Why shouldn't the Bose-Einstein value be used to calculate birthday probabilities?

Can you please expound and simplify the embolden phrase below?

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 $\dfrac{n^k}{3!}$ 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 $\dbinom{n+k-1}{k}$. But the ordered birthday lists are equally likely, not the unordered lists [Emphasis mine], so the Bose-Einstein value should not be used in calculating birthday probabilities. $\square$

Blitzstein. Introduction to Probability (2019 2 ed). p 20.

posted almost 3 years ago

CC BY-SA 4.0

DNB‭

-107 reputation 73 1 -105 128