What's the probability of rolling the same number twice with a pair of dice?

There are a number of questions similar to this, but I'm asking about rolling two dice twice, not one die two times. Or I guess you could think of it as 4 dice total, with each pair distinguishable.

So Person A rolls two dice (d6) and gets some result between 2 and 12, with 7 being the most likely. Now Person B rolls two dice. What are the odds that Person B rolled the same number as Person A?

I've been trying to work this out for some time now, but it's been a while since I've taken Probability. (Do we have to do some conditional probability stuff here?)

Well, just go about it in the simplest way possible. There is no need for conditional probability here because the second roll is independent of the first.

If $X_1$ is the first roll and $X_2$ is the second one, the probability of them being the same is:

$$P(X_1=X_2) = \sum_{n=2}^{12} {\left[ P(X_1=n) P(X_2=n) \right]} = \sum_{n=2}^{12} {\left[ P(X=n)^2 \right]}$$

because the probability functions are identical

Note that the value of $P$ depends on $n$ because some numbers are more probable than others when you have two dice. This is the only difference with the similar problem of single die rolls.

• I'm getting 101/648. Sound about right? – Cory G. Sep 30 '14 at 18:08
• I get something else. – Henno Brandsma Sep 30 '14 at 18:17
• I think the number in Prometheus' reply is correct so you should be getting $\frac{73}{648}$ – patatahooligan Sep 30 '14 at 18:43
• Ahh, typo. Got it. Thanks! – Cory G. Sep 30 '14 at 19:06

I guess by 'number' you mean 'sum'. My answer is based on this guess.

The probability of rolling a sum of $s$ is $1/36$ times the number of ways to write $s$ as an ordered sum of two numbers from $1$ to $6$. If we let this number be $n(s)$, we have

$$n(2)=1,\ n(3)=2,\ n(4)=3,\ n(5)=4,\ n(6)=5.\ n(7)=6,$$ $$n(8)=5,\ n(9)=4,\ n(10)=3,\ n(11)=2,\ n(12) = 1.$$

The probability of rolling $s$ twice is $(n(s)/36)^2$, so the probability of rolling a number twice is

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

Because both pairs of dice are independent, call their sums $S_1$ and $S_2$, this is just $P(S_1 = 2)\cdot P(S_2=2) + P(S_1 = 3)\cdot P(S_2=3) + \ldots + P(S_1 = 12)\cdot P(S_2 = 12)$

And the probabilities of $S_i=2,3,\ldots,12$ is are classical ($\frac{1}{36}$ for 2 and 12, $\frac{2}{36}$ for 3 and 11, up to $\frac{6}{36}$ for sum 7).