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

Here’s the part to focus on: if you are wrong, the host will **certainly** offer you the correct door. That’s it.

If you’re wrong, the host’s door will be right **for sure**. Convince yourself of this.

So, there are 3 doors. That means the chances you get it wrong are 2/3, yes? And we know, if you get it wrong, the host will certainly, definitely offer you the correct door.

So with a chance of 2/3, the host is offering you the correct door.

If you only focus on this, I think it makes perfect sense.

It might help though if you imagine there are a thousand doors. If there are a thousand doors, the chances that you get the right door are very, very low. Its 1/1000 that you chose the right door. Right?

So then, with a chance of 999/1000, **the host is offering you the right door**. So you should take the door being offered by the host, its almost certainly the correct one.

Make sense?

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