What is the probability to find exactly $4$ trios of people that have the same birthday? (from $k$ people)

Question

What is the probability to find exactly $4$ trios of people that have the same birthday? (from $k$ people). I was asked to solve this using the Inclusion–exclusion principle. Can anyone please help in doing this??

Answer 1

We can use the sample space of ordered $k$-tuples of integers between $1$ and $365$, inclusive. Then finding the probability amounts to counting the number of such $k$-tuples that have exactly $4$ numbers that appear in $3$ entries each. I'll take your question to mean that the following $16$-tuples do not count (in other words if a date is shared by more than $3$ people then the group does not count):

$$(1,1,1,2,2,2,3,3,3,4,4,4,4,7,8,9)$$

$$(1,1,1,1,1,1,2,2,2,3,3,3,4,5,6,7)$$

$$(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,5)$$

$$(1,1,1,1,2,3,4,5,6,7,8,9,10,11,12,13)$$

Note that in the last example, there are $4$ ways to choose a trio from the four people who have the same birthday. I won't count this as a success.

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Clearly for $k$ less than $12$, the answer is $0$. Notice also that for $k$ greater $4\cdot3+361\cdot2=734$ the probability should be zero as well, by the Pigeonhole principle. than For $k$ greater than or equal to $12$, we start by calculating the number of elements of our samples space, which is just $$365^k\tag{1}$$

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Now the number of ways to choose the $4$ numbers that will be the shared birthdays of the trios is $$\binom{365}{4}$$

Order these numbers from least to greatest. The number of ways to choose the three people who share the lowest-numbered date is $$\binom{k}{3}$$

The number of ways to choose the three people who share the second-lowest-numbered date is $$\binom{k-3}{3}$$

The number of ways to choose the three people who share the third-lowest-numbered date is $$\binom{k-6}{3}$$

The number of ways to choose the three people who share the largest-numbered date is $$\binom{k-9}{3}$$

Thus the number of ways to choose trios and their shared dates is the product

$$\binom{365}{4}\binom{k}{3}\binom{k-3}{3}\binom{k-6}{3}\binom{k-9}{3}\tag{2}$$

Notice that $k-9$ will be less than three if $k$ is less than $12$, so intuition that the probability should be zero when $k <12$ appears mathematically.

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Now we must count the number of ways to assign dates to the remaining people. We have $361$ dates and $k-12$ people. We can imagine the dates as boxes and the people as balls so that we must distribute $k-12$ balls in $361$ boxes such that no box has more than $2$ balls in it.

If we use stars and bars we may imagine $k-12$ stars and $360$ bars so that (without the restriction for at most $2$ balls in each box) the number of distributions is $$\binom{k+348}{360}$$

Now we can subtract the number of distributions with $3$ balls in at least one box. There are $361$ ways to choose the box, and the number of ways to distribute the remaining $k-15$ balls among the $361$ boxes is, once again using stars and bars,

$$361\binom{k+345}{360}$$

But we've overcounted the number of distributions with $3$ balls in at least two boxes. We choose the $2$ boxes, then distribute the remaining $k-18$ balls. This is $$\binom{361}{2}\binom{k+342}{360}$$

If we continue this process of overcounting and adjusting we get

$$\binom{k+348}{360}-361\binom{k+345}{360}+\binom{361}{2}\binom{k+342}{360}-\binom{361}{3}\binom{k+339}{360}+\cdots$$

Using the notation from this question, this sum is $N(k-12,361,2)$. It matches with Brian M. Scott's formula

$$N(k-12,361,2)=\sum_{i=0}^{361}(-1)^i\binom{361}i\binom{k+360-3i}{360}\tag{3}$$

We're allowed to have the upper limit of the sum be $361$ because we take $\binom{a}{b}=0$ if $a<b$.

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Our final answer is the product of $(2)$ and $(3)$, divided by $(1)$. The probability that there are exactly $4$ trios of people with the same birthdate in a group of $k$ people, then, is $$\displaystyle\frac{\displaystyle{\binom{365}{4}\binom{k}{3}\binom{k-3}{3}\binom{k-6}{3}\binom{k-9}{3}\sum_{i=0}^{361}\left[(-1)^i\binom{361}i\binom{k+360-3i}{360}\right]}}{365^k}$$

Hm.