Why does switching doors in the Monty Hall Problem increase odds: 2 doors, 50-50

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I have read through around 10 articles and webpages on this problem, and still don’t understand. I’ve run simulations and yes, switching does get you better odds, but why?

In: Mathematics

35 Answers

Anonymous 0 Comments

Here’s a way of explaining it that helped me see it:

Imagine you pick the left-hand door, and then Monty says “*you can either stick with the left-hand door, or you can choose to have the prize if it’s behind either the middle or right hand door*”. Obviously you would switch because you are now being offered 2/3 odds instead of 1/3 odds.

So after you’ve switched and your choice is locked in, Monty says “*OK, go ahead and see if you’ve won, but to save you time opening doors, I’m going to let you know it’s not behind the middle door*”.

When you switch around the order like this it is obvious that the decision to switch is correct. But there is no actual difference between this and the original problem – either way Monty is offering you a 2/3 choice – the only difference is *when* he tells you which door (out of the two that you didn’t originally pick) does not have the prize.

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