# Same birthday question: 2 different variants have same answer?

How many people must there be before the chances that someone has the same birthday as you do is at least 0.5? How many people must there be before the chances that at least two people have a birthday on September 6 is at least 0.5?

For the first one, I get: 1 - Prob.(no one has same bday) >= 0.5

1 - (364/365)^n >= 0.5

n <= ln (0.5) / (ln (364/365)

n ~= 253

The second one, I get: 1 - Prob.(no one has same bday) - Prob.(1 person has birthday on 9/6) >= 0.5

1 - (364/365)^n - ((1/365)((364/365)^(n-1))) >= 0.5

((364/365)^(n-1))(364/365 + 1/365) <= 0.5

(364/365)^(n-1) <= 0.5

n is also ~= 253??

Lots of other sources say the answer to the second one should be 613, which I don't understand.

• Notice that even the probability that $n$ people have a different birthday each is not $\left(\dfrac{364}{365}\right)^n$. – user228113 Feb 24 '16 at 7:35
• The answer to the second question is the answer to the first question divided by $365$. – barak manos Feb 24 '16 at 9:25

The first question has a very well known solution (try searching for "birthday paradox"), so I will go straight to the second one.

For the second question: Suppose we label the people $p_1,\dots,p_n$ and begin by imposing that $p_1$ has its birthday on 6th September (let's call $B$ the set of people having their birthday on 6th September and write $p_1\in B$). Then impose that $p_2$ has its birthday on 6th September too. This has probability $\frac{1}{365}\cdot\frac{1}{365}$.

The next case to consider is (while keeping $p_1$'s birthday on 6th September) that $p_3$ has its birthday on 6th September and $p_2$ does not have its birthday on 6th September, which has probability $\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}$. The case after this is to exclude $p_3$ as well and set $p_1\in B, p_2\notin B, p_3\notin B, p_4\in B$ which has probability $\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}$, and so on.

It is necessary to exclude $p_2\in B$ to avoid counting twice cases like $B=\{p_1,p_2,p_3\}$ for instance, which are taken into account in the first case. Like this, the probability of two people having birthday on 6th September and one of them being $p_1$ is $$\tag{1} \frac{1}{365}\cdot\frac{1}{365}+\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}+\ldots+\frac{1}{365}\cdot\frac{364}{365}\cdot\stackrel{n-2}{\ldots}\cdot\frac{364}{365}\cdot\frac{1}{365}$$ where in the last sum there are $n-2$ of the $\frac{364}{365}$ fractions.

This can be rewritten as $$\tag{2} \bigg(\frac{1}{365}\bigg)^2\bigg(1+\frac{364}{365}+\ldots+\big(\frac{364}{365}\big)^{n-2}\bigg)$$ which in turn can be transformed into $$\tag{3} \bigg(\frac{1}{365}\bigg)^2\bigg(\frac{1-\big(\frac{364}{365}\big)^{n-1}}{1-\frac{364}{365}}\bigg)=\frac{1}{365}\bigg(1-\big(\frac{364}{365}\big)^{n-1}\bigg)$$ using the geometric progression formula.

Now that we dealt with all the cases involving $p_1\in B$, we impose $p_1\notin B$. We start by counting the cases where $p_2\in B$: first $p_2,p_3\in B$; then $p_2\in B, p_3\notin B, p_4\in B$ and so on.

The probability of this has an expression similar to (1), except that there is a $\frac{364}{365}$ at the beginning of each term (due to the condition $p_1\notin B$) and there is one term less because there is one person less to worry about (namely $p_1$): $$\tag{1'} \frac{364}{365}\bigg(\frac{1}{365}\cdot\frac{1}{365}+\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}+\ldots+\frac{1}{365}\cdot\frac{364}{365}\cdot\stackrel{(n-2)-1=n-3}{\ldots}\cdot\frac{364}{365}\cdot\frac{1}{365}\bigg)$$ Accordingly, (1') is equivalent to $$\tag{2'} \bigg(\frac{1}{365}\bigg)^2\cdot\frac{364}{365}\bigg(1+\frac{364}{365}+\ldots+\big(\frac{364}{365}\big)^{n-3}\bigg)$$ and $$\tag{3'} \bigg(\frac{1}{365}\bigg)^2\cdot\frac{364}{365}\bigg(\frac{1-\big(\frac{364}{365}\big)^{n-2}}{1-\frac{364}{365}}\bigg)=\frac{1}{365}\cdot\frac{364}{365}\bigg(1-\big(\frac{364}{365}\big)^{n-2=(n-1)-1}\bigg)$$ We can expand a bit more and get $$\tag{4'} \frac{1}{365}\cdot\bigg(\frac{364}{365}-\big(\frac{364}{365}\big)^{n-1}\bigg)$$ This process goes on until there are $n-2$ people excluded from $B$ and the only possibility is $p_1,\dots,p_{n-2}\notin B, p_{n-1},p_n\in B$. Here, the version of (1) is just $$\tag{1''} \bigg(\frac{364}{365}\bigg)^{n-2}\bigg(\frac{1}{365}\cdot\frac{1}{365}\bigg)$$ as there must be $(n-2)-(n-2)=0$ copies of the $\frac{364}{365}$ in the last term. This (1'') is clearly equivalent to $$\tag{4''} \frac{1}{365}\cdot\bigg(\big(\frac{364}{365}\big)^{n-2}-\big(\frac{364}{365}\big)^{n-1}\bigg)$$ Now, to compute the total probability two people being in $B$ we do $(3) + (4') + \dots + (4'')$, which yields $$\tag{5} \frac{1}{365}\cdot\bigg(1-\big(\frac{364}{365}\big)^{n-1}+\frac{364}{365}-\big(\frac{364}{365}\big)^{n-1}+\dots+\big(\frac{364}{365}\big)^{n-2}-\big(\frac{364}{365}\big)^{n-1}\bigg)$$ We see there are $(n-1)$ occurrences of $\big(\frac{364}{365}\big)^{n-1}$ and the other terms form a geometric progression for which we use the same formula as before. We get $$\tag{6} \frac{1}{365}\cdot\bigg(\frac{1-\big(\frac{364}{365}\big)^{n-1}}{1-\frac{364}{365}}-(n-1)\big(\frac{364}{365}\big)^{n-1}\bigg)$$ Expand $$\tag{7} 1-\big(\frac{364}{365}\big)^{n-1}-\frac{(n-1)}{365}\big(\frac{364}{365}\big)^{n-1}=1-\big(\frac{364}{365}\big)^{n-1}\bigg(1+\frac{(n-1)}{365}\bigg)$$ Finally, impose (7) $\geq 0.5$, from where you can see that $n\geq 613$