The story as it goes is well known: You are in a TV game, the host (called Monty) tells you the rules…

“There are 3 doors, one holds a flashy car, the other two goats… You may choose wisely…”

Easy… You have 1 in 3 options to win the price. You look at the doors, all look the same, no noises, no clues… Monty is not helping you… So you choose…done!But this not all… Cheeky old Monty has a final twist up his sleeve… He goes an opens one of the doors you didn’t choose only to reveal a goat, and then asks:

“Last chance! Do you stick or swap doors?”

So what do you do?

Which option does give you more chances to win? Or does it not matter? Think about it, and write it down…

So it turns out that the answer is SWAP! (You get a 2 in 3 chance of winning)

There are so many ways you could work this out, but this simple situation caused a big stir back in the early 1990, with people arguing both ways despite the mathematical proof already known at the time. But when it comes to simple things that seem that contradict reason, sometimes a proof is not enough and one needs to understand why is it the way it is…

Common sense has the ability to do this with ease in this case:

Let’s play a slightly different game. Let’s say that after you picked you initial door, Monty offers you the option to swap your door for the other 2 doors altogether… answer now is clear, right? But why is it different to choose the 2 doors in which we now for sure that one has a goat, than to choose only the door of those 2 that does not has that goat? (Monty by opening the door does just that) Reality is that in both games you only loose by swapping if the 2 other doors hold goats, i.e if your initial door holds the car, i.e 1 in 3 times and win otherwise, i.e 2 in 3 times…

Extra information on *The Monty Hall problem:*

*Monty Hall tested!*

A twist on the Monty Hall problem

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