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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?

 Mar 16, 2019
 #1
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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 

 

Geno3141

 Mar 16, 2019
 #2
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Hi Geno      laugh

Long time, no see.

I hope you've been well.

We have missed you.    And I /we are glad you have put in an appearance.

Why didn't you log in?

Melody  Mar 16, 2019
 #3
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Actually, your answer was wrong. Here is the solution that AoPS gave.

 

There are a total of $6^3=216$ possible sets of dice rolls. If one of the re-rolled dice matches the pair we set aside and the other two form a pair, we will have a full house. But we will also have a full house if all three re-rolled dice come up the same.

Consider the first case. There are $3$ ways to pick which of the three dice will match a pair, and then $5$ ways to pick a value for the other two dice so that they form a pair (but don't match the first three dice), for a total of $3\cdot 5=15$ possible outcomes, plus the outcome that all five dice match.

In the second case, we need all three dice to match each other. There are $5$ ways to pick which value the three dice will have so that they don't match the first pair, plus the outcome that all five dice match.

So there are a total of $15+5=20$ ways to get a full house without all five dice matching, added to the possibility that all five dice match, which makes $21$ ways to get a full house. So, the probability is successful outcomes/total outcomes= 21/216= 7/72

 Mar 16, 2019

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