As I watched the miners get taken out behind the woodshed yesterday, for yet another trading session, I started to think about how sure I was the bottom was in.

Not so, it seems.

My mind started to track back to other times I was wrong, and how I teach others to think when things are going against them.

One of the most important things I learned was the Birthday Problem. It’s a rather famous statistics problem I didn’t learn in any of my stats classes at either Villanova or London Business School. And you wonder why I think an overabundance of degrees is a waste?

Let me take you back two-and-a-half years, in the middle of a graduate session in New York City.

The class I had been teaching with bosom bud Andy had been going well. The kids were from a smorgasbord of the best schools in America. We had Harvard, Michigan, Washington University in St. Louis, Penn State, Colorado, Colorado State, and my trusty alma mater, Villanova, among others.

And they were lovely young people, as well.

Andy was teaching about data and statistics when I chipped in with my favorite statistics problem.

It blew their minds, as it did mine when I first learned about it.

The Question

When I interjected in Andy’s class (with his permission, of course), I asked one question:

What is the probability that at least two of you have the same birthday?

To be clear, we’re just looking for day and month, not year. So my birthday is December 20, 1974. We’d consider December 20, 1990, a match because we’re not looking at the year.

There were 229 students, 2 teachers, and 5 bank support staff in the room, for a total of 236 people.

Now, if you’ve never heard of this famous problem before, you may offer something like “1 in a million,” as I did the first time I heard it. It’s not even close to the correct answer, as you’ll see.

To their great credit, the kids didn’t offer any outlandish answers like I did.

Most guesses were from “very low” to “50%” to “63%” to “very high” and “above 90%.”

I was impressed.

But with that many people in the room, the probability that at least two people have the same birthday is 99.99999%.

Don’t believe it? I’ll prove it to you.

The Birthday Problem Explained

Before I show you the math, let me tell you what happened in class.

After I asked the question and got those excellent answers, I said, “Now, let’s have some fun.”

I then asked, “Who has a January birthday?”

About 25 people raised their hands.

Then and there, I knew we’d have a match.

I pointed to a young intern and asked, “What date were you born on?”

“January 17th,” she replied.

“Any of you with your hands up born on the 17th?” I asked.

Another intern raised his hand.

The class clapped uproariously.

I said, “Wow! The rabbit came out of my hat on the first try!”

The kids marveled.

Since I already knew the math, I wasn’t surprised at all.

The Math Behind The Birthday Problem

Alright, let's dive deeper into the math behind the Birthday Problem. Don't worry, I promise it won't be as daunting as it sounds!

The probability is easier to calculate if we consider the opposite: what is the probability that all people in a group have different birthdays?

First, consider an empty room and the first person who walks in. The probability that their birthday is not shared with anyone else (since they're alone) is 100% or 365 out of 365.

Now, the second person who comes in can have any of the remaining 364 days as their birthday, without sharing it with the first person. So the probability for the second person is 364/365.

As the third person enters, they must avoid the two existing birthdays, so their probability is 363/365.

As you continue this process, the probability for each new person declines.

So, for a group of 23 people, you'd multiply these individual probabilities together:

Let P = Probability.

P(all different) = (365/365) x (364/365) x (363/365) x ... x (343/365)

This comes out to around 0.4927, or a 49.27% chance that all 23 people have different birthdays.

But remember, we want the probability that at least two people share a birthday. So, we subtract the "all different" probability from 1:

P(at least one shared) = 1 - P(all different) = 1 - 0.4927 = 0.5073, or 50.73%

So, surprisingly, in a group of just 23 people, there's over a 50% chance that at least two people share the same birthday!

This counterintuitive result makes the Birthday Problem so fascinating.

For you traders out there, this means “take the bet!” if there are 23 people in the room, since the probability is in your favor.

By the time we reach 57 people in a room, the chance is just over 99% that two people will have the same birthday.

So with over 230 people in the room, the call was a no-brainer.

Here’s a chart of the Birthday Problem outcomes:

Credit: Sean Ring

I knew of the Birthday Problem because I read a book called Chance by Amir D. Aczel a long time ago.

Aczel was a statistics professor who wrote that little book because he was so frustrated at the innumeracy displayed daily. I share his opinion.

Aczel dedicated an entire chapter to the Birthday Problem because it’s so counterintuitive that even statisticians have trouble with it.

So don’t fret if you’re still trying to understand it. Professional mathematicians are still trying to understand it.

But more than that, it’s a reminder not to use our guts when it comes to financial decision-making. It’s essential to think of the statistics behind our choices.

Wrap Up

Today’s short Rude is a reminder we don’t know it all, and we may never will.

But that’s okay because every advantage acquired brings us closer to the life we want.

Just being aware of this paradox puts you in a special club.

May knowing this increase your intellectual curiosity and eagerness to learn.

Have a wonderful and restful weekend!