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My question is about when not to use Bose-Einsten statistics (otherwise known as stars and bars). Of course it shouldn't be used when order matters or you can not sample with replacement. However, while reading Blitzstein's textbook it is noted:

"The Bose-Einstein result should not be used in the naive definition of probability except in very special circumstances"

where the $P(A)$ under the naive definition is defined to be $\frac{|A|}{|S|}$.

It then proceeds to give an example:

"As another example, with $n = 365$ days in a year and $k$ people, how many possible unordered birthday lists are there? For example, for $k > = 3$, we want to count lists like (May 1, March 31, April 11), where all permutations are considered equivalent. We can't do a simple adjustment for overcounting such as $n^k /3!$ since, e.g., there are 6 permutations of (May 1, March 31, April 11) but only 3 permutations of (March 31, March 31, April 11). By Bose-Einstein, the number of lists is ${n + k -1}\choose k$. But the ordered birthday lists are equally likely, not the unordered lists, so the Bose-Einstein value should not be used in calculating birthday probabilities."

This doesn't make sense to me. Of course, the ordered birthdays are equally likely. Why wouldn't the unordered birthdays also be equally likely? Could someone explain this to me and potentially provide another example for me to see the limitations of Bose-Einstein? Thank you in advance!

  • The ordered throws of a pair of dice are equally likely. The unordered throws are not. – lulu Sep 07 '23 at 13:13
  • "Where the $P(A)$ under the naive definition is defined to be $\frac{|A|}{|S|}$" which is as you say incredibly naive. That is guaranteed to be true only when it is known ahead of time that the elements in the same space $S$ are known to be equally likely. Most circumstances of using stars and bars, those outcomes are not equally likely. – JMoravitz Sep 07 '23 at 13:13
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    "Why wouldn't the unordered birthdays also be equally likely" There may be some exotic circumstance where you force them to be, but under most reasonable interpretations of the problem they wont. Under most reasonable interpretations of the problem, your "first" person will have a randomly selected birthday... then your "second" person will also have a randomly selected birthday independently selected from the first, and so on. This setup is going to lead to the $n^k$ different ordered outcomes to be equally likely and not the stars-and-bars outcomes being equally likely. – JMoravitz Sep 07 '23 at 13:17
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    Flip two identical and indistinguishable fair coins: you get $2$ Heads or $2$ Tails or $1$ of each, corresponding to the stars and bars ${2+2-1 \choose 2-1}=3$ possibilities. Are the probabilities $\frac14,\frac14, \frac24$ or $\frac13,\frac13,\frac13$? – Henry Sep 07 '23 at 13:19
  • "Why wouldn't the unordered birthdays also be equally likely?" Because they didn't occur on a scale where quantum effects predominate. It's the same reason why it's incorrect to treat two particles in a Bose-Einstein condensate as if "one each of heads and tails" were twice as likely as "two heads", only viewed from the other end of the scale. – David K Sep 07 '23 at 19:42

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