Thursday, June 22, 2006

The Monty Hall Paradox

[ARG] [Thinky Stuff]

Okay, apparently everyone in the world is aware of the Monty Hall Paradox except for me. Warning: If you're a Perplex City player, the link above and the discussion below will spoil card #174 Jaunty Paul for you, which is what this is coming out of. But.. if you're not a probability nut, and I mean a nut, and you haven't seen this before, you might do yourself a favour to read on anyway.

It goes like this: You're on a gameshow, and you're asked to choose between three doors. Behind two of the three doors are goats. Behind the other door is a yacht. You get to choose one door, and you win whatever is behind it. You pick door A. The gameshow host steps foward before you open the door, and says, "Are you sure about that choice? Do you want to change your mind?" He opens door C, and shows that there is a goat behind door C. The gameshow host always opens a door and reveals a goat at this stage in the program, regardless of whether the player has chosen the correct door.

The question: What are your chances of winning the yacht if you stick with door A? What are your chances if you switch to door B?

So my first answer was, "That's easy, it's a 1 in 2 chance no matter whether you stick with A or go with B." Wrong.

So my second answer was, "Well, you could say it's a 1 in 3 chance, no matter which you choose. But it's still even odds." Wrong.

The correct answer is: you have a 1 in 3 chance if you stick with door A, but your odds improve to 2 in 3 if you switch doors. To put it another way, you are more likely to win the yacht if you switch. A player who always switches at this point in the routine will win the yacht 2 out of 3 times.

That is, on your first guess, you have a 2 out of 3 chance of being wrong, so when the choice is offered, and you have (more likely than not) chosen the wrong door, it makes sense to jump ship to the other one.

Huh? Arrgh! My brain hurts.

The Wikipedia entry gives a good illustration:
It may be easier to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The player picks a door; 99% of the time, the player will pick a door with a goat. Thus, the chances of picking the winning door at first are very small: only 1%. The game host then opens 98 of the other doors revealing 98 goats and offers the player the chance to switch to the only other unopened door. On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. At this point a rational player should always switch.
Well, yes, that sounds good when you say it like that, but I still don't see how it squares up with the fact that there are two doors, and the yacht is just as likely to be behind either one. The probability that I've chosen the wrong one doesn't affect the probability that a given door hides the yacht.

And this is why I parted ways with maths in the eighth grade and have never seen eye to eye with it since. Oh, and I solved the damn Perplex City card. Grr.


That guy said...
This comment has been removed by a blog administrator.
GregT said...

In the immortal words of Strong Bad: DELETED!

I like a good flame war, but I'm not going to have one with someone I know in real life who isn't identifying themselves on their comments.

That guy said...

Is that supposed to make me contrite or something?

AJ said...

Okay there's obviously something going on there...

Entertaining story, if obvoious solution. Mocking you would seem redundant yet praise would seem hollow. I don't know what to say except, "props".

Stu said...

What's maddening for me is that I once *knew* how this works - but now your confusion is my confusion.


Andrew said...

I'll try a different explanation. I'm going to lay out a series of logical facts (I don't know if you're much into logic though, that's somewhat mathy as well).


There is a 2/3 chance that you pick the wrong door.

If you pick the wrong door then switch, you will get the right door. This takes some thinking, but logic it out and you'll realize it's true.

Therefore, if you pick the wrong door then switch there is a 2/3 chance. However your odds of initially picking the right door is only 1/3.

The trick is that what you select does change the probability.

To put it another way...

Let's say door B is the right one.

You have door's A B C.

If you choose A, then he reveals door C and B is left. That means that switching is the right move. This is a 1/3 chance where switching is correct.

If you choose B then he reveals either A or C. However switching now would be incorrect, this is a 1/3 chance as well.

If you choose C then he reveals A. Switching now would be correct, this is a 1/3 chance.

By summing up both opportunities where switching is correct you have 2/3 and the one opportunity where switching is incorrect is 1/3.


Hope that helps some. Oh, and what is your Perplex City ID? I'm way behind on my cards but I play when I get the chance, figure I might as well add you to my tracker.

Duncan said...

The first time I saw this problem was actually on an episode of Numb3rs. It was the one bit of math in the whole run of the series so far that I really didn't get. But the more that I think about it (and it takes some thinking) the more that I see it.

Or you can just accept that there are people out there who are much smarter than you (at least at some things) and their answers are, in fact, right. I do this a lot.

GregT said...

I think I get the explanation as Andrew delivered it (which also appears on the Wiki entry). And I understand that if you run a simulation it proves the statistics - the person who switches will actually find the yacht more often than the one who doesn't. But I still can't line it up with what common sense tells me, or see where common sense is going wrong.

My Perplex City tag is GregT. I think I'm about rank 1300 odd, or 20 or 30 somethingth in Australia.

Andrew said...

Well, you're doing way better than I am. I really need to get on the ball (I think I'm in the 7000's somewhere).

Jason O said...

It doesn't line up with "common sense" because the logic is counter-intuitive.

Think about it, there are three doors. A, B, and C.

You pick door A and the host opens door C and reveals a goat behind it. The common reaction is going to be "Well I didn't pick door C, so there is a chance that door A was the right choice!"

Your first reaction doesn't play the odds. The reaction is based on an assumption that your odds that door A was the right choice is still the same or even better, when in fact it is worse.

Actually, when I think of it that way, then the paradox makes sense. I couldn't fathom it, even though the math made sense, until I thought about what the common reaction would be.

Essentially, the assumption that you picked the right door is made on the thought that you have a 1 in 2 chance of winning, when in fact it never changed from 1 in 3. You already made the choice before they revealed the goat behind door C. All that doors is confirm door C is the wrong choice. Switching to door B is a 2 in 3 shot while sticking with door A is the same as before, 1 in 3.

Unfortunately, you still have a 33% chance of guessing wrong.

Don said...

Wellity Wellity... What about roulette? You bet on Black four times in a row.. and four times Red comes up... what is the probability of another Red coming up on the next roll? Which colour is the smart money on?
Logic will tell you that Black is a good bet.. but in fact each roll has a one in two (50%) chance of landing on red. Is it not the same in this example?! Every time a door is revealed you make a NEW and UNIQUE choice.. even if that choice just happens to be the same as the choice you previously made.. you have the same chance of getting the prize... now... although my internet university masters in game theory is as worthless as the electrons it's written on.. my ramblings hopefully make sense to somebody!! also... Hi Greg! Long time, no see!! ;)

Anonymous said...

An important point is that the host must show the losing door on purpose.
And because he shows the losing door on pupose, choosing a door then switching is the same as picking the two doors you didn't pick.
That is if you pick door one, he purposely shows door two was a loser, and you switch to door three.
The result is the same as if you had been able to choose doors two(which he had to choose) and three to begin with.

Chris - Sheffield - Yorkshire said...

"I still don't see how it squares up with the fact that there are two doors, and the yacht is just as likely to be behind either one"

If this statement was correct, then there would be particular sense in swapping, of course. But the statement is wrong. The door you chose first had a one third chance of being the right one, and it still does !

Try and imagine the case of a hundred doors. You pick one, and Monty opens 98 of the others. You now have two doors to choose from. But they are not equally likely to contain a yacht. You will nearly always win by swapping (the exception is the 1 time in 100 that you origianlly chose the right door).

The whole point of this puzzle is to demonstrate that if there are options you do NOT always have a 50/50 chance of picking the right one. In fact, it is only 50/50 when you have no useful information at all on which to make your choice ! Any useful information makes the choice not 50/50 FOR THE HOLDER OF THAT INFORMATION ! In this case you know that your original door only had a 1/100 chance of being right.