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Punker
03-03-2003, 12:03 AM
Take the following statistics:

3700 hands
AA/KK 24 times
When receiving AA/KK, win 14 times.
Assume average 8 players to table.

Within statistical expectation? Or not?

Homer
03-03-2003, 05:29 PM
First Question

Probability of receiving AA/KK in a given hand = = (8/52) * (3/51) = 24/2652 = .905%

Expected number in 3700 hands = .00905*3700 = 33.5

Standard deviation in 3700 hands = sqrt (3700*.00905*(1-.00905)) = 5.76

Percentage of time you will be dealt AA/KK less than 24 times over the course of 3700 hands = normsdist((24-33.5)/5.76) = 4.95% (Approximately 1/20)

Second Question

You won 53% of the time when dealt AA/KK. The expected win rate depends on how tight/loose the table you are playing at is. If on average you see the flop with four opponents, then 53% is about right (Reference (http://mentorms.best.vwh.net/poker/HE_Value.htm)).

-- Homer

Punker
03-03-2003, 06:06 PM
Thank you for your response, but I have a few more questions...

Is your reception of AA/KK normally distributed? Isn't it binomially distributed?

And according to your reference, 53% is accurate if I go to showdown against 4 opponents every time, not if I see the flop with 4 opponents, isn't it?

BruceZ
03-04-2003, 12:21 AM
Is your reception of AA/KK normally distributed? Isn't it binomially distributed?

Yes it is binomially distributed, but Homer Simpson has answered question 1 correctly. The standard deviation he computes as sqrt (3700*.00905*(1-.00905)) = 5.76 is the standard deviation of the normal distribution which is correctly used to approximate this binomial distribution. This is different from the standard deviation of the binomial distribution itself which is just sqrt(3700*.00905).