# Find the probability that at least two people out of $k$ people will have the same birthday

The Birthday problem. Find the probability that at least two people out of $$(k=5)$$ people will have the same birthday.

The usual approach would be to use $$p=1-\frac{P_{365,5}}{365^5}$$ However, I want to know what went wrong with my approach: $$p=\Pr (A_5\cup A_4\cup A_3\cup A_2)$$ where $$A_i$$ is the event where $$i$$ persons have the same birthday so $$p=\frac{1}{365^5}(365^5+365^4 \cdot 364+365^3\cdot 364 \cdot 363+365^2 \cdots362)$$ I used $$365^5$$ since $$5$$ people may have the same birthday, therefore we get a value from $$365$$ with repetition. Furthermore, $$4$$ people can have the same birthday so we pick $$365^4 \cdot 364$$ and so on.

This is obviously wrong since $$p>1$$ however may I ask what went wrong with my thinking?

• Suppose the year had $7$ days instead of $365$, then try to use the same argument. It should become more obvious where/why it's wrong.
– dxiv
Dec 26, 2021 at 6:51
• Think about what your expressions are trying to count. Of all the $365^5$ possible outcomes in the denominator, how many assign everyone the same birthday? Try to reason everything out based on the multiplication principle.
– Karl
Dec 26, 2021 at 6:52
• @Karl I used the thinking that we have $365$ balls where I will pick $5$ times and I can also pick the same ball for at least $2$ persons. So i simply added the probabilities for the union of the events of $j$ persons having the same birthday/ball Dec 26, 2021 at 6:59
• Try to compute probability that they have different birthday and the result will easily follow. You should get something like $1 - (1-\frac{1}{365})*(1-\frac{2}{365})*...(1-\frac{k}{365})$. What is interesting in this formula is that if k is around 23 the result is already close to 50%. Not intuitive at all. Dec 26, 2021 at 7:16
• Say, the case when 4 people have the same birthday - you have to pick this day - 365 possibilities, then, once the day is fixed, you have to choose 4 people (out of 5) gives you 5 and then, once the day and people are fixed you have to assign birthday to the 5th guy which gives you 364 possibilities. So the total cases (when 4 birthdays are the same) is 365*5*364 Dec 26, 2021 at 7:24

First, let’s calculate the probability that nobody has the same birthday. Since birthdays are independent, we can use the formula that $$P(A\cap B) = P(A)*P(B)$$.

Let $$A_{i,j}$$ be the event that the $$i$$th and $$j$$th person don’t share the same birthday. We then want to calculate $$P(A_{1,2}\cap A_{1,3}\cap A_{1,4}\cap A_{1,5}\cap\ldots\cap A_{4,5})$$

That is, the probability that none of them share the same birthday is the probability that for every two different people we pick, they have a different birthday. By independence, we get that this is equal to $$P(A_{1,2})*P(A_{1,3})*ldots*P(A_{4,5})$$

Clearly, $$A_{i,j}$$ has the same probability every time—it’s always $$\frac{364}{365}$$. Thus, we can simplify this to $$\left(P\left(\frac{364}{365}\right)\right)^m$$ where $$m$$ is the total number of different people we can compare. Since we compare $$2$$ people at once, and we have $$5$$ people total, we have $$m = {5\choose 2} = 10$$ ways of choosing two distinct people.

Thus,

Our probability that nobody has the same birthday is $$\left(\frac{364}{365}\right)^{10}$$ The probability that AT LEAST $$2$$ people share the same birthday is the complement probability of nobody having the same birthday. Thus, $$1 -\left( \frac{364}{365}\right)^{10}$$
is the probability you’re looking for.

However, your way of computing it doesn’t work out. It seems like you’re trying to first compute the probability that all 5 people have the same birthday + the probability that 4 people have the same birthday + etc etc. I’m not sure exactly what you tried to do in your method, though. But basically, you can look up how to calculate the probability of exactly $$k$$ people sharing the same birthday out of a group of $$n$$, and add that total for each $$k\in\{2, 3, \ldots, n\}$$

Let $$X$$ denote the number of people with the same birth day .

Well $$X$$ can be either $$0$$ , or it can be $$2,3,4,5$$. Because $$X=1$$ does not make sense

Then $$X=0,2,3,4,5$$

$$P(X=0)=\frac{\binom{365}{5}\cdot 5!}{365^{5}}$$

$$P(X=2)=\frac{\binom{5}{2}\binom{365}{1}\binom{364}{3}\cdot \frac{4!}{2!}}{365^{5}}$$

and so on

Then $$P(X\geq 2)=1-P(X<2)=1-\frac{\binom{365}{5}\cdot 5!}{365^{5}}$$