Two pair o' dice

The probability of rolling a sum $S$ is $\dfrac{1}{36}$ times the number of ways $n(S)$ to write $S$ as an ordered sum of two numbers from $1$ to $6$ . Using stars and bars, we have

$\displaystyle \begin{aligned} n(2) & = \binom{0 + 2 - 1}{2 - 1} = 1 \\ n(3) & = \binom{1 + 2 - 1}{2 - 1} = 2 \\ n(4) & = \binom{2 + 2 - 1}{2 - 1} = 3 \\ n(5) & = \binom{3 + 2 - 1}{2 - 1} = 4 \\ n(6) & = \binom{4 + 2 - 1}{2 - 1} = 5 \\ n(7) & = \binom{5 + 2 - 1}{2 - 1} = 6 \\ n(8) & = \binom{6 + 2 - 1}{2 - 1} - 2 = 5 \\ n(9) & = \binom{7 + 2 - 1}{2 - 1} - 4 = 4 \\ n(10) & = \binom{8 + 2 - 1}{2 - 1} - 6 = 3 \\ n(11) & = \binom{9 + 2 - 1}{2 - 1} - 8 = 2 \\ n(12) & = \binom{10 + 2 - 1}{2 - 1} - 10 = 1 \end{aligned}$

The probability of rolling $S$ twice by the product rule is $\left(\dfrac{n(S)}{36}\right)^2$ , or

$\displaystyle \frac{1}{36^2} \left(2 \sum_{k \ = \ 1}^{5} k^2 + 6^2\right) = \frac{1}{36^2} \left(\frac{5(5 + 1)(10 + 1)}{3} + 36\right) = \frac{146}{36^2} = \boxed{\dfrac{73}{648}}$

Let $P(n) =$ Probability they both have a total of $n$ .

Then,

• $P(2) = (\frac{1}{36})^2$
• $P(3) = (\frac{2}{36})^2$
• $P(4) = (\frac{3}{36})^2$
• $P(5) = (\frac{4}{36})^2$
• $P(6) = (\frac{5}{36})^2$
• $P(7) = (\frac{6}{36})^2$
• $P(8) = (\frac{5}{36})^2$
• $P(9) = (\frac{4}{36})^2$
• $P(10) = (\frac{3}{36})^2$
• $P(11) = (\frac{2}{36})^2$
• $P(12) = (\frac{1}{36})^2$

These are computed as follows... For example, for $P(5)$ each person has 4 ways they can roll a 5 out of 36 possible rolls, so each person has a $\frac{4}{36}$ chance of rolling a 5. So,

$P(5) = (\frac{4}{36})^2$

The others were calculated similarly.

So, the total probability is given by:

Probability $= \sum P(n) = \boxed{0.113}$