How many people must there be in a room before the probability that someone has the same birthday as you do is at least \(1/2\)? How many people must there be before the probability that at least two people have a birthday on July 4 is greater than \(1/2\)?
This problem differs from the classic birthday paradox because we’re now asking about matching a specific birthday rather than finding any two people with matching birthdays.
A
Think of this as repeatedly asking “Does this person share my birthday?” For each person, the probability they have your birthday is \(1/n\) (where \(n = 365\)), so the probability they don’t is \((n-1)/n\).
If we have \(k\) people in the room (besides yourself), the probability that none of them share your birthday is:
\[\begin{align*} \Pr\{\text{no match}\} &= \left(\frac{n-1}{n}\right)^k \end{align*}\]Therefore, the probability that at least one person shares your birthday is:
\[\begin{align*} \Pr\{\text{at least one match}\} &= 1 - \left(\frac{n-1}{n}\right)^k \end{align*}\]We want this probability to be at least \(1/2\):
\[\begin{align*} 1 - \left(\frac{n-1}{n}\right)^k &\geq \frac{1}{2} \\ \left(\frac{n-1}{n}\right)^k &\leq \frac{1}{2} \\ k \ln\left(\frac{n-1}{n}\right) &\leq \ln\left(\frac{1}{2}\right) \\ k &\geq \frac{\ln(1/2)}{\ln((n-1)/n)} \\ k &\geq \frac{-\ln 2}{\ln(1 - 1/n)} \end{align*}\]Using the approximation \(\ln(1-x) \approx -x\) for small \(x\):
\[\begin{align*} k &\geq \frac{-\ln 2}{-1/n} \\ k &\geq n \ln 2 \\ k &\geq 365 \times 0.693 \\ k &\geq 253 \end{align*}\]For an exact calculation with \(n = 365\):
\[\begin{align*} k &\geq \frac{\ln 2}{\ln(365/364)} \\ k &\geq \frac{0.693147}{0.002747} \\ k &\geq 252.5 \end{align*}\]Therefore, we need 253 people (besides yourself) in the room.
B
This is subtly different. Now we’re asking: among \(k\) people, what’s the probability that at least two have July 4 as their birthday?
Let \(X\) be the number of people with July 4 birthday. Then \(X\) follows a binomial distribution with parameters \(k\) and \(p = 1/365\).
The exact probability calculation using binomial distribution is:
\[\begin{align*} \Pr\{X \geq 2\} &= 1 - \Pr\{X = 0\} - \Pr\{X = 1\} \\ &= 1 - \binom{k}{0}\left(\frac{1}{365}\right)^0\left(\frac{364}{365}\right)^k - \binom{k}{1}\left(\frac{1}{365}\right)^1\left(\frac{364}{365}\right)^k \\ &= 1 - \left(\frac{364}{365}\right)^k - k\left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^k \\ &= 1 - \left(\frac{364}{365}\right)^k\left(1 + \frac{k}{365}\right) \end{align*}\]We want:
\[1 - \left(\frac{364}{365}\right)^k\left(1 + \frac{k}{365}\right) > \frac{1}{2}\]This doesn’t have a closed form solution, but we can solve numerically. Testing values:
For \(k = 612\): \(\begin{align*} \Pr\{X \geq 2\} &= 1 - (364/365)^{612}(1 + 612/365) \\ &\approx 1 - 0.169 \times 2.677 \\ &\approx 1 - 0.452 \\ &\approx 0.548 > 0.5 \end{align*}\)
For \(k = 611\): \(\Pr\{X \geq 2\} \approx 0.497 < 0.5\)
Therefore, we need at least 612 people for the probability to exceed \(1/2\).