The Monty Hall Problem

[Rowan]

I am still unsure of the math being sugguested that claims you have a higher chance of picking the correct door, despite there being only 2 to choose from.
to put this simply and for those who cant be bothered watching the video;

>you are given a selection of 3 doors.
>behind one of those doors, is the prize.
>the other 2 doors is nothing
>select the door you think the prize is behind (33.3% chance of success)
>host reveals a door that has nothing leaving only your selection, and the other door unopened.
>gives you the option of changing your mind and selecting the other door.


Apparantly, math and logic says that the odds of it being in the door you did not select, is higher than the one you intially picked because when you picked it before one of the doors with nothing was eliminated from the equation, you had a 33% chance of success. By now having the option to choose between 2, you increase your odds of because now you are only picking between 2, but ONLY if you choose the door you didnt pick.

the video explains in more detail, but I cannot fathom it.

[Heartless Angel]

It's pretty simple. Because he always reveals one of the wrong doors, AFTER you have chosen a door, choosing a wrong door to begin with means only the right one will remain when you make your switch. The only way you can win without switching is to pick the right door on the first attempt, which is a 1 in 3 chance. Switching, you win in any case in which you chose the wrong door to begin with, which is a 66% chance.

EDIT: More simply put, because the wrong door is removed only after your selection, you're not actually gambling between the remaining two doors, you're only gambling on your original choice. Switching, you're betting against your original choice, staying, you're betting on it.

EDIT 2: Electric Boogaloo: An even simpler way to think of this, consider what the switch is actually asking. By removing a wrong door, he is leaving you two possibilities for judging your first guess, right, and wrong. He's not actually giving you a new choice of doors here, he's just asking you if you think your first was right. Switch wins if my first guess was wrong, stay wins if my first guess was right. Because your first was just a blind guess on a 1 in 3 chance, it's most reasonable to bet against yourself.

[Alpha]

You pick your first door. You have a 1/3 chance of being right.

That means the other doors, together, have a 2/3 chance of being right.

Then Monty Hall opens one of the two doors you didn't choose. The door he picks must have a zonk behind it (a bad door). Switching to the unopened door still carries that 2/3 probability of being correct, it's just now been concentrated into that remaining door. This is true only because Monty must open a door that does not have the prize behind it. The video gives an example with 100 doors, and Monty opening 98 of them to reveal zonks. In that case switching has a 99/100 chance of being the right decision.



Numberphile video: Monty Hall Problem - Numberphile - YouTube

I've lost whole days watching Numberphile lol, careful.

[Pete]

While perfectly logical, it still is hard to wrap your mind around. Wouldn't the idea be the same, with a 2/3 chance of the prize being your door (1) or the goat (2), and a 1/3 chance of it being door 3?

Isn't it still a 1/3 chance regardless, since there's only one prize and three doors?