You are on a gameshow, and you can pick between three doors. In one of them there is a car, and in the two other doors there is a goat.

You pick one door, and the gamehost opens another door where there is a goat inside (the gameshow host knowns where the car is, so that he can open a door without the prize) And you get the option if you want to change your door (after the host has opened an "empty" door), or keep the original you choosed in the first place...

It should be obvious that you should change door:
because if you dont have the option of changing door after the host opens an "empty" door, You have 1/3 = 33.3333...% chance of winning the prize.
But when the host "eliminates" one door, and you get the option of picking another door out of two doors. You have now two options and, one is containing a prize, so your chances now should be 1/2 = 50 % ?

But I've heard that the answer to the puzzle is 2/3 = 66.6666... % chance of winning a car after the host gives you the option of changing, and you change.

One way to think about this is lets say you can pick from 1000 doors, and the host opens 998 doors. Then it is two doors left. Then it becomes alot clearer that you should change door.

But can someone prove/explain why you have 2/3 of winning a car with three doors and not 1/2?

Can you explain it this way?
When you pick one out of the three doors you have 1/3 = 33.333...% equity
And when the gameshow host "eliminates" one door you get 1/3 = 33.333...% more equity when you change?
therefore you got 2/3 = 66.666...% total equity?

It's not true that the probability for the option of switching is 1/2.

Assuming the gameshow host will always show another door with a goat inside, the probability for the option of switching is 2/3. You picked the right door on your initial choice with probability of 1/3.

Suppose there are n>=3 doors in general and you choose a door and the gameshow host shows (n-2) of the other (n-1) doors and each of them have a goat inside. Clearly, if n is very large, you should switch; in any case, the initial choice had probability of 1/n of being correct and it should be clear that the probability for the option of switching is (n-1)/n since the probability that you were correct initially did not change.

I think the easiest way to see it is like this:

Imagine 3 trials in which you must switch and 3 trials in which you cannot switch. Now imagine the goats are behind A and B with the prize behind C. On each of the three trial types, you must start with a different door.

We get the following results:

Switch Trials
Pick A - Host Opens B - Switch to C - Result = Win
Pick B - Host Opens A - Switch to C - Result = Win
Pick C - Host Opens A or B - Switch to A or B - Result = Lose

No Switch Trials
Pick A - Host Opens B - Stay with A - Result = Lose
Pick B - Host Opens A - Stay with B - Result = Lose
Pick C - Host Opens A or B - Stay with C - Result = Win

Notice that on the switch trials we win 2/3 times. In the no switch trials we win 1/3 times.

This video also does an excellent job of explaining it: http://www.youtube.com/watch?v=mhlc7peGlGg

You are inconsistent with yourself. Why would you switch if you think your chances are 50%? That would mean that either remaining door has the same probability.

The first quote above is correct. Your chances if you can't switch are 1/3, so obviously that is what they are if you don't switch. Therefore the chances if you do switch must be 2/3 since these must sum to 1.

This is why these are worth doing again and again. Different people often either ask a slightly different question, or in this case, make a slightly unique kind of mistake.

I knew the answer to this puzzle, but I couldnt clearly see why the answer was 2/3 of winning the prize and not 1/2.

I didn't mean that both doors had the same probability... I meant that if you stayed with your original door (didn't switch) your chances of hitting the prize is always 1/3. But my logic said that if you changed you had 1/2 chance of hitting the prize and not 2/3 which it is.


But I now see it more clearly that the chances for picking the prize is 2/3 if you switch... Thanks.

I think the easiest way to understand the 2/3 chance is the following:

We all agree on having a 1/3 chance for your initial pick and now the remaining group of two doors together has a chance of 2/3 of containing the prize and since one of those two is opened/removed you are now simply able to go with the whole 2/3-group.

The following is only a little more info to understand the logic behind it, it's not really necessary, so the above post should be sufficient:

The thing one has to keep in mind for that, is that picking a goat-door from the two remaining doors-group is always possible for our one prize-scenario and hence doesn't add information about anyone of the two door-groups as a whole (of course it adds info about the specific door opened).
This not adding info about the groups means it doesn't change the chances for one of the two groups.

If on the other hand there were two equal prizes and he always opens one if he can it would add information about the groups as a whole and then we'd always have a 100% pick.(Then there would be two cases A he opens one, B he can't open one. In both cases you gained knowledge about the remaining group as a whole A it was a 50%-group B it was a 100%-group)

Your thought contains an implicit reshuffling of goat and prize only that would make it 50%.

"Can you explain it this way?"
I would consider the explanation to be wrong**, so my answer would be no but the much more important thing to get out of this is:
Whenever an explanation leaves a gray area/doubts for yourself you should never be satisfied with giving this explanation. So good that you asked imo.

**(the numbers add up nicely by "chance" for this specific scenario)

Why I would call it wrong:

Since the numbers add up we might guess that something like adding the eliminated equity seems right. But it's not exactly the way you described it.

Consider the following scenario 1 prize 3 goats (1 goat-door opened) rest the same.

Now again the groups have 1/4 chance for our first pick 3/4 for the remaining group and what happens now when one goat-door gets opened?

(According to your explanation it would have to be 1/4+1/4 = 50%.)

the 3/4-group still has 3/4 but only two doors so the group-probability get's evenly spread among the remaining doors so it's (3/4) / 2 = 3/8 = 0.375 for each of those two.
25% +1/2*25% = 37,5%
The probability get's split up over the group, due to the same logic as in the explanation two posts above.

Not reading the above but when my mathematical explanation doesn't work for people I explain this to, I usually try telling them to think of this simple question.

Imagine it was the same question except there's one million doors. The host then tells you 999,998 which are wrong, do you switch now? (What are the chances of you picking the right one originally? Hint: 1/1000000)

Not correct logic (perhaps). Consider this variation. The host opens a door at random, and it turns out that there was no prize in the door shown. What's the chance your door has the prize?

The entire issue is whether the host opens the door "at random". Obviously if he does then you still have no advantage by switching. It's only when he uses his knowledge that you gain new information to change your chances.

You are correct that the logic presented by wull above is 100% wrong.

Here is how I finally convinced a PhD mathematician. It reflects the conversation I had with him.

MC – I don’t know yet where the car is (this is the twist - but be patient).

Now, contestant – please pick a door.

C – I pick door 1

MC – Ok- if I offered you the chance to switch to door 2, would you do so?

C – Don’t see why I should. I’ll stay with 1

MC – What if I offered you the chance to switch from door 1 to door 3

C – Same answer

MC – OK – what if I offered you the chance to switch to both doors 2 and 3. If the car is behind either, you win

C – You don’t know where the car is, right?

MC – Right

C – OK – in that case I would switch

MC – But wait, what if I told you that at least one of the two doors had a goat behind it.

C – I know that; there are 2 doors with goats, so if you offer me 2 doors, at least one has a goat

MC – So you would switch even knowing that at least one has a goat

C – Of course.

Assistant comes over and whispers something to MC - where the car is

MC – Ok, now what if I told you that door 2 had a goat. Would you still switch to doors 2 and 3

(This step makes the problem identical to the original posed – the fact that the MC knows which door has the car before the contestant chooses does not change anything since he will always open a door with a goat and then offer a switch)

C – Yes, why not. You did not tell me anything that changes the probabilities – only which of the 2 doors has a goat and we know that at least one of them does. What you offered me was a choice of 2 doors vs. my original choice of one door giving me a 2 out of 3 chance of winning. All the other info you gave me is extraneous.

Was that a typo?
Of course it's absolutely valid, when I answer to an OP then I expect that OP to be the given scenario and that I don't have to repeat everything from that OP and the host removing a door means of course removing a goat-door, since it's explicitly stated as the given scenario and like totally obviously not opening a random door.

If you/someone doesn't believe that I implied it just read my second post where I explicitly mention it.

What I can agree on is that the "always can and always will show a goat-door for the two-door-rest-group, not adding probability-information about the group as a whole"-isn't obvious to everyone and is perhaps better explicitly stated.

In other words it's better added to the first post to make it easier to understand, but anyone who was willing to read my 2nd post would have read it already.

It was not a typo. This statement is wrong:

If one of those doors is removed randomly, even if it happens to be a goat door, then the "group" is now worth only 1/3, exactly the same as the door already selected (and has no benefit for switching). It only retains it's 2/3 value if the host uses foreknowledge to *always* remove a door with 0% chance of containing the prize.

Even this isn't sufficient to guarantee that it retains its 2/3 value. Suppose that when the prize is behind your door (A) he always opens door B. He only opens door C when the prize is behind door B. In that case when he shows you that door B has a goat, A and C now each have a 50% probability of having the prize, even though the host is using his foreknowledge to always remove a door with a 0% chance of containing the prize. To make switching preferable, we also need to know that if the prize is behind our door A, then the host chooses one of the other 2 doors to open with equal probability. The intended correct solution to the problem depends on that condition which is almost never stated.

When this latter condition is not in effect, opening door B only removes the possibility that the prize is behind door B, which represents 1/3 of all cases. But when this latter condition is stated, opening door B not only removes the 1/3 of cases where the prize is behind door B, but it also removes 1/2 of the cases where the prize is behind door A. Those are the cases where the prize is in A and he opens door C. This is the critical step that leaves only half as many cases where the prize is behind door A as compared to being behind door C, making door C the 2:1 favorite.

Definitely not, for one goat-door being removed and only two remaining we do not only have 1/3 for both remaining doors. For this (and this random chosing is a totally different scenario btw) I'd guess it to be 1/2 and 1/2. In contrast to my earlier posts 50/50 here is more of a guess.

I won't try to explain it to you further, there are enough posts here doing so already. You said that merely the fact of removing a goat door from the two unpicked doors, gave you a 2/3 chance to win if you switch, because those two doors had 2/3 chance combined. That statement was wrong, and is still wrong. Given only those conditions, there is no advantage to switching.

There is a reason why I said "opened/removed"
Opening alone doesn't mean I can't take the door. If it's just opened I can still take it. The reason why I explicitly added /removed is because this is a 100% process where the host knows where the goat is and always shows us a goat-door which then is a 100% removal (since we aren't taking a goat-door obv.) due to knowledge as clearly stated in the op:

I agree that with only writing removed it might be hard for the reader to see all implications but I made an immediate follow-up post addressing this issue.
I can only agree on calling it wrong if we take my "opened/removed" as isolated from the removal-process of the op and as isolated from my immediate follow-up post which is too much isolation for my taste.

While math itself is 100% black or white language isn't.

You are misunderstanding my post, I think we both understand something else by what you wrote here:

You said "If one of those doors is removed randomly, even if it happens to be a goat door, then the "group" is now worth only 1/3, exactly the same as the door already selected (and has no benefit for switching)."
When we are keeping the removal process from the op which includes showing then this means two doors where there is definitely a prize behind one of them add up to only 66% price behind them which is obviously wrong.

I think you not only meant to change it to random removal but also to not showing then the 1/3 part is true.
I think sticking to the op-scenario would make it easier for us to understand each other.

Best regards,
Wull

I didn't notice this sentence earlier. Obviously the group can't be 1/3 and the door already selected be 1/3 because these have to sum to 1. In this case, say where the wind randomly blows open door B to reveal a goat, then A and C each have a probability of 1/2 of having the prize.

Yes, of course I know this but the comment was in the context of the post it was referring to, and meant that the chance was equal to the door already picked and not increased. It was confusing the way I wrote it.

Yeah that's what I meant, but I think spadebidder perhaps just means killing the door like removing without showing, with the chance of the price being behind the killed door as well, then 1/3 would be true.
But of course for the case we know the killed door was a goat-door it can't be 1/3 for both.

Cheers