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