### The Monty Hall ‘Paradox’

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The Monty Hall problem is simple:

You find yourself on a game show called “Let’s Make A Deal.” The game is very simple. There are three doors: door #1, door #2, and door #3. Behind one door is a million dollars. The other two doors contain worthless joke prizes. All you have to do is pick which door you want to open, and you get whatever is behind it. But you only get to open one door. By simple math, then, you obviously have a 1 in 3 chance of picking the correct door and becoming an instant millionaire.

You pick a door. As soon as you tell Monty (the gameshow host) what door you want to open, he stops and says, “Okay, you’ve made your choice. Now, I’m going to do what we always do here on this game. I’m going to open one of the other two doors for you that I know has a booby prize.” And he does so. Then he asks, “Okay, now, would you like to stay with your original guess, or would you like to switch to the other door that’s still closed? You only get one shot, so do you want to stay with your original choice, or switch?”

Here’s the question: is there any compelling reason to switch doors?

The answer is surprising!

You’re better off, by 2:1, if you switch doors. It’s completely counter-intuitive (I still have trouble convincing myself :-), but it’s true.

The comment that clinched it for me is this one, by Geoff Richards :

If you switch after Monty eliminates one wrong door, then you always swap your choice (if your first choice would have won, you change to a loosing one, and vice verse). And since there are initially two wrong doors to choose from your initial choice is more likely to be wrong, so switching reverses those odds and makes you more likely to win.

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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.

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

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.

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.

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

switchingdoors 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 toswap, in which case Claudia’s analysis above appears correct.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?

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

Thanks for the link!

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 “Iâ€™m going to do what we always do here on this game.”) In this case, it’s always a mistake to switch.