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If we normalize those terms, don't we get 1/4 , 1/2 , 1/4 as the probabilities, since ##|HT\rangle## and ##|TH \rangle## are indistinguishable?With classical statistics, i.e., where the particles are distinguishable, there are four possible outcomes:

##HH, HT, TH, TT##

Since all four outcomes are a priori equally likely, the probability for HH is 1/4. This is applicable to tossing

macroscopic coins, where quantum effects are negligible.

With Bose–Einstein statistics, where the allowable states must be symmetric under exchange, there are only three possible outcomes:

##HH, \quad (HT+TH)/ \sqrt{2}, \quad TT##

Consequently, the probability for HH increases to1/3.