the Monty hall problem



The Monty hall problem is (from wiki)

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

So the answer is supposed to be switching doors as that gives you 2/3 chance opposed to not switching which only gives you a 1/3 chance, but I can’t seem to wrap my head around Why?

In: Mathematics

The key point to make is that the host can NOT reveal the winning door. Thus, by opening one of the “dummy” doors, some information is given by the host.

It’s easy just to write out the possibilities.

1.) You pick the prize door. host reveals a dummy at random.

2.) You pick one dummy door, host reveals the other dummy door. You swap and win.

3.) You pick the second “dummy” door, host has to reveal the first “dummy”. You swap and win.

There’s a 1/3 chance you initially picked the correct door. If you opt not to change doors, it remains a 1/3 chance.

There’s a 2/3 chance you initially picked the wrong door. Picking the wrong door forces the host to open the other wrong door, which means the remaining door has to be the correct one. In other words, you have a 2/3 chance of winning if you choose to switch doors. Picking the wrong door initially and then switching guarantees that you win.

I disagree with all previous answers. sorry to the teacher…. This is a probability question there are 9 alternatives. Not a fractions question

TLDR; The host cannot choose a car and cannot choose the contestants door. Also there are more goats than cars. The combination of these two result in a 2/3 probability that both people chose goats. so it is better for the contestant to change their choice.