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heavybody
10-28-2004, 02:03 PM
In a 3-handed hold-em game what are the odds that one of the players will be dealt a pair(any pair)? I tried to figure this out but it is very hard for me. If it isn't too much trouble explain how you solved it.
Thanks heavybody

jimdmcevoy
10-28-2004, 04:47 PM
The odds of recieving a pair are 1 in 17. The easiest way to understand this is no matter what the first card you recieve is, there are 3 other cards in the deck that you can recieve as your second card out of 51. So the chance of getting a pair is 3 out of 51, which is 1 in 17.

Therefore the chance of not getting a pair is 16 out of 17.

So the chance of the 1st person getting a pair while the 2nd and 3rd does not is (1/17)*(16/17)*(16/17). The same goes for the 2nd person getting a pair while the other two do not, and the same again for the 3rd person getting a pair while the other two do not.

So in the end the chance that one and not more than one person will get a pair is:

3*(1/17)*(16/17)*(16/17) = 768/4913 = 15.6%

This is actually only an aproximation. I didn't take into account the fact that if you deal out 2 cards that do not make a pair, the probability that the next two cards make a pair actually goes down, it is 134/2450 which is a little less than 1 in 18, not 1 in 17. If you really really want me to work out the exact answer I'll do it, but I don't think you'll be that interested in the statistics (I'm not)