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Monty Hall Problem - Solution is false

I have been thinking about this problem for a while, and no matter how much I think about it -- I remain convinced that the answer provided is incorrect.

As was suggested, when Monty Hall exposes a door, the assumption is that from that starting point, it is better to switch, because it provides the contestant with a 2/3 of a chance to win.

It’s only good for your ego, and is statistical false at that moment you must make a choice.

Monty Hall did not just expose the door; he REMOVED that door from the pool of choices.  Utilizing the third (and exposed) door is a statistical red herring.  The contestant in the end only had 1 out of 2 chance to choose the right door.

It is the same thing with "Who wants to be a Millionaire?" game show.  The 50/50 equation does not make your chance any greater than 50%, because in the end, you have to choose from two choices for your 'Final Answer', no matter how positive your phone in friend may be.

Saturday, September 27, 2003

Yes!  Another thread on this topic!

Ok.  Is the MHP the same as going 50-50 on WWTBAM?  No.  Understanding how they are different is critical.

Tuesday, September 30, 2003

Find a friend, and play the game. Enlightenment will follow.

Jim Lyon
Tuesday, September 30, 2003

Hmm, T.J.

Are you saying that you think using the 50:50 option on Millionaire makes no difference to your chances? (assuming that you know nothing about the correct solution of course).

David Clayworth
Friday, October 3, 2003

It all depends on whether you know 2 things for certain: a) MH will *always* open a door after you choose, and b) the door he opens will always be an incorrect one.

If both of those conditions are met, then, indeed, having an a priori strategy of always switching is the far better option, because you effectively get to choose 2 of the 3 initial choices rather than 1.

Of course, in the real Let's Make A Deal, neither of these 2 conditions are necessarily true (MH often offers some third other choice, and even occasionally opens the winning door after you choose, based on the criteria of what he thinks will make the most interesting show).  Therefore, one can't draw any conclusions in the "real" problem.

Ray Trent
Tuesday, October 7, 2003

The solution is:

Your first choice is based on 1 correct out of 3 possibilities: a 66.666 percent chance of failure.

After M.H. has exposed a bogus door, there are two choices left, the one you originally went with at 66.6% risk, and the remaining door.

With only 2 doors remaining, your risk is reduced to a mere 50% at this point, so you should switch every time.

Paul Jensen
Tuesday, October 14, 2003

Everyone on here is basing the calculation on the initial status, not the final and critical deciding moment.

I strongly believe that the actual statistics should only be derived from the real data that is present at the moment you must make a choice, not a moment you flipingly choose, assuming that the factors are true (that a door will always open, and it'll be a losing choice).

In the end, you are given 2 choices -- the one you chose, and the one you didn't.  This is where statistics should be applied, not the initial starting point.

Thursday, October 23, 2003

To the gentleman that asked about that the 50:50 puzzle -- it actually improves your chance...

From 25% to 50% due to the choices being available at that time.

Just as Monty taking away a door improves your chance from 33% to 50%.

Yes, it is fun to statistically derive the possibilities from the start -- which is why they love to throw around the high number of probabilities of a chess move over 50 moves.  It explicitly ignores the current moment issue where you are at the 49th move, and you have 2 choices, one to escape checkmate, and another to not.  Reaching that situation is a very low probability due to the choices that is offered, but once you reach that moment, you have 2 choices left, and a 50% chance that you'll survive for that move.

Timing is very important, in my opinion, when it comes to statistics.  The moment of the final choice is where the calculation shall apply.

Thursday, October 23, 2003

Dear TJ,
            This has been said a million times before, but imagine there are a thousand doors. You choose one door at random; MH then removes 998 empty doors, leaving just the one you had chosen out of a thousand, and the one he chose out of the other 999, knowing full well that the 998 he discounted were losers.

Stephen Jones
Sunday, October 26, 2003

The given solution is correct.  Maybe this will put it in terms the fifty-percenters can understand:

There are only two ways you can win:
A) Pick the prize door then "stay", or
B) Pick an empty door then switch

B works because Monty Hall rules out the other empty door for you.  It should be obvious that A works, and also that switching from the prize door or staying with an empty door will fail.  It should also be obvious that you can make your decision whether to stay or switch *before* Monty reveals an empty door (after all, that's what we're doing now).

So if you want to win, you have to attempt A or B.  There's no other way.  Now, which do you think is easier?  Since there are twice as many empty doors as there are prize doors, you're twice as likely to succeed by attempting plan B.

Bill Hees
Sunday, November 23, 2003

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