#1
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Not sure how useful this would be to know, but I can\'t figure it out
what is the probability that two people get dealt the same hand at a full 10 handed table?
for instance one gets AT and the other gets AT should i be using combinatorics or what? any help would be greatly appreciated |
#2
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Re: Not sure how useful this would be to know, but I can\'t figure it out
Yes, you should be using combinatorics.
After you look at your two cards, there are 50*49/2 = 1,225 different hands a single opponent might hold. If you have a pair, there is 1 way for another opponent to have the same cards. If you have two different suited cards, there are 3 ways for an opponent to have the same hand. If you have two different unsuited cards, there are 7 ways for an opponent to have the same hand. To compute exactly the probability of 1 out of 9 opponents has the same hand is a bit tedious (but not too hard if you want to do it). A good approximation is to use 1 - (1-X)^9 where X is the probability of a single opponent having the same hand. |
#3
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Re: Not sure how useful this would be to know, but I can\'t figure it o
thank you very much for pointing me in the right direction
much appreciated! |
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