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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?)

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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.

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

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

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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).

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