## Solutions to Birthday Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

Problem 1:

What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?

Solution 1:

The probability of the second person having a different birthday from the first person is $\frac{364}{365}$. The probability of the first three persons having different birthdays is $\frac{364}{365} \times \frac{363}{365}$. In this way, the probability of all n persons in a room having different birthdays is $P(n) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+1}{365}$. For the value of n, when P(n) falls just below 1/2 is the least number of people in a room when the probability of at least two people having the same birthday becomes greater than one half (that is, more likely than not). Now, one can make the following table: $\begin{tabular}{|c|c|}\hline N & P(n) \\ \hline 2 & 364/365 \\ \hline 3 & 0.9918 \\ \hline 4 & 0.9836 \\ \hline 5 & 0.9729 \\ \hline 6 & 0.9595 \\ \hline 7 & 0.9438 \\ \hline 8 & 0.9257 \\ \hline 9 & 0.9054 \\ \hline 10 & 0.8830 \\ \hline 11 & 0.8589 \\ \hline 12 & 0.8330 \\ \hline 13 & 0.8056 \\ \hline 14 & 0.7769 \\ \hline 15 & 0.7471 \\ \hline 16 & 0.7164 \\ \hline 17 & 0.6850 \\ \hline 18 &0.6531 \\ \hline 19 & 0.6209 \\ \hline 20 & 0.5886 \\ \hline 21 & 0.5563 \\ \hline 22 & 0.5258 \\ \hline 23 & 0.4956 \\ \hline \end{tabular}$

Thus, the answer is 23. One may say that during a football match with one referee, it is more likely than not that at least two people on the field have the same birthday! 🙂 🙂 🙂

Problem 2:

You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?

Solution 2:

The probability of the first person having a different birthday from yours is $\frac{364}{365}$. Similarly, the probability of the first two persons not having the same birthday as yours is $\frac{(364)^{2}}{(365)^{2}}$. Thus, the probability of n persons not  having the same birthday as yours is $\frac{(364)^{n}}{(365)^{n}}$. When this value falls below 0.5, then it becomes more likely than not that at least another person has the same birthday as yours. So, the least value of n is obtained from $(\frac{364}{365})^{n}<\frac{1}{2}$. Taking log of both sides, we solve to get $n>252.65$. So, the least number of people required is 253.

Problem 3:

A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

Solution 3:

For the nth person to earn a free entry, first $(n-1)$ persons must have different birthdays and the nth person must have the same birthday as that of one of these previous $(n-1)$ persons. The probability of such an event can we written as $P(n) = [\frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+2}{365}] \times \frac{n-1}{365}$

For a maximum, we need $P(n) > P(n+1)$. Alternatively, $\frac{P(n)}{P(n+1)} >1$. Using this expression for P(n), we get $\frac{365}{365-n} \times \frac{n-1}{n} >1$. Or, $n^{2}-n-365>0$. For positive n, this inequality is satisfied first for some n between 19 and 20. So, the best place in the queue to get a free entry is the 20th position.

More later,

Nalin Pithwa.

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