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#2
09-05-2002, 05:23 PM
 Guest Posts: n/a
Re: A Question on rec.gambling.poker

I'll take a stab at this one:

10300 hands.

Here's my thought process...

1) The problem is asking for 'the number of hands that you must be dealt for the probability to be at least 50% that all 1326 hands on your list have been marked'. This seems to be the same as asking: 'On average, how many hands will it take for you to mark off all of the hands?'.

2) The average number of hands you need dealt to go from zero marked off to 1326 marked off is the same as adding the average number of hands dealt to go from 0 to 1, from 1 to 2, from 2 to 3, ..., from 1324 to 1325, and from 1325 to 1326.

3) Assume you've just marked your xth hand off the list. The number of subsequent hands it'll take you to mark off the (x+1)th hand is governed by a geometric distribution with probability of success

p = (N - x) / N,

where N is the total number of hands, x is the number of hands you already have marked, and (N - x) is the number of hands you haven't yet marked.

Asking for the average number of hands it'll take you to mark off the (x+1)th hand is the same as asking for the mean of the geometric distribution with this p. As the mean of the geometric distribution is

mu = 1 / p,

the average number of hands it'll take to mark off the (x+1)th hand is N / (N - x).

4) So, the average number of hands needed to mark off half of the list is the summation of N/(N-x), as x runs from 0 to 1325, where N = 1326.

This works out to 10299.72 or so, so you need 10300 hands on average.

Is this anywhere close? What's the real answer? Tell me, I want to know.

PP