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I have problem with understanding the difference between two basic tasks below:

You have $2$ fair coins. You flip them over once. Find the probability of getting $2$ tails if:
a) Coins are distinguishable
b) Coins are undistinguishable

The first problem is clear for me. We have $$ \frac{\mbox{1 case}}{\mbox{number of functions from {0,1} to {0,1}}} = \frac{1}{4}$$

But I have problem with second one. In my opinion there is no difference between them. If I take $2$ coins, let say that they are green and blue, I got probability $0.25$ but if I paint the same coins on the black I should get $\frac{1}{3}$ (this is statement from the book)...

Imo there is still $0.25$ because in b) we have:
$p_1 = \mbox{probability of 2 tails} = 0.25$
$p_2 = \mbox{probability of 2 heads} = 0.25$
$p_3 = \mbox{probability of tail, head} = 0.5$ because we have 2 times bigger change to get { tail,head } than {head, head}. So the answer is $$ p_1 = 0.25 $$

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    The probability is indeed $\frac{1}{4}$ in both cases. Do not fall into the trap of thinking that $Pr(A)=\dfrac{|A|}{|S|}$ in all cases where $S$ is the sample space. That is simply not true much of the time. It will happen to be true when the sample space is "equiprobable", meaning each outcome in the sample space is equally likely to occur. There are only two outcomes to playing the lottery, you win or you lose, but the probability of winning certainly isn't $\frac{1}{2}$. – JMoravitz Oct 16 '19 at 17:53
  • I recommend reading my answer here to the question of "Why is flipping a head then a tail a different outcome than flipping a tail then a head?" – JMoravitz Oct 16 '19 at 17:54
  • Ok, so it seems to be error in my book... thanks @JMoravitz –  Oct 16 '19 at 17:55
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    You are right, it's hard to believe an actual published book suggests that you get 1/3 in the indistinguishable case. It's a common beginner's mistake to treat the 3 distinct outcomes as equally likely, but, as you noted, the HT result is twice as likely as the other two. What does the book propose as the probability of 6,6 when you throw two identical dice? If it says 1/21, then it's wrong but at least it's consistent. – Ned Oct 16 '19 at 17:56
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    Just for fun: the correct answer is $1/3$ if these coins were photons (bosons) and obey Bose-Einstein statistics. You can find this exact "two B-E coins" situation discussed as a toy example here and here. However, I agree with the other comments that it is unreasonable to interpret "undistinguishable coins" to mean "entites that obey B-E statistics". Whether you can distinguish them or not, "coins" are not bosons. – antkam Oct 16 '19 at 20:03

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