Rolling two doubles with 4 and 5, 6 sided dice

Question

I am playing a game that gives bonuses for rolling doubles and was wondering what the probability of rolling 2 sets of doubles on one roll with 4, 6 sided dice. Also, what the probability would be with rolling 5, 6 sided dice. As long as you can show 2 doubles that's all you need so this would include 4 of a kind and full houses as well.

Thanks!!

Answer 1

First I present a solution for the four dice problem. We compute probability by finding the number of desired outcomes over the number of total outcomes.

Case 1: there is a four-of-a-kind.

There are clearly only six desired outcomes in this case.

Case 2: there are two distinct doubles.

Call the doubles $aa$ and $bb$. Let us suppose without loss of generality that $a > b$. Then there are $\binom{6}{2} = 15$ ways to choose the values of $a$ and $b$. Once we have chosen the values, there are $\binom{4}{2} = 6$ ways to have two of the dice come up $a$ and the other two come up $b$. Hence there are a total of $6\cdot15=90$ desired outcomes.

That gives 96 desired outcomes out of $6^4 = 1296$ total possibilities, for a probability of $\frac{96}{1296} = \frac{2}{27}$.

Now let's look at the five dice problem. If we start with two pairs $aabb$ and need to add a fifth die, we either get a full house or two pairs and a single.

Case 1: full house

Let our full house be $aaabb$. There are 6 ways to select $a$ and 5 ways to select $b$. (Unlike the above, in this case we have to treat $a$ and $b$ as distinguishable, since there are more $a$'s than $b$'s.) Then there are $\binom{5}{3} = 10$ ways to arrange the $a$'s and $b$'s among the five rolls. That makes a total of $6\cdot5\cdot10 = 300$ full houses.

Case 2: two pairs and a single

Let the outcome be $aabbc$. This time since we don't want to overcount $a$'s and $b$'s we suppose $a > b$. Then there are $\binom{6}{2} = 15$ ways to choose the values of $a$ and $b$ as in Case 2 of the four dice case. Then there are 4 ways to choose $c$. Finally there are $\frac{5!}{2!2!1} = 30$ ways to arrange the $a$'s, $b$'s, and $c$'s, for a total of $15\cdot4\cdot30 = 1800$ ways.

In total we get $300+1800 = 2100$ desired outcomes out of $6^5 = 7776$ possibilities, for a probability of $\frac{175}{648}$.

Answer 2

Another way to think about this problem is as follows... Suppose we label the dice 1,2,3,4.

In order to have two distinct doubles we have 6 choices for the first die and 5 choices for the 3rd die.

That fixes our set up in forms like

1,1,3,3

Now we can ask, how many permutations of the string above are there? We know from elementary counting that there are 4!/2!2! Which is 6 total ways. However we have over counted because we have counted all permutations of

1,1,3,3 and 3,3,1,1 by the construction above. Therefore we need to divide by 2.

Our final answer becomes 6*5*3 = 90