Your First Instinct is Wrong: The Monty Hall Problem (and its logical solution)
Your First Instinct is Wrong: The Monty Hall Problem (and its logical solution)
A complex problem with a simple solution you will get wrong.
Presented in a game show in the 60s-70s and named after its host, the Monty Hall problem is a famous probability puzzle that has baffled mathematicians and game show contestants alike for decades.
In this post, I'll explain the problem, its solution, and why it still confuses people to this day.
The Problem
In the final round of the show, the player was presented with three doors. Behind one of the doors was a valuable prize, such as a car, while the other two doors hid worthless booby prizes, such as a goat.
The player was asked to choose one of the doors. Then, before opening the chosen door, the host, Monty Hall, would open one of the other two doors to reveal a goat.
At this point, the player was given a choice: stick with their original door or switch to the other unopened door.
So, what's the optimal strategy? Should the player switch or stay with their original choice?
First Instinct
Let’s say there are three doors: A, B and C.
After a player has chosen a door (let’s say door A), Monty opens one of the closed doors (say B) and reveals the goat behind it. This gives the player the impression that now there are two possible doors (doors A and C) behind which the prize is: a 50/50 probability.
After all, there are only two unopened doors left, so it seems like the chances are equally split between them.
Answer
This intuition is wrong. In fact, the probability of winning if the player switches is 2/3, while the probability of winning if they stick with their original choice is only 1/3.
This may seem counterintuitive, but it can be explained simply by applying a bit of logic.
Explanation
Event A: The player chooses a door
When the player makes their initial choice, there is a 1/3 chance that they choose the door with the prize and a 2/3 chance that the prize is behind one of the other two doors.
Essentially, 2 out of 3 times, the player will choose the door which doesn’t have the prize behind it.
Let’s say the player chose door A. Then, we know that 2 out of 3 times, the prize will be behind one of the closed doors i.e. doors B and C.
Event B: Monty opens door C and gives the player the option to switch to door B
Monty opens door C and reveals that there is no prize behind it.
From event A’s conclusion, we knew that 2 out of 3 times, the prize would be behind doors B or C. Moreover, Monty just opened door C and revealed that there is no prize behind it. So the prize has to be behind door B.
Therefore, the optimal strategy is to always switch doors.
The source of confusion in the Monty Hall problem is not taking all events into consideration.
The probability of two events is 50-50 when you know nothing about them.
In this problem, the player is supposed to make an informed decision while considering everything that happened.
The Monty Hall problem is a great example of how our intuition can sometimes lead us astray when it comes to probability and statistics. Despite its simplicity, it has been the subject of much debate and confusion over the years. However, with a little bit of math and logic, the solution becomes clear.
In essence, it's often best to take a deep breath and thoughtfully consider the problem at hand before jumping to any conclusions.