Happy Birthday!

For today’s post, I thought I’d return to statistics. Remember the Monty Hall problem I talked about last year? If not, do a search on “Monty” on the Blog page. That was an example of statistics defying common sense. Today’s post is another one of those.

This post is about the probability of any two people in a group of people in a room having the same birthday. But let’s simplify this scenario with an equivalent one. Suppose we have a random number generator that generates a number between 1 and 365, including 1 and 365 – kind of like a 365 faced die. Let’s say we “roll” this die twice. What is the probability that the two numbers generated are the same, the successful event?

As is often the case in statistics, it is easier to look at the probability of the unsuccessful events. You can then subtract that from 1 to get the probability of the successful event since

Probability of Success + Probability of Failure = 1 or

Probability of Success = 1 – Probability of Failure

since one or the other must happen. Remember that a certainty in probability is “1”, absolutely no chance is “0” and other probabilities are between those two numbers. For example, the probability of flipping a heads is 0.5. Please see my posts on probability for a review if needed.

So if we roll this 365-faced die twice, the first roll sets the number and the chance of the second roll matching that number is 1/365 and the chance of not matching that number is 364/365. This probability is

${1}\hspace{0.33em}{-}\hspace{0.33em}\frac{364}{365}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{00274}$

Very small! I wouldn’t bet on that happening. This is equivalent to the probability that two random people have the same birthday. Please bear with me here, but an equivalent expression that takes into account that we are rolling the die twice (or have two people in the room) is:

${1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{\frac{{2}\times{1}}{2}}{=}\hspace{0.33em}{1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{1}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{00274}$

That expression in the exponent, (2×1)/2, is how to calculate the number of pairs that have a chance of being a success. Since we are just rolling the die twice (or there are just two people in a room), we only have 1 pair. If we roll the die 3 times, there are (3×2)/2 or 3 pairs of numbers to compare. Note that this exponent is generated by multiplying the number of rolls by one less, then dividing by 2. So for three rolls (3 people in a room), the chance of two numbers being the same are

${1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{\frac{{3}\times{2}}{2}}{=}\hspace{0.33em}{1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{3}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{00}{82}$

Well that appeared to have increased the odds a bit. Let’s roll the die 10 times or have 10 people in a room. There are (10×9)/2 0r 45 pairs that have a chance of being the same. So the probability in this case is

${1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{\frac{{10}\times{9}}{2}}{=}\hspace{0.33em}{1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{45}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{1161}$

That means that in a group of 10 people, you have slightly better than an 11% chance that any two people have the same birthday. That really increased the chances with just a few more people! You can keep doing this for any number of rolls (people) using the formula

${1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{\frac{{n}{(}{n}{-}{1}{)}}{2}}$

where n is the number of rolls or number of people in a room. If you let n = 23, you will find that the chance of any two people having the same birthday is

${1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{\frac{23(22)}{2}}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}{-}\hspace{0.33em}{\left({\frac{364}{365}}\right)}^{{253}\hspace{0.33em}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5005}$

You have better than a 50% chance that in a room of 23 people, two of them will have the same birthday! Mathematically, this is so because you have 253 pairs to compare, or 253 opportunities of a success. What a surprise!

Posted on Categories Pre-VCE, Statistics