AA seven times in 5.5 hours

[barrett]

A friend of mine was recently dealt AA seven times in 5.5 hours at a B&M tournament (and won all 7 times). I was curious about the odds of such a run but I'm not sure on the general strategy for solving the problem.

Assume about 35 hands dealt per hour.

[elitegimp]

facts:
* the odds of being dealt AA = 1/221
* playing 35 hands per hour for 5.5 hours = 192.5 hands (we'll use 192 for the calculations)

So you have an event that happens 1/221 of the time (on average), and you want to know the odds that _exactly_ 7 of 192 trials of this event occur. That's (192 choose 7)*[(1/221)^7]*[(220/221)^(192-7)] = 2.85e-5 or 0.00286% of the time.

For more info on why, google "Bernoulli trial" (and event is a Bernoulli trial if it has only two outcomes, with a probability of <i>p</i> that one outcome occurs and a probability <i>1-p</i> that the other outcome occurs) and "binomial distribution" (a distribution is binomial if you want to know the probability that if you run a Bernoulli trial with a known <i>p</i> value <i>n</i> times, exactly <i>x</i> of those trials will be the outcome associated with <i>p</i&gt. Hope this helps.

Two little PSs
* this does not include the probability that he wins all 7 [img]/images/graemlins/smile.gif[/img]
* if you want to know the probability that he has AA <b>at least</b> 7 times, you calculate the probability of each outcome (7,8,9,...,192 times) and sum them, or sum the probability of getting AA 0, 1, 2, 3, 4, 5, or 6 times over that stretch and subtracting from 1. According to my calculations, the odds he gets AA 7 or more times in 192 hands is 3.199e-5 or 0.003199%

[barrett]

Thank you very much!