Miscalculating Probability of At Least $2$ People Having The Same Birthday

Question

Regarding the problem: choosing 23 people randomly, show that there is greater than a $50$ percent chance that at least two of them will have the same birthday. What is the error in the way I'm trying to calculate this probability:

The equation one must of couse be familiar is that any probability expresses the ratio of the number of favorable outcomes to the number of total outcomes. The number of total outcomes is easily seen to be $365^{23}$. The number of favorable outcomes should be calculated then. What we favor is "at least two people with the same birthday"; and this could mean either exactly $2$ people, or exactly $3$ people, or ..., exactly $23$ people. Nevertheless, we severally count them. For exactly $2$ people (and similarly for all other counts), we break the counting procedure into two tasks: how many ways can $2$ people be combined, and how many ways can two people have the same birthday. The first task is evident to be $23\choose 2$, and the second $365$. Hence, by the multiplication principle, exactly 2 people can have the same birthday in $23\choose 2$$\times 365$ ways. Similarly, exactly $3$ people can have the same birthday in $23\choose 3$$\times 365$ ways; and similarly, exactly $23$ people can jave the same birthday in $365$ ways. So adding all the favorable outcomes, and doing some factoring, we have: $365\times ($$23\choose 2$$ + $$23\choose 3$$ + ... + $$23\choose 23$$)$. Since $n\choose 0$$+$$n\choose 1$$+ ... + $$n\choose n$$ = 2^n$. We can simplify the result into $365\times (2^{23} – 23 – 1) = 365\times (2^{23} – 24)$. So this, the number of favorable outcomes, divided by the number of total outcomes ($= 365^{23}$) is going to be $365\times (2^{23} – 24)\over 365^{23}$$ = $$2^{23} – 24 \over 365^{22}$. However, this is a very small number, and nowhere is it close to $0.5$. I would appreciate guidance; thank you in advance.

Answer 1

The probability of one pair sharing a birthday and the other $21$ having different birthdays is $$\dfrac{{23 \choose 2} \times 365 \times 1 \times 364 \times 363 \times \cdots \times 344}{365^{23}} \approx 0.3634.$$ Your numerator implicitly only had the terms ${23 \choose 2} \times 365$, so was far too small. Not requiring the others to have different birthdays would cause other types of miscounting.

As you say, there might have been three people sharing a birthday (and the others all different), which would have had probability $\frac{{23 \choose 3} \times 365 \times 1 \times 1 \times 364 \times 363 \times \cdots \times 345}{365^{23}} \approx 0.0074.$ Or perhaps two pairs of people each sharing birthdays and the others all different with probability $\frac{\frac12{23 \choose 2}{21 \choose 2} \times 365 \times 1 \times 364 \times 1 \times 363 \times \cdots \times 345}{365^{23}} \approx 0.1109.$ Those two cases deal with the possibility that there are $21$ birthdays between the $23$ individuals with a combined probability of about $0.1183$.

Then you also need to consider $20, 19, \ldots, 1$ birthdays between the $23$ individuals, which gets more complicated though for a combined probability of only about $0.0256$.

It is easier just to do the subtraction from the probability of $23$ distinct birthdays: $1- \frac{ 365 \times 364 \times 363 \times \cdots \times 343}{365^{23}} \approx 0.5073$ which is the same as $0.3634+0.1183+0.0256$.

Answer 2

1. If you are enumerating everyone's birthday in the denominator, then for each term in the numerator, you need to multiply by the possible birthdays of the people who don't match. So in the $365 \choose 2$ term for instance, you need to be multiplying by $365^{23-2}$ to account for everyone else.

2. This still won't quite do it, because your method isn't counting scenarios like "two pairs of people each of whom have the same birthday (but different between the pairs)".