# Probability that no two of them celebrate their birthday in the same month

If there are 12 strangers in a room, whats the probability that no two of them celebrate their birthday in the same month??

## Try

Im trying to solve this by using the binomial distribution. Let $$X$$ be the number of couples that celebrate their birthday in the same month. we want to calculate $$P(X=0)$$.

Notice that this is binomial because each couple is a trial and we get either they have same birthday or no. The number of couples is $${12 \choose 2 } = 66$$ and the probability of success is

$$P = 12 \frac{1}{12} \cdot \frac{1}{12} = \frac{1}{12}$$

Threfore, according to the binomial, we have

$$P(X=0) = {66 \choose 0} \frac{1}{12}^0 (\frac{11}{12} )^{66} = \boxed{0.003}$$

which differs from the book answer which is $$0.00005$$. What is the mistake here? To me it seems correct my answer to this problem.

• The error is that the events you are considering (person $a$ and $b$ don't share a birthday, person $a$ and $c$ don't share a birthday, etc...) are not independent events which is a requirement to use the binomial distribution in the first place. This should be obvious considering that if $a$ and $b$ share a birthday and $b$ and $c$ share, so too must $a$ and $c$. Oct 17, 2018 at 4:22

• The first one can have any birth month out of 12. The probability that no two in the room share a birth month is $$\frac{12}{12}$$.
• The second can have any birth month but that of the first stranger. The probability of this is $$\frac{11}{12}$$.
• The third can have any birth month but that of the first two. The probability is $$\frac{10}{12}$$, etc.
We continue until the last stranger, where the associated probability is $$\frac1{12}$$, then multiply to get the correct result of $$\frac{12!}{12^{12}}=0.000537$$.