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I came across a variation of the birthday problem asking "in a room of $4$ people what is the probability that at least $3$ of them share the same birthday".

I was unsure of the answer and thought that it would be P($3$ share the same birthday) + P($4$ share the same birthday), which equals: $1\cdot\frac{1}{365^2} + 1\cdot\frac{1}{365^3}$, and this comes out to be around $0.0000075$%.

However my friend said that he thinks to correctly calculate the answer, the probability of $4$th person not having the same birthday should be included in the calculation somewhere.

What would be the correct probability of at least $3$ out of $4$ people sharing the same birthday, and how could you extend the problem to work out the probability of at least "$x$" out of "$y$" people having the same birthday?

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  • $\begingroup$ Your friend is correct. $\endgroup$
    – drhab
    Dec 5, 2020 at 13:08

2 Answers 2

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i) $3$ share the same birthday

Number of ways to choose $3$ people out of $4$ having the same birthday $ = { 4 \choose 3}$

The probability of $3$ of them having the same birthday (and $4$th must have different birthday otherwise this will include cases when all $4$ of them have the same birthday).

So $\displaystyle P(3) = { 4 \choose 3} \frac{364}{365^3}$

ii) All $4$ share the same birthday

$\displaystyle P(4) = \frac{1}{365^3}$

Desired probability $ = P(3) + P(4)$.

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Let me go for the general case.

For $i=1,2\dots,365$ let $E_i$ denote the event that at least $x$ persons have birthday on day $i$.

Then to be found is: $$P\left(\bigcup_{i=1}^{365}E_i\right)$$ and we can do that with inclusion/exclusion and symmetry, leading to:

$$\cdots=\sum_{k=1}^{365}(-1)^{k-1}\binom{365}kP\left(\bigcap_{i=1}^kE_i\right)$$ Then what remains is finding expressions for $P\left(\bigcap_{i=1}^kE_i\right)$ for $k=1,2,\dots,\lfloor\frac{y}{x}\rfloor$.

This because, if $k$ exceeds $\lfloor\frac{y}{x}\rfloor$ then $P\left(\bigcap_{i=1}^kE_i\right)=0$, so we could also write:$$\cdots=\sum_{k=1}^{\lfloor\frac{y}{x}\rfloor}(-1)^{k-1}\binom{365}kP\left(\bigcap_{i=1}^kE_i\right)$$

In your case ($x=3$ and $y=4$) we have $\lfloor\frac{y}{x}\rfloor=1$ so it reduces to: $$365P(E_1)=365\left[\binom43\left(\frac1{365}\right)^3\frac{364}{365}+\left(\frac1{365}\right)^4\right]=\binom43\left(\frac1{365}\right)^2\frac{364}{365}+\left(\frac1{365}\right)^3$$

Quite a job to find these probabilities $P\left(\bigcap_{i=1}^kE_i\right)$ but by small $\lfloor\frac{y}{x}\rfloor$ there is hope.

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