The Monty Hall Problem I just dont get it.

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Why does the probability “lock in” and not change if you remove a variable but doubles for the other one?

I read it multiple times and it still doesnt make any logical sense to me.

In: Mathematics

33 Answers

Anonymous 0 Comments

You pick a door, let’s just say door 1.

What are the chances you guessed the door with the prize? 1 out of 3.

That means there’s a 2 out of 3 chance you guessed wrong, correct?

So now Monty reveals the prize was not behind door 2.

The chances that you guessed wrong initially *are still* 2 out of 3. By switching, you’re taking the odds that you initially guessed wrong.

Imagine it was 10 doors. You had a 90% chance of guessing the wrong door at the beginning.

Monty opens *eight* of the other doors.

The chance you guessed wrong initially is *still* 90%, and you’re being asked “Do you *really* think you’re *that* lucky to have guessed right? Or is it starting to get suspicious that I, the host of the show who knows where the prize is, haven’t opened this *one* other door?”

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