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SGS
04-13-2005, 04:35 AM
Yea so I'm stuck on this problem even though it seems as it should be fairly elementary. Suppose we play poker with 5 dice, what is the probability we roll exactly one pair? It seems to me that it should be 6 times 5 choose 3 dividede by 6^5. But this gives me a small number that can't be right I feel my numerator is misisng something. Any help? Thanks in advance.

SGS

gaming_mouse
04-13-2005, 04:47 AM
Without looking at your solution, I get (for exactly 1 pair):

6*(5 choose 3)*(1/6)^2*(5/6)(4/6)*(3/6)= 0.462962963

BruceZ
04-13-2005, 09:11 AM
[ QUOTE ]
Yea so I'm stuck on this problem even though it seems as it should be fairly elementary. Suppose we play poker with 5 dice, what is the probability we roll exactly one pair? It seems to me that it should be 6 times 5 choose 3 dividede by 6^5. But this gives me a small number that can't be right I feel my numerator is misisng something. Any help? Thanks in advance.

SGS

[/ QUOTE ]

6*C(5,3)*C(5,3)*3!/6^5 =~ 46.3%

There are 6 possible pairs, times C(5,3) ways to choose the other 3 numbers. But then you have to multiply this by C(5,3) ways to choose the 3 dice that get the non-pair (same as C(5,2) ways to pick the 2 that get the pair) times 3! ways to assign the 3 non-pair numbers to the 3 non-pair dice.

Or you can do it the way gaming mouse did it:

6*C(5,3)*(1/6)^2*(5/6)(4/6)*(3/6)=~ 46.3%

Where 6 is the number of pairs, but this time the first C(5,3) is the number of ways to choose which 3 dice get the non-pair. Then (1/6)^2 is the probability of getting the pair on the pair dice, and (5/6)*(4/6)*(3/6) is the probability of getting all different numbers on the 3 non-pair dice.