Should you swap? Quiz show probability question?

Question

Ok, here's the question:

You're on a quiz show and there are 3 doors, A, B and C. Behind one of the doors is the prize, behind the other two are nothing. You make your choice, let assume A. The host then says, "the prize is not behind door C" and he gives you the option of swapping your choice.

Do you make the swap?

Please explain your answer with maths if possible but not essential.

Answer 1

I've heard adaptations of this puzzle before.  But I think when it has been presented on Yahoo! Answers, there is a piece of the puzzle missing.

You are given a free choice of three identical doors, and no clue as to which has the money behind it.  You pick the correct door one in three times.

One of the doors is eliminated, and the prize is now behind one of the two remaining doors.  (If you picked A, and they eliminated C for you, the prize is either in A or B).

At this point, in a completely fair game, your odds are now 50/50.

But there is no logical reason to swap here.  You have not been given any new information about doors A or B.  One of them has the prize behind it, and both have an equal chance of being the winner.  So I see no compelling reason to swap.

If someone can prove the need to swap, or tell me what I am missing about the question, I'd be really interested.

Answer 2

It's the Monty Hall Dilema.  The base assumption is that the door that is revealed must be known to not be the correct door.  That has to be the case because if it weren't, you could be shown the prize and there wouldn't be much of a game.  Switching doubles your odds, even if it doesn't seem intuitive.

Answer 3

If he said that it is not behind C, than that still means it could be behind A. I would stick with A, cuz the host is just trying to get me to switch and pick B, and B probably has nothing.

And for the math part. 3 doors means a 33% chance of getting the right answer. Eliminate door C and you now have a 50% chance of picking the correct door. Is that all you were looking for?

Answer 4

Absolutely you should swap.  When you make your initial decision you have a 1/3 chance of being right, and a 2/3 chance of being wrong.  The host knows where the prize is, and will always show you a door that doesn't have it, no matter if your choice is right or wrong.  The thing is that once the door is shown your initial probabilty of being correct DOES NOT CHANGE!  It's still only 1/3.  Which means that the probability of it being behind the other door is 2/3.  So you should always switch, as it doubles your chances of winning.  Here's a fun little applet that both explains this problem and lets you try it out as many times as you like.

Answer 5

my logic goes with ZCT on this one: once a door is eliminated, you can't include the eliminated door as part of the odds. it's no longer 2 in 3, it's now 1 in 2.

However, the best explanation I've ever seen is from wikipedia: what if the two unchosen doors were combined, and then offered as a switch choice?

Pick from A, B, or C, A is selected.

Choice offered is now: Keep A, or switch for B _and_ C?

Here, it is obviously better to switch, and you clearly get the 2 out of 3 odds as theorized.

Why is the answer to this question always switch? If you work out the probabilities, it's something like this:

1 in 3 chance of any door having the car

1 in 3 chance of the choice being the car

~Choose A, Car A

*Choose B, Car A

*Choose C, Car A

*Choose A, Car B

~Choose B, Car B

*Choose C, Car B

*Choose A, Car C

*Choose B, Car C

~Choose C, Car C.

There are 9 possible outcomes. If you don't switch, you can only win the car 3 out of 9 times (~3 correct choices and 6 incorrect, unswitched choices.) If you always switch, you will win the car 6 out of 9 times (*6 incorrect choices, and switching these will result in 6 correct choices, and 3 car losses.) This is the stated correct answer.

But, the problem with the above analysis is that the 3rd door is a red herring. If we exclude the third door from the math, we end up with a 50/50 choice. because of the way probability is calculated, 2/3 is the correct answer, but the in actuality, if you have a choice between two things, it's always 50/50, regardless of how many previous choices have been eliminated. (see wikipedia about how this problem relates to the Deal or No Deal gameshow.)

2 doors with opportunity to switch:

~Choose A, Car A

*Choose B, Car A

*Choose A, Car B

~Choose B, Car B

Sticking with the choice will yield the prize ~2 out of 4 times, switching will yield the car *2 out of 4 times.

One more thought: What if the two doors without the prize were combined as one door? If A has the car, essentially B and C would be the same choice (one or the other would be eliminated.)

~Choose A, Car A

*Choose B or C, Car A

*Choose A or C, Car B

~Choose B, Car B

*Choose A or B, Car C

~Choose C, Car C

6 possible choices. ~3 of them will yield the car, *3 of them will yield the car after switching, and thus be perfectly 50/50.

Answer 6

Yes you should swap.

Basically it is a probablilty calculation.

in the begining you have a 33% chance of being right.

By eliminating one door and not swapping you still keep your 33% chance.

If you change you raise your chances to 50%.

That's the theory anyway

Answer 7

I'd stick with A.

The fact he's told me the prize is not in C only means that my odds of winning with box A have shortened to evens from a 1 in 3 chance, but the same can be said for box B.  If I change to box B and I am wrong even after my first hunch that A was right, then I'll be left to deal with the fact I had the prize and lost it. If it's in box B and I stick with box A then I'll just know it wasn't meant to be. C'est la vie.

I'd be the same on Deal or No Deal, if I ever got down to the final box and was offered a swap I'd not swap, I'd not be able to deal with knowing I had the prize and gave it away.