The Monty Hall problem

The following simple Maths problem has proven to be subtle, controversial, paradoxical and idiosyncratic. All this aside, the result is nonetheless factual and makes good television.

So you’re on a game show where you’ve been shown 3 doors. Behind one of the doors is a prize. Behind the other 2 are goats. You get to choose a door. Let’s say you choose door 1. The host then opens one of the doors you didn’t choose to revel a goat.

You’re now asked if you want to stay with your fist choice of door or would rather switch doors. What do you do? Your first thought is to stay door 1. After all, it has worked out so far. Since there are 2 doors now, you reason that it’s a 50/50 shot for the prize.

Right?

Wrong!!!

The best strategy for this game is to switch every time.

• The player who’s strategy is to switch every time will only lose when door they initially selected had the prize behind it.

• Since the odds of choosing the prize on the first move are one in three. The odds of losing the game when you switch every time are also one in three.

• This means that a person who switches every time will win two-thirds of the time. This is double the odds of winning of the person whose strategy is to stay every time.