Chance of Double 1 or Double 6

Question

I'm trying to fight out what are the chances of double 1 or double 6 I know that for 2 six-faced dice (aka 2D6) the probability would be 2/36. What are the chances for 3D6? 4D6? 5D6? and etc?

Answer 1

Assume you have $nD6$ , i.e., that you throw n six-faced die . I am assuming also you want the probability of having exactly a double 1 or a double six. Then you have nP2 ways of having two die with the same outcome, each with probability $1/6$. The rest of the die must land in a face other than the repeated value, so all other $n-2$ outcomes will have probability $5/6$. Then, when you throw all $n$ die, you can have two equal outcomes in $nP2$ ways, EDIT the other $(n-2)$ throws can happen in $(6-1)^{n-2 }=5^{n-2}$ ways (the other outcomes can each be anything other than the repeated value) and the remaining $n-2$ outcomes each will have probability $5/6$ for a probability of $$nP2 (5^{n-2})(1/6)^2 (5/6)^{n-2} $$

Answer 2

The way I would approach it is given $n$ six-sided dice, we need $2$ of them to show either both ones or both sixes, and $n-2$ of them to show anything but what is showing on the pair or the other double.

There are $n \choose 2$ ways to pick the ‘magic’ $2$ dice out of the $n$. These dice have a $\frac{1}{18}$ chance of showing either double 1's or double 6's: $\frac{1}{3}$ chance of picking a 1 or a 6 on the first throw, and a $\frac{1}{6}$ chance of the second matching the first.

In the remaining $n-2$ dice, we have two possibilities. One is that the remaining dice all show something other than 1 or 6. The second is that there is at most one 1 or one 6 is seen—not the one selected for the ‘magic’ pair.

Case 1 should have probability ${{n-2} \choose {n - 2}}\left(\frac{4}{6}\right)^{n - 2}$

Case 2 should have probability ${{n-2} \choose 1}\left(\frac{1}{6}\right) {{n-3} \choose {n - 3}}\left(\frac{4}{6}\right)^{n - 3}$

The sum of these two cases is: $$ \left(\frac{2}{3}\right)^{n - 2} + \left(\frac{2}{3}\right)^{n - 3} \frac{n - 2}{6}\\ =\left(\frac{2}{3}\right)^{n - 3}\left(\frac{2}{3} + \frac{n - 2}{6}\right)\\ =\left(\frac{2}{3}\right)^{n - 3}\left(\frac{n + 2}{6}\right) $$

If correct, the probability would be: $$ {n \choose 2}\frac{1}{18}\left(\frac{2}{3}\right)^{n - 3}\left(\frac{n + 2}{6}\right) $$