Mathematics of Poker - trying to solve a problem

[Beavis68]

I am trying to use the variance and normal distribution equations on page 23 and 24 to come up with a normal distribution on being dealt aces (or any specific pair) over x number of hands.

I am trying first to figure out the variance for a single deal using the normal distribution example on page 24.

This is what I come up with:

Vaces = 1/221*(1-1/221)^2+220/221*(0-1/221)^2

I dont know how to insert the special symbols for the generic equation they use.

[tarheeljks]

shouldn't you be using the variance for a binomial distribution?

[CT11]

I agree with tarheeljks. A Normal distribution is often more analytically tractable but you shouldn't try and use it for everything. You could come up with a similar answer using the Central Limit Theorem with involves the normal distribution but that would be a lot more work.

Assuming you are asking for the variance of the total number of pocket aces delt over x hands. Then it would be x*p*(1-p) where p here is 1/221.

EDIT: you also need a more for the CLT like large x.

[Beavis68]

thanks.

can you clue me in on why this is a binomial distribution instead of normal? Is it because it id constrained on the negative side at 0?

[Beavis68]

actually, that gives me the same answer which is a SD of less than 1 for 221 hands. That seems too low to me. That would mean that there is less than a .15% chance of getting 4 or more AA in 221 runs.

[tarheeljks]

This is the definition of the binomial distr. from Wolfram Mathworld:

The binomial distribution gives the discrete probability distribution P_p(n|N) of obtaining exactly n successes out of N Bernoulli trials (where the result of each Bernoulli trial is true with probability p and false with probability q==1-p).

a bernoulli trial is an experiment that has two outcomes. the most basic example is flipping a coin b/c you either get heads or tails. however most experiments can be set up to qualify as bernoulli trials. wikipedia offers the example of rolling a die 10 times and recording the number of sixes.

in your experiment the Bernoulli trial is being dealt a two card hand. DUCY?

[tarheeljks]

[ QUOTE ]
actually, that gives me the same answer which is a SD of less than 1 for 221 hands. That seems too low to me. That would mean that there is less than a .15% chance of getting 4 or more AA in 221 runs.

[/ QUOTE ]

calcs?

[AaronBrown]

When the probability of success is low, the Poisson distribution is a better approximation than the Normal for a binomial.

You expect to get X/221 pairs of Aces over X hands. Call this M. For a Poisson, the variance equals the mean. The probability of getting no pocket Aces is exp(-M). The probability of getting K pocket Aces is M^K*exp(-M)/K!.

Over 221 hands, for example, the exact probability of getting zero pocket Aces is (220/221)^221 = 0.3670. The Poisson approximation is exp(-1) = 0.3678. The probability of getting three pocket Aces is 0.6117. The Poisson approximation is 0.6131. The exact standard deviation is 0.9977. The Poisson approximation is 1.0000.

[Beavis68]

Ok thanks a lot guys. I have a math background but my stats is only in application at work.

Does this mean because this is not a normal distribution the typical sigma values wont apply? Like 99.7% of all points being within +/-3sigma?

That is what is throwing me off because otherwise my numbers match.

[AaronBrown]

Yes, that is correct. If the distribution does not have the Normal shape, the standard Normal confidence intervals will not apply.

If you deal enough hands, the distribution of the number of pocket Aces will begin to look more and more Normal.