Clarification of a point from lecture today (distinguishable dice vs. indistinguishable dice)

[Kevin Lin]

There was a point of confusion today in class, regarding whether we should consider dice to be distinguishable or indistinguishable.

Suppose we roll two dice. First, suppose that they are distinguishable. (For example, suppose one die is red and the other is blue.) This means that there are 36 possible outcomes, all equally likely:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Each of these outcomes has an equal 1/36 probability of occurring.

Now suppose that the dice are indistinguishable. Then this means that we will treat a roll that results in (i, j) to be the same as a roll that results in (j, i). So the possible outcomes are now:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 3), (3, 4), (3, 5), (3, 6),

(4, 4), (4, 5), (4, 6),

(5, 5), (5, 6),

(6, 6)

However, in this case, not all outcomes have the same probability! The outcomes that are of the form (i, i) will have probability 1/36; all other outcomes have probability 2/36 = 1/18. For example, let's look at the outcome (3,5). Even though a roll (3, 5) and a roll (5, 3) are indistinguishable and so lead to the same outcome, there are still two ways that this outcome can happen. Therefore, the probability of the outcome (3, 5) is 2/36 = 1/18.

Now let's compute the probability that the sum of the two numbers is equal to 7 when we roll two dice:

In the case that the dice are distinguishable, this probability is equal to the probability of getting (1,6) + the probability of getting (6,1) = 1/36 + 1/36 = 1/18.

In the case that the dice are indistinguishable, this probability is equal to the probability of getting (1,6) = 1/18. Same answer!

Kevin