Comments on: The Monty Hall ‘Paradox’

By: Reid Ellis

Reid Ellis — Tue, 30 Nov 1999 00:00:00 +0000

I just think of it this way: the first choice you made had a 33% chance of being correct. Switching to the other door increases your chance to 50%. Pretty simple, really.

By: Claudia

Claudia — Tue, 30 Nov 1999 00:00:00 +0000

You can look at it a much simpler way.

Suppose you chose door one. Monty opens door three and there is a goat. So you swap to door two. YOU WIN!!!

Suppose you choose door two. Monty opens door three. You swap and YOU LOSE!!!!

Suppose you chose door three. Monty opens door one you swap and YOU WIN!

These are the only possible senarios and in 2 of them you win.

Just forget everything you know and it makes sense

By: Big Bob

Big Bob — Tue, 30 Nov 1999 00:00:00 +0000

There are two scenarios:

1. Pick the correct door(1/3 of the time). Monty opens either of the other doors. You switch, you lose.

2. Pick an incorrect door (2/3 of the time). Monty opens the only other losing door. You switch, you win.

By: Michael

Michael — Tue, 30 Nov 1999 00:00:00 +0000

Here’s the way I look at it (it’s sort of a before/after point of view):

If there are three cards to choose from we know that there is a 33% chance of winning. But, when big Monty reveals a goat door, that eliminates one possibility from the scenario, because the contestant knows not to pick that door. So then there are two doors left. Forgetting all context, one knows that 50% of all those two doors wins – switching or no, thats the new probability of winning. – but this isn’t my point of view, because this alone still doesn’t seem right to me.

I view it as an equation, if you will, in which a variable, the door chosen by the contestant, is to be identified as winning or not. In the beginning, there are three possibilities: losing door A, losing door B, and winning door; therefore, we are 33% sure of the door’s identity. When Monty reveals a losing door (A or B, like we care), there are only two possibilities left – we’ll say losing door B and the winning door. Now the equation is simplified a step and we’re 50% sure of the door’s identity. Finally, Monty reveals the chosen door (simultaneously eliminating the other door, making only one possibility), and we know (100% sure) that it’s either the winner or loser B. No switching is required because that doesnt change the number of possibilities available, it just kinda symbolizes that one of the two possibilities was chosen.

By: James in LV

James in LV — Tue, 30 Nov 1999 00:00:00 +0000

This paradox is great…but Claudia’s post shows how a person walked right into it’s trap. After Monty has opened one of the doors, you have a new choice to make. Therefore, all odds are reset to the new decision at hand. It really isn’t a matter of switching doors either because you can simply say you haven’t even selected a door at all yet. Since a new choice has to be made among two doors, it’s simply a 50% chance. The trap is assuming that you have to swap, in which case Claudia’s analysis above appears correct.

By: Michael H.

Michael H. — Tue, 30 Nov 1999 00:00:00 +0000

James’ comment reflects the common intuition, that after Monty opens a door to reveal a booby prize, the two remaining doors are equivalent (i.e. there’s a 50% chance with either door). That intuition is wrong, which is why this is called a paradox.

Big Bob had it right. Your original choice is only right 1/3 of the time, so it makes sense to switch after Monty opens one of the other doors. The odds are 2/3 that the money is behind one of the doors you didn’t pick. Monty gives you some really useful information by eliminating one of those doors. The odds are still 2/3 that the money is behind one of the other doors, but now one of those doors has been eliminated for you.

Consider this: if it were equally good to switch or stick with your original door, then you could stick with your door and be right 50% of the time. That means that, presented with 3 doors, you can pick the right one 50% of the time. Nice work, Kreskin!

Or consider this: Instead of 3 doors, there are 100. You pick one, then Monty opens 98 of the remaining 99 doors, to reveal nothing but booby prizes. Would you stick or switch to the other unopened door?

By: Michael H.

Michael H. — Tue, 30 Nov 1999 00:00:00 +0000

There’s a very nice explanation of this paradox — and even a chance to play the game — at:

http://math.ucsd.edu/~crypto/Monty/montybg.html

By: Harald

Harald — Tue, 30 Nov 1999 00:00:00 +0000

Thanks for the link!

By: T. Bishop

T. Bishop — Tue, 30 Nov 1999 00:00:00 +0000

Consider the possibility that Monty is malicious, and only gives the option of switching if the original guess was correct. (He’s lying when he says “I’m going to do what we always do here on this game.”) In this case, it’s always a mistake to switch.