitting it's target or being miles off. Or between an arbitrage being profitable or not. I learned just enough statistics to get an idea of how powerful the more advanced tools can be. Thanks.

PairTheBoard

[AaronBrown]

I really does make a difference, and not just with missiles and billion-dollar arbitrages.

A casino owner wants to do a roulette promotion that will really get some attention. He has a deal where you buy 36 chips for $36 and have to place them on single numbers. The first person to win X times out of the 36 gets $1 million. He wants to makes sure that the game will last a while, so he gets a lot of customers, and that the expected value of the payoff is less than the profit from the expected roulette bets.

He's had one statistics course. He knows the expected value of a bet in double-zero roulette is 35*(1/38) - 1*(37/38) = -0.05263. The standard deviation is 5.7626. Over 36 spins multiply the expected value by 36 (-1.8947) and the standard deviation by 6, the square root of 36, to get 34.5757.

His old stat textbook has a Normal distribution table in the back. That says only 1 time in 10 million do you get an observation more than 5.2 standard deviations above the mean. That's a profit of $170.98 (-1.8947 + 5*34.5757) in this case. So he says he'll give $1 million to the anyone who wins 6 or more times out of 36, for a profit of $180. That costs him less than an expected dime per attempt, and he's making $1.89, so he can afford that.

People come from all over, 25,000 the first day. They play an average of 4 times each, for 100,000 attempts. He has to pay off 33 times, for $33 million. Unfortunately for the rest of us, he goes bankrupt and cannot continue the game.

The real chance of winning 6 or more times betting on single numbers 36 times in roulette is 1 in 3,049; not 1 in 13,941,629.

[PairTheBoard]

Right. Although this example would not meet the common guideline for using the Normal Aproximation, Np and N(1-p) to be at least greater than say 5. The casino owner must have missed the lecture on when to use the Poisson aproximation for the Binomial.

PairTheBoard