A different birthday problem

At an event I attended yesterday, the icebreaker was to line up in order of our birthdays. There were about 800 people in attendance. Of course, a number of people found their “birthday buddy.” The organizer also announced that one person in the group had a birthday that day, and we all sang “Happy Birthday” to her.

I was surprised that there was only one person in the group with that birthday. “What are the odds?” I wondered.

Which reminded me of this exchange from The Big Bang Theory:

Sheldon Cooper: What are the odds that two individuals as unique as ourselves would be connected by someone as comparatively workaday as your son?
Beverly: Is that a rhetorical point, or would you like to do the math?
Sheldon Cooper: I’d like to do the math.

In this case, the math isn’t hard.

For each person in the room, there are two possible outcomes: It is their birthday, or it isn’t. Statisticians call this a “Bernoulli Trial.” We will count “It is their birthday” as a success. Without accounting for leap years, the probability of success is 1/365.

If you conduct a number of Bernoulli Trials and count the successes, the result follows the Binomial Distribution. The Binomial Distribution has two parameters: $n$, the number of independent experiments, and $p$, the probability of success for each experiment.

Let’s let $x$ represent the number of successes in our birthday experiment; that is, the number of people who have today as their birthday when we do the experiment. The random variable $X$ follows the binomial distribution with parameters $n$ and $p$, and we write this as:

$$X \sim B(n,p)$$

The mean of a Binomial distribution is simply $n * p$, in this case 800 * 1/365 = 2.19. If we repeated this experiment day after day, we would expect the average value of x to be 2.19.

Already, the outcome of x=1 doesn’t seem as improbable as I first thought.

The probability of getting exactly x successes in n trials is given by the probability mass function:

$${Pr}\left (X=x  \right ) = \binom{n}{x}p^x\left (1-p  \right )^{n-x}$$

Plugging our values into the formula, we get: $${Pr}\left (X=1  \right ) = \binom{800}{1}\frac{1}{365}^1\left (1-\frac{1}{365}  \right )^{800-1}$$

(An aside about the notation: $\binom{n}{x}$ is read as “n Choose x,” and a scientific calculator has a key for this. It is equivalent to $\frac{n!}{x!(n-x)!}$ )

Having set up the equation, we can grab our calculator and plug-and-chug to get

${Pr}(X=1)\approx 0.2448$

In other words, the odds of getting 1 birthday in that crowd is about 1:3.

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Gerald Belton
Statistician, Adjunct Instructor
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