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Jun 25th, 2009, 09:49 AM
#2
Hyperactive Member
Re: Probability question
I'm a little confused. You're using two decks with 20 cards each with only Ace, K, Q, J, 10 for a total of 40 cards and the hands are 12 cards?
I'm a little rusty but assuming the above I would do:
Number of Hands that have 3 of a kind of 4 different cards
= (8,3) (5,1) (8,3) (4,1) (8,3) (3,1) (8,3) (2,1)
Where (n,m) = number of ways of choosing m items from n choices (nCr)
This equals 56 x 5 x 56 x 4 x 56 x 3 x 56 x 2 = 56^4(5!) = 1180139520
Now you have to subtract how many of those hands contain the three 10's which is 5.
So we are left with 1180139515
Total Hands = (40,12) = 5586853480
Probability = 1180139520 / 5586853480 = 0.211
There must be something wrong here either in the assumptions about the deck/hands or in my calculations since this is way higher than you observed. Can you clarify how many cards are in the deck and how many cards are in a hand?
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