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Q. Two fair dice are rolled. What is the probability that the sum of the numbers on the top faces is 8.

A. Case 1: Distinguishable dice

$\Omega=\left \{(x,y) \mid 1\leq x\leq 6 \wedge 1\leq y\leq 6 \right \}$.

$S_{8}=\left\{(2,6),(3,5),(4,4),(5,3),(6,2) \right\}$ $$P\left ( S_{8}\right )=\frac{\left | S_{8} \right |}{\left | \Omega \right |}=\frac{5}{36}$$

Case 2: Indistinguishable dice

$\because (m,n)$ and $(n,m)$ are the same outcome here,

$\Omega=\left \{(x,y) \mid 1\leq x\leq y\leq 6 \right \}$.

$S_{8}=\left\{(2,6),(3,5),(4,4)\right\}$ $$P\left ( S_{8}\right )=\frac{\left | S_{8} \right |}{\left | \Omega \right |}=\frac{3}{21}=\frac{1}{7}$$ However I've read elsewhere that the probability in both cases is the same which makes sense too since if we were to perform this experiment with colored dice in front of a normal and a color blind person, they'd come up with the same probabilities. Does distinguishability not matter here?

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Distinguishability is irrelevant, for more or less exactly the reason described.

To give another example, if we flipped two indistinguishable coins simultaneously, we should get one heads and one tails about half the time, not one third of the time. Even though there are only three distinguishable outcomes, the outcomes are not of equal probability (as they would be in the case of distinguishable coins), so we cannot simply take the number of desired outcomes divided by the number of possible outcomes in such a case.

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By indistinguishable not all outcomes have the same probability. The probability on e.g. $(2,6)$ is twice the probability of $(4,4)$. This because outcome $(2,6)$ can be reached in $2$ ways:

die1 gives $2$, die2 gives $6$, or

die1 gives $6$, die2 gives $2$.

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