Three doors

Originally posted by castlerook
Fair enough. Again though, I think it's fairly easy to remove any ambiguity by saying something like "after the contestant picks a door, the host always reveals a door that does not contain the prize" or something like that.

Edit: Just clicked on your profile--loved it 🙂

Thanks. 🙂

But I agree, it would remove ambiguity.

dood: I don't agree but I think we've took this far enough. Be well.

This was discussed back in January at:

http://www.redhotpawn.com/board/showthread.php?threadid=59662

The answer is that if you switch, you win 2/3 of the time, but if you stay with your original choice, you only win 1/3 of the time.

__________________________________________________


Say you pick door #1

If car is behind:
1 - switch and you lose
2 - switch and you win (because host would open #3)
3 - switch and you win (because host would open #2)

The key is that the host knows where the car is, and always opens a losing door.


If you stay with # 1, however:

If car is behind:
1- stay and you win
2 - stay and you lose
3 - stay and you lose

If you switch, you win 2 out of 3 times.
If you stay with door #1, you only win 1 out of 3 times.


So yes, you should switch.

Originally posted by TheBloop
This was discussed back in January ...
So yes, you should switch.

Thanks

Very pleased to have stimulated such debate. Sorry to those who yawned because it had been posed before.

This one did the rounds (no doubt not for the first time) when I was a maths undergrad at Oxford University - I didn't realise then that it was a real life game show. My peers and I debated it at length and struggled to agree, which suggests it is not easy, or at least not intuitive, although now it seems very clear to me that sticking gives you a 1/3 chance and switching 2/3, for reasons you explain quite lucidly. But then I've had 15 years to figure it out.

Another way to look at it is:

1. At the start, you have a 1/3 chance of getting it right first guess. This is easy to see.

2. Sticking strategy will work if you are right first time, not if you are wrong first time, so 1/3 chance of success.

3. Switching will work if you are wrong first time, not if you are right first time, so 2/3 chance.

Therefore, sticking gives 1/3 probability of win, switching 2/3. Put that way, it seems quite clear to me that this is the answer. But, again, I've had 15 years to figure it out. .

Originally posted by TheBloop
This was discussed back in January at:

http://www.redhotpawn.com/board/showthread.php?threadid=59662

The answer is that if you switch, you win 2/3 of the time, but if you stay with your original choice, you only win 1/3 of the time.

__________________________________________________


Say you pick door #1

If car is behind:
1 - switch and you ...[text shortened]... times.
If you stay with door #1, you only win 1 out of 3 times.


So yes, you should switch.

Im not getting it...

There are 3 doors, 33% chance you will pick the right one

you pick a door
the host opens another door to find it empty
that leaves 2 doors with a 50% chance that the prize is in either one
why does switching doors improve the odds?

Your saying that when the host open the empty door, that it improves the odds for the door you didnt pick, but it dosnt improve the odds of the one you did pick? why is that?

Ok there are two scenarios.

Lets say that there are three doors

Alpha
Bravo
Charlie

Scenario 1:
Lets say the prize is in Alpha
Lets say contestant picks Alpha
Host opens Bravo.
if you didnt switch you would get the prize
if you switch to Charlie, you lose,
So if you originally picked the door where the prize lay behind, you got a 50 50 % chance of winning.

Now Scenario 2:
Prize is in Bravo
contestant picks Alpha
Host opens Charlie
if you switch doors, you win
if you dont switch doors, you lose
50 % chance you will win.

Now to those who thinks that there can be another scenario, you wrong, if the prize is in door charlie, then it will still be scenario 2 (just different names of doors)

Now if the Host would pick random out of the two remaining doors, then thats when switching doors would have 2/3 chance of getting the correct door, however, the door the host picks is ALWAYS the empty door.

So no matter if you switch doors or not, Its a 50% chance of getting the prize

oh, and btw, on the "Ask Marilyn" site, she said 9 out of 10 readers were disagreeing with her, and most of them were from universities... hmmm I wonder why?

Actually I know exactly why

When the question was posed to her, it said that the host opened up the door, and the door was empty. Marilyn is thinking that the host picked the door on random, and it was empty. However, if there was the condition that the Host WILL ALWAYS PICK the goat. Then she was wrong, or using different terms.

We're getting killed by language here, not mathematics!

Let's not forget that probability is an exact science. Controversies arise when problems are not precisely stated. I dare say that true mathematicians will go to great pains to make sure the problem is precisely stated.

There are two DIFFERENT problems in play here

Originally posted by flyUnity
Now to those who thinks that there can be another scenario, you wrong, if the prize is in door charlie, then it will still be scenario 2 (just different names of doors)

Then you have a 2/3 chance of being in scenario 2 and a 1/3 chance of being in scenario 1.

For anyone wishing to add in "the answer is switch 66% chance of winning", that idea has already been confirmed, so please don't post it again. Nevertheless, I feel that my proposal, while not definitely confirming a 100% stick, should at least signify that the puzzle needs more debate over what the probabilities truly are, taking into account the mind of the gameshow host.

Debate that I really do not want to go further into, but think the discussion between myself and palynka covers the main ideas - read over it and decide for yourself whether you favour 66% switch, 100% stick or somewhere in between.

Originally posted by flyUnity
oh, and btw, on the "Ask Marilyn" site, she said 9 out of 10 readers were disagreeing with her, and most of them were from universities... hmmm I wonder why?

Actually I know exactly why

When the question was posed to her, it said that the host opened up the door, and the door was empty. Marilyn is thinking that the host picked the door on random, and ...[text shortened]... make sure the problem is precisely stated.

There are two DIFFERENT problems in play here

I think you're absolutely right that part of the confusion stems from imprecise language (see my earlier posts), but you have it backwards. Only if we assume that the host will always pick the goat will switching give a 2/3 probability of winning.

The best way I know to illustrate this (rather than give an abstract proof) is with the following, equivalent scenario: remove 3 cards from a deck, 2 kings and an ace. We're going to play a game where I lay the cards in a row, face down, and you pick one. You win if you end up picking the ace. Now, after you pick a card, I'm going to look at the two cards you didn't pick, and I will always then turn over a king. Note that I can do this regardless of whether the card you picked is a king or ace. Now, I give you the option of staying with your original card, or switching to the other card that I didn't turn over.

Find some cards and try the above several times, and it will soon make sense why switching is better.

Originally posted by flyUnity
...Your saying that when the host open the empty door, that it improves the odds for the door you didnt pick, but it dosnt improve the odds of the one you did pick? why is that?

No...the original problem states that after your original door selection, the host always opens a losing door on purpose. That's really the key here, the fact that the host knows where the good prize is, and always opens a losing door after your first selection.

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 the host had someone else enter the game AFTER he opened one of the losing doors, then THAT person's odds would be 50/50 to choose the right door... but that's a different question.

The keys here are: 1) you're being asked "should you switch?" after the host opens a losing door .... and (2) you know the host always opens a losing door after your original pick.


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.

Hope this helps!

And, FYI, here is a link to the web site of Marilyn herself, and she explains the problem as well as listing a bunch of letters she received to argue the issue...

http://www.marilynvossavant.com/articles/gameshow.html

Originally posted by TheBloop
And, FYI, here is a link to the web site of Marilyn herself, and she explains the problem as well as listing a bunch of letters she received to argue the issue...

http://www.marilynvossavant.com/articles/gameshow.html

Ok, I finally got it lol.

Such debate on this.

The key, though, lies in the fact that the host KNOWS where the prize is, and can always pick a losing door, regardless of whether your initial guess was wrong or right.

The original post was not abiguous. He took the long way home to lay out the rules, but he covered the bases. This is clearly the age old "Monty Hall problem". The math is absolute that if the player switches when given the option, his chances of winning are 2/3, vs. 1/3 if he does not switch doors. There isn't any debate. If you believe there is a debate, then you are failing to understand how this problem works (I know, I used to be in that camp, and argued vehemently...but was wrong).

EDIT- If you think that sticking with your original guess is a good plan, follow the below link. Do the game 100 times (or until you are sick of it) by either swithching every time or sticking every time. See for yourself what the results are.

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

EDIT 2- Just ran through it for fun. Results after 200 tries:
Switching: 57 wins out of 100 tries
Not Switching: 22 wins out of 100 tries.
This is really a small sample, but it is easy to see how the results are stacking up...

Originally posted by BLReid
The original post was not abiguous. He took the long way home to lay out the rules, but he covered the bases. This is clearly the age old "Monty Hall problem". The math is absolute that if the player switches when given the option, his chances of winning are 2/3, vs. 1/3 if he does not switch doors. There isn't any debate. If you believe there is a debate, th ...[text shortened]... ies.
This is really a small sample, but it is easy to see how the results are stacking up...

Read my last post. This idea has already been covered. It is not the subject of any debate.