Each of five, standard, six-sided dice is rolled once. Two of the dice come up the same, but the other three are all different from those two and different from each other. The pair is set aside, and the other three dice are re-rolled. The dice are said to show a "full house" if three of the dice show the same value and the other two show the same value (and potentially, but not necessarily, all five dice show the same value). What is the probability that after the second set of rolls, the dice show a full house?
There check out that link...
That problem has been posted before.
IF, on the off chance have a wifi prob or are terribly afraid of links (or just lazy)
To explain NinjaAnswer's answer:
Of the three dice rolled, one may match the original pair and two are different, and this can occur three ways: ( 1/6 x 5/6 x 5/6 ) + ( 5/6 x 1/6 x 5/6 ) + ( 5/6 x 5/6 x 1/6 ) = 75/216
[ where 1/6 is the probability of matching the first two and 5/6 is the probability of not matching the first two ]
Or, two may match the original pair and one is different:
( 1/6 x 1/6 x 5/6 ) + ( 1/6 x 5/6 + 1/6 ) + ( 5/6 x 1/6 x 1/6 ) = 15/216
Or, all three may match the original pair: ( 1/6 x 1/6 x 1/6 ) = 1/216
Or, all three may be the same, but not match the original pair:
( 5/6 x 1/6 x 1/6 ) = 5/216
[ where 5/6 is the probability of getting a new number and 1/6 is the probability of matching that new number ]
Adding these together: 75/216 + 15/216 + 1/216 + 5/216 = 96/216 = 4/9 "
none of above math stuff belongs to me