Do they? Suppose, for simplicity, we used two 2-sided dice (i.e., coins).
Then: 1/4 of the time we get a head and a tail (we get both behind the divider 1/2 the time, and we pull both in front of the divider 1/2 the time), 3/8 of the time we get two heads, and 3/8 of the time we get two tails. I don’t know anything about quantum mechanics, but I thought the “indistinguishable” stats we were going for would make these all equal.
You’re right. For distinguisable coins, you get probabilities for HH, TT and non-matched coins of 1/4, 1/4, and 1/2. For indistinguishable coins, you’d want the probabilities to be 1/3 each. For the method I came up with, the probabilities are 6/16, 6/16, and 4/16, which are closer to the indistinguishable case, but not quite there. [Emily Litella]Never mind[/Emily Litella]