Bill Snedden

As a some-time statistician, I love this problem because it illustrates so nicely how probability can confound our intuition.

I have found that a neat way to make the otherwise unintuitive solution more intuitive is to increase the original sample space to dramatize the result.

Let's say that I have one million lottery tickets on a table. I tell you that only one of them is a winner and allow you to choose one. I think we'd all agree that you have a 1:1,000,000 chance of picking the winning ticket. A pretty small chance, right?

Now, after you've picked that ticket, I reveal that 998,000 of the remaining tickets are losers. Then I ask you if you'd like to exchange the ticket you picked for the one remaining ticket.

How do you like them odds?!

Regards,

Bill Snedden

Quantum Ninja

I find a visual representation helps:

X = door containing the prize
O = empty door
- = a door that's been revealed to be empty and cannot be chosen

The three possible configurations are:

1 2 3
-------
X O O
O X O
O O X

Now let's say you initially pick door 1. As you can see by looking down column 1, you have a 1/3 chance of winning at this point.

After the host opens an empty door, the table looks like this:

1 2 3
-------
X - O
O X -
O - X

Now let's compare the two strategies using the above table:


Stick with your initial choice
1st row: you win
2nd row: you lose
3rd row: you lose
Probability of winning with this strategy: 1/3

Switch to the other door
1st row: you lose
2nd row: you win
3rd row: you win
Probability of winning with this strategy: 2/3

This helped me understand it, anyway.

Tenpudo

You forgot a possibility:

1 2 3
-------
X O -

That bumps it back to 1/2 with either strategy...

DaveGE

No, because X 0 - and X - O have a chance of 1/6th each.

Clutch

Bill, agreed! I always teach the problem in my Critical Thinking classes, because it wonderfully underscores my general message: Get in the habit of not trusting your gut feelings.

I also love it because I told it to my father-in-law, an emeritus stats prof, over dinner a few years back, and he immediately told me the odds were .5 after one door was opened!

The next night at dinner, he apologized.

And yes, the million-door version makes it a no-brainer. That second unopened door waaayyyy down the line, there, looks pretty darn suspicious...

Darwin26

... well all testing aside, Monty is a really wonderful congenial man.

Aria

I had always thought the "solution" contained an assumption that doesn't play out very well.

It treats the unopened doors as the same thing, essentially, when they are two different doors. So, each door should be accounted for individually, and every layout (and order) should be accounted for. Only if you ignore order, do you come to this conclusion.

Also, do a test with cups with a friend. Hell, we could organize an online door picking test.

Armchair dissident

I'm not sure I understand this analogy - it doesn't appear to equate to to the original problem. If you revealed the 999,998 tickets that lost, then the analogy would be better - and it would still be better to switch. (apologies if this is what you meant!).

The reasoning in this case is thus: Of the original selection, I have a 1 in a million chance of winning. Once you discard the 999,998 cards that loose, I now would appear to have a 1 in 2 chance that my card wins, and the same chance that your card wins - except I don't: If I hold the winning card, then you have discarded 999,998 cards entirely at random knowing that I have the right card, and can then present me with one random wrong card. But this will only happen once in every million times the game is played. Of the remaining 999,999 times the game is played, you will have the card, and I loose.

This is covered extensively in a book "Reckoning with Risk" by Gerd Gigerenzer (ISBN: 0-140-29786-3) which I can highly recommend

Farren

DAMN I just got it. But I think most of the explanations (on the linked sites too) are too wordy and confusing.

Once you've picked a door, the probability is skewed by the fact that the gameshow host can only give you (probablistic) information about the other two doors.

Mageth

Perhaps a better way to say that is, if the prize is behind the door you chose, it doesn't matter which of the two "booby prize" doors are revealed.

I think Bill's extreme example is the best way to show why it must be the case that switching is the best option.

This is difficult to understand because it's not intuitive. After a door is chosen, and before a door has been revealed, there is a 1/3 chance that the prize is behind the chosen door and a 2/3 chance that the prize is behind one of the other two doors. Opening one of the two doors does not change those original probabilities - there is still a 1/3 chance that it is behind the chosen door and a 2/3 chance it is behind one of the two unchosen doors (obviously, the unopened door). Intuitively, it may seem that it does change the probablity, but in reality it does not.

It is the same odds as it would be if neither of the two doors was opened and you were offered the chance to switch to the other two unrevealed doors. Opening one of the doors does not change the probability.