Three Doors on a Gameshow Problem

If you switch, you win two out of three games...so yes, you should switch.

For example:

Assume the car is behind door #1.



If you pick (and then switch):
Door #1, the host opens, say #2. If you switch (to #3), you lose.
Door #2, the host opens door #3, if you switch (to #1) you win
Door #3, the host opens door #2, if you switch (to #1) you win.

If you switch, you will win if the car is behind EITHER one of the other two doors (2 out of 3 chance). But if you DON'T switch, you only win if the car is behind your original choice (1 out of 3 chance).


You can apply the exact same logic to any of the 3 doors you would originally choose regardless if the car is behind #1, #2 or #3.

The key is that the host knows where the car is, and ALWAYS opens a losing door after your choice.

Here's a post I made when this subject came up a couple of years ago...this covers all possibilities:



To see all of the possibilities:

You choose door #1:

If prize is behind door #1, the host will open either #2 or #3. In either of those two cases, if you switch, you lose, because the prize is behind #1.

If the prize is behind #2, the host will open #3... if you switch (to #2, after the host opens #3), you win.

If the prize is behind #3, the host will open #2...if you switch (to #3, after the host opens #2), you win.

So if you choose #1, the only way you win by "staying" is if the prize is behind #1...but if you switch, you win, regardless of if the prize is behind #2 or #3... you win either way.




Now, let's say you pick #2:

If prize is behind door #2, the host will open either #1 or #3. In either of those two cases, if you switch, you lose, because the prize is behind #2.

If the prize is behind #1, the host will open #3... if you switch (to #1, after the host opens #3), you win.

If the prize is behind #3, the host will open #1...if you switch (to #3, after the host opens #1), you win.

So if you choose #2, the only way you win by "staying" is if the prize is behind #2...but if you switch, you win, regardless of if the prize is behind #1 or #3... you win either way.




Finally, let's say you pick #3:

If prize is behind door #3, the host will open either #1 or #2. In either of those two cases, if you switch, you lose, because the prize is behind #3.

If the prize is behind #1, the host will open #2... if you switch (to #1, after the host opens #2), you win.

If the prize is behind #2, the host will open #1...if you switch (to #2, after the host opens #1), you win.

So if you choose #3, the only way you win by "staying" is if the prize is behind #3...but if you switch, you win, regardless of if the prize is behind #1 or #2... again, you win either way.



In any case, the only way you win by "staying" is if the prize is behind your original choice (1/3 chance) ...but if you switch, you will win if the prize is behind EITHER of the other two doors (2/3 chance).

If you do switch, you win if the prize is behind either of the doors you didn't originally pick...but if you stay with your original choice, you only win if you picked the correct door in the first place, which is only a 1 in 3 chance. That's why you should switch.


This, btw, is known as the "Monty Hall Problem", named after the host of the American Game Show "Let's Make A Deal" which ran in the 60s and 70s

An explanation of the problem, complete with diagrams, can be found at:

http://en.wikipedia.org/wiki/Monty_Hall_problem


As this article will point out, the 2/3 odds of winning by switching only applies in certain specific situations (i.e. host behaviors). So the 2/3 odds may not apply in all "3 door" problems..it depends on what the host does after the contestant has selected a door.


That thread is at:

http://www.redhotpawn.com/board/search.php?threadid=0&authorname=TheBloop&dayfrom=1&monthfrom=2&yearfrom=2001&dayto=9&monthto=10&yearto=2009&page=5