Background
Many people are familiar with the so-called Birthday "Paradox" that, in a room of 23 people, there is a better than 50/50 chance that two of them will share the same birthday. In its more general form for n people, the probability of no two people sharing the same birthday is p(n)=365!365n(365−n)!. Similar calculations are used for understanding hash-space sizes, cryptographic attacks, etc.
Motivation
The reason for asking the following question is actually related to understanding a specific financial market behavior. However, a variant on the "Birthday Paradox" problem fits perfectly as an analogy and is likely to be of wider interest to more people with different backgrounds. My question is therefore framed along the lines of the familiar "Birthday Paradox", but with a difference, as follows.
Situation
There are a total of 60 people in a room. Of these, it turns out that there are 11 pairs of people who share the same birthday, and two triples (i.e. groups of 3 people) who have the same birthday. The remaining 60−11⋅2−2⋅3=32 people have different birthdays. Assuming a population in which any day is equally likely for a birthday (i.e. ignore Feb 29th & possible seasonal effects) and, given the specified distribution of birthdays mentioned above, the questioner would actually like to understand how likely (or unlikely) it is that these 60 people were really chosen randomly. However, I am not sure if the question posed in that way is actually answerable at all. When I posed this question on another site (where it was left unanswered), I was at least advised to re-state the question in a slightly different way, as follows below.
Question
If 60 people are chosen at random from a population in which any day is equally likely to be a person's birthday, what is the probability that there are 11 days on which exactly 2 people share a birthday, two days on which exactly 3 of them share a birthday, and no days on which 4 or more of them share a birthday?
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