# Birthday probability question

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?

• Your friend is correct. Dec 5, 2020 at 13:08

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

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.