I am deeply angry about the way I have been taught (for a given value of teaching) statistics.
In my first years of uni, statistics was a class where a bored woman would read slides projected in a wall. Not even good slides: they had mostly formal definitions of probability distributions. They explained nothing. It was just things we needed to memorize.
And because of the abysmal state of my statistics classes it took me an unreasonable time to actually understand how the Monty Hall problem works.
You might know about it? The one about the three doors, a car and two goats.
The very basic explanation: you are in a TV contest and you are offered a choice of three doors. Behind one of them, there's a car, behind the other two there are goats. After you choose one door, the presenter opens one of the remaining doors showing a goat, and gives you the choice to change your original choice or not.
Should you switch? Yes, you should. These days a lot of people know you should. But they have no idea why.
And this makes me angry because the explanation is incredibly simple for such a brainfucking problem, and gives you very nice peek into how probability works.
Ultimately, the Monty Hall Problem is a sequence of four choices:
So we're going to build a little table to solve this. Starting with choice 1, where the car actually is:
| where it is | weight |
|---|---|
| 1 | 1/3 |
| 2 | 1/3 |
| 3 | 1/3 |
To the left, what choices have been made, to the right the probability of those choices happening in sequence, now for step 2:
| where it is | first choice | weight |
|---|---|---|
| 1 | 1 | 1/9 |
| 1 | 2 | 1/9 |
| 1 | 3 | 1/9 |
| 2 | 1 | 1/9 |
| 2 | 2 | 1/9 |
| 2 | 3 | 1/9 |
| 3 | 1 | 1/9 |
| 3 | 2 | 1/9 |
| 3 | 3 | 1/9 |
Up until here everything is good, but in next step there's a twist: the presenter gets to make a choice only if you picked the right door; otherwise, they can only open the other wrong door. That means the weights become unbalanced.
| where it is | first choice | presenter | weight |
|---|---|---|---|
| 1 | 1 | 2 | 1/18 |
| 1 | 1 | 3 | 1/18 |
| 1 | 2 | 3 | 1/9 |
| 1 | 3 | 2 | 1/9 |
| 2 | 1 | 3 | 1/9 |
| 2 | 2 | 1 | 1/18 |
| 2 | 2 | 3 | 1/18 |
| 2 | 3 | 1 | 1/9 |
| 3 | 1 | 2 | 1/9 |
| 3 | 2 | 1 | 1/9 |
| 3 | 3 | 1 | 1/18 |
| 3 | 3 | 2 | 1/18 |
To complete the table we just add the change/not change choice which you always are allowed to do so it just splits every path into two with half the weight.
| where it is | first choice | presenter | change | win? | weight |
|---|---|---|---|---|---|
| 1 | 1 | 2 | Y | 🐐 | 1/36 |
| 1 | 1 | 2 | N | 🚗 | 1/36 |
| 1 | 1 | 3 | Y | 🐐 | 1/36 |
| 1 | 1 | 3 | N | 🚗 | 1/36 |
| 1 | 2 | 3 | Y | 🚗 | 1/18 |
| 1 | 2 | 3 | N | 🐐 | 1/18 |
| 1 | 3 | 2 | Y | 🚗 | 1/18 |
| 1 | 3 | 2 | N | 🐐 | 1/18 |
| 2 | 1 | 3 | Y | 🚗 | 1/18 |
| 2 | 1 | 3 | N | 🐐 | 1/18 |
| 2 | 2 | 1 | Y | 🐐 | 1/36 |
| 2 | 2 | 1 | N | 🚗 | 1/36 |
| 2 | 2 | 3 | Y | 🐐 | 1/36 |
| 2 | 2 | 3 | N | 🚗 | 1/36 |
| 2 | 3 | 1 | Y | 🚗 | 1/18 |
| 2 | 3 | 1 | N | 🐐 | 1/18 |
| 3 | 1 | 2 | Y | 🚗 | 1/18 |
| 3 | 1 | 2 | N | 🐐 | 1/18 |
| 3 | 2 | 1 | Y | 🚗 | 1/18 |
| 3 | 2 | 1 | N | 🐐 | 1/18 |
| 3 | 3 | 1 | Y | 🐐 | 1/36 |
| 3 | 3 | 1 | N | 🚗 | 1/36 |
| 3 | 3 | 2 | Y | 🐐 | 1/36 |
| 3 | 3 | 2 | N | 🚗 | 1/36 |
At this point you can just solve the problem by adding the weight of every line where you get a car while changing or not changing:
What do you know. The numbers support this: the chance of getting the car while changing is twice the chance of getting a car while not changing. And if you play the game enough times, you will find that the actual results match these probabilities, so science works.
So next time, remember to not change, and you get to win a goat. It's a lot cuter. and better for the environment.