If you’ve never heard of the Monty Hall Problem before, here is a brief summary.
You are in a game show where you are faced with three closed doors. Behind one of the doors is a brand new car and the other doors hide a goat each. You then choose a door in an attempt to win the car but before the door is opened the host (Monty Hall) opens one of the remaining doors he knows contains a goat. You are then given the opportunity to choose whether to stay with your original selection or switch to the other, remaining closed door.
Intuitively most of us tend to think that it doesn’t really matter because we think there is a 50/50 chance either way but the reality is that statistically you have a greater chance of winning the car if you switch. In fact you stand a 2/3 chance of winning with the switch strategy as opposed to a 1/3 chance with the stay strategy.
If you remain unconvinced feel free to try it out for yourself with a friend or a die and a piece of paper.
Anyway, the point of this post is to say that in the two years or so that I’ve known about this I’ve never heard a concise and easy-to-understand explanation for why this is so. But now I have one…
You have a 2/3 chance of selecting a goat instead of a car. This means that two thirds of the time Monty Hall will reveal the last remaining goat which means that two thirds of the time if you use a switch strategy you will win.
To highlight how this works, imagine the same scenario but with 1000 doors and only 1 car. You have a 1/1000 chance of selecting the car or a 999/1000 chance of selecting a goat. If Monty Hall opens 998 other doors with goats behind them it is extremely likely (999/1000) that the last remaining door will have the car. (Thanks A3!)
You have two strategies: switch or stay. Using the stay strategy, if you accidentally choose a car first off (1/3 chance) you will win but if you choose a goat (2/3) you’ll lose. Using the switch strategy you’ll reverse these odds giving you statistically a 66.6% chance of winning every time.