Probability Puzzle courtesy of Nassim Nicholas Taleb

A couple of weeks ago Nassim Taleb posted the following probability puzzle on Twitter.

Probability puzzle Nassim Taleb

I love puzzles so I took a stab at this using two different approaches. You can download my Excel spreadsheet solution here.

Approach 1: Probability Theory

I recognized the problem as being an example of binomial probability which I used previously in the birthday problem. We want to know the probability of having x successes from N trials where the probability Π of success in any one trial is constant. It’s a little trickier because there are 16 squares on the grid, so to begin I simplify it by calculating the probability of getting three darts in one specific square.

Note: It wasn’t clear to me whether we should calculate for exactly three darts in a square or at least three darts in a square. I did both as the extra computation is trivial.

x = 3 ; N = 8 ; Π = 1/16 = 0.0625 – plug these numbers into the following formula and you should get 0.0099, that is the probability of getting exactly 3 darts in a specific square, e.g. the top left square.

To compute the probability of getting exactly three darts in any square, simply multiply by 16 and you get 0.1584. Now to extend this and compute the probability of getting at least three darts in any square, simply repeat the above for x = 4, 5, 6, 7, 8 and sum the answers together. Or sum the probabilities for x = 0, 1, 2 and subtract from 1. Either way you should get an answer of 0.1723, that is a 17% chance of getting at least three darts in any square.

Approach 2: Simulation

Let’s say you understand the problem well but you’re not so hot at probability (that’s ok by the way, probability is hard, I’ve regularly seen experts make mistakes!) BUT you can code. If you can code you can randomly simulate the problem over and over again and just see how the results pan out. Heck, you don’t even need code per se, I did this in Excel – download my spreadsheet here. The simulated results matched the theory pretty well – 0.1542 and 0.1699 for exactly 3 darts and at least 3 darts respectively.

Closing Thoughts

Doing fun puzzles like this is a great way to stay sharp statistically, think of it like a work out in the gym. And even if you don’t solve the puzzle, trying to solve it before peeking at the solution is how you expand your statistical knowledge. I recommend looking at all the replies to the initial tweet to see how some people got it wrong and how others got it right. And remember there can be multiple possible solution methods. I’ve spoken previously about the importance of tackling problems in more than one way in the Monty Hall Problem. It’s good to have these different approaches in your arsenal because:

  1. You never know when you’ll need them.
  2. Our knowledge is solidified when we understand a problem from different angles.
  3. If we’re not 100% confident in our answer, another approach can serve as a validation.
  4. Depending on the audience you may have to try different explanations to ensure they understand.

PS: Things can get a little funky when you start considering the cases where there are three darts in more than one square. I ignored that here but it does explain the slight difference between the theoretical solution and the simulated solution.


The birthday problem

How many people would you need in a group before you could be confident that at least one pair in the group share the same birthday?

One day, back in Smurfit Business School, our statistics lecturer challenged us to a bet. He predicted, confidently (smugly even), that at least two of us shared a birthday. He bet us each the princely sum of €1. I glanced around me and I counted close to 40 students in the room. Being the savant that I am, I also know there are approximately 365 days in a year, and so I thought, you’re on! I mean, even allowing for some probability magic: 40 people, 365 days, this is free money!

I soon learned this was the famous birthday problem and although I was beginning to feel cocky as we got half way through my classmates’ birthdays, our teacher ultimately prevailed. It turns out that in a group of just 23 people the probability of a matching pair of birthdays is over 50%!

I hope this spreadsheet and the explanation below will help you understand why this is so.

  • We need at least 2 people to have any chance of having a matching pair. This is trivial. Person A has a birthday on any day. The probability of Person B matching is 1/365.
  • With 3 people, there are three possible matches: A matches B, A matches C or B matches C.
  • With 4 people there are 6 possible combinations (count the edges in the little diagram shown here).four people and six combos You might spot a pattern by now. In mathematics these are known as combinations. After a while counting manually becomes tedious but, thankfully, for any given number of people we can use the combination formula to see how many possible combinations exist – jump to column B in the spreadsheet for a closer look.
  • The probability for any one of these combinations being a matching pair is 1/365. Think of that like a bet: each individual combination is a bet with a 1/365 chance of winning. How many of these bets would we have to place to get at least one win.
    • Here’s a neat little probability trick for answering an “at least” type question. Compute the probability of not winning at all, i.e. precisely zero wins, and subtract that value from 1.*
  • Column C in the spreadsheet uses the binomial distribution formula to compute the probability of a specific number of wins from a given number of bets where each bet is independent and has an equal probability of success.
  • In our case we want to compute the probability of precisely zero wins and subtract this value from 1. This gives us the probability of at least one win.

birthday chart

In the results, we can see that 23 is the magic number where the probability of at least one match exceeds 0.5. Remember there were close to 40 in my class so my teacher knew at a glance that his probability of finding at least one pair was close to 0.9 … and there were enough suckers in the room to cover his lunch!


* This little problem inversion trick can be generalized further to any occasion when we are faced with a difficult question. If you’re struggling, try inverting the question. Having difficulty predicting fraud? Maybe try predicting “not fraud”! It sounds trivial, silly even, but inverting a problem can get you out of a mental rut. For a famous example, see how statistician Abraham Wald used this technique to help the Allies win WW2.