# Not The Monty Hall Problem

This is what people who get the MontyHallProblem wrong think the MontyHallProblem is.

There are three doors: One of them has a prize. You pick one. Another contestant picks another one. The door picked by the other contestant is opened, and reveals a non-prize.

The answer to this problem, unlike the MontyHallProblem, is that it doesn't make any difference. The fact that a door opened at random didn't contain a prize increases the probability that the door you picked does contain a prize, from 1/3 to 1/2.

This is because, if you picked a prize, the probability that another door opened at random has a prize is zero, but if you didn't the probability that another door opened at random has a prize is 1/2.

In table form: you pick A; prize is in P; the other fellow picks F.

The probabilities given below are a-priori probabilities that the tabulated outcome will occur, given your random choice and the other contestant's random choice.

```         A      B       C      Pr
1       P      F       .      1/3 * 1/2       win
2       P      .       F      1/3 * 1/2       win
3       F      P       .      1/3 * 1/2       other contestant wins
4       .      P       F      1/3 * 1/2       lose
5       F      .       P      1/3 * 1/2       other contestant wins
6       .      F       P      1/3 * 1/2       lose
```
3 and 5 can be ruled out, as the other Contestant didn't in fact pick the prize, but he might have done.

That leaves equal probabilities of winning and losing if you don't switch.

Compare with MontyHallProblem: You pick A; prize is in P; host shows you H.

As before, the probabilities given are a-priori. Given that the prize is in A, the chance of you picking B and the host showing you C are 1 in 3.

```         A      B       C      Pr
1       P      H       .      1/3 * 1/2        win
2       P      .       H      1/3 * 1/2win
3       .      P       H      1/3lose
4       .      H       P      1/3lose
```
The probability of winning if you don't switch is now only 1/3, while if you do switch it's 2/3. Because the host knows where the prize is, there is no randomness in which door he picks when you don't pick the prize: he must pick the only other door which doesn't have the prize.

The two problems look the same, the only difference is in whether the door that is shown to you is picked at random or not.

That is why it is difficult.

The above is correct, but already covered in MontyHallSolution.

AndrewMcGuinness wrote: Because the host knows where the prize is, there is no randomness in which door he picks when you don't pick the prize: he must pick the only other door which doesn't have the prize. See, this is where the MontyHallProblem goes wrong. It doesn't specify why Monty opens that door, or why he chooses that particular one. Maybe he picked one totally randomly. Maybe he seriously doesn't know where the prize is.

I believe that's why people tend to get the "wrong" answer to the MontyHallProblem. It's not really a math or probability problem, it's more like a ReadingBetweenTheLines? problem. Once it's established that Monty knows a priori that the door he's about to open doesn't have the prize, your calculation as a contestant has to change. The problem as typically stated leaves this constraint vague or implied at best.

Exactly. If you're telling the problem as a "brain-teaser", you need to state that the Host is deliberately opening an empty box, but not draw too much attention to it so that you can "fool" your victim. If you just want to explain the maths, you need to be very explicit that it is a crucial point. If you don't make the point clear, there's no point going on about "poor mathematical ability" of someone who doesn't get it - the answer isn't wrong.

Being familiar with the problem, I didn't notice that the MontyHallProblem page leaves the question vague. It ought to be corrected at the top to make the point clear.

True, but you haven't appreciated the history. The gaffe could have been blown straight away in the way you've described, but it was far more interesting not to do that, but instead introduce the spoiler by way of a modified question, as done in MontyHallSolution.

I may be getting this wrong, but from the description: There are three doors: One of them has a prize. You pick one. Another contestant picks another one. The door picked by the other contestant is opened, and reveals a non-prize.

The case where the other player won is already eliminated, otherwise the other player has already won and there is nothing to choose. The simple fact is (1) you picked a door, (2) someone picked and opened another door, and it is empty, (3) now do you change your choice?

Does it matter whether that someone is Monty or another player? I don't think so - the fact is, that door is revealed to be empty. So the math is still the same: When you picked one door, the chance for the prize to be in the other two doors is 2/3 (in other words, the chance that you are wrong initially is 2/3). Opening one of them showing it empty does not change the fact that your change of being wrong initially is 2/3. The difference now is you can "flip" to having 2/3 chance of being right but picking the remaining door (which you cannot do when there are still 2 remaining doors). -- OliverChung

Yes, it does matter (assuming the other player doesn't know where the prize is). If you can't follow the explanation in this page, try the explanation in MontyHallSolution.

Yes, OliverChung, you hit the nail on the head: it does matter, but it's easy to assume it doesn't. It matters because when a door opened at random is empty, that gives you extra information about whether your first guess was likely to be correct, because if you guessed wrong, a random door would be more likely to have a prize than if you guessed right. OTOH, if a door known to be empty is opened, that doesn't add information about whether your first guess is correct, as you already knew there was at least one empty door. Therefore the BayesTheorem? explanation in MontyHallSolution works. -- AndrewMcGuinness

Here's another NotTheMontyHallProblem that may help explain why people make the wrong choice:

Suppose you pick one door out of three, knowing that there is one prize behind one of the doors.

Then the host (the "Monty Hall"-like guy) has free choice to open another door or do nothing. We can assume (or suspect) that he knows which door holds the prize.

If he knows that you selected a losing door, he'd just do nothing, letting you lose.

But if he knows that you selected the winning door (and that he loses if you win), then he'd be highly motivated to convince you to change your choice. So he opens a losing door and "gives you another chance to chose."

In this scenario, if he's offering to let you change doors do not do it - because he'd only show you a losing door if your first pick was a winner. In fact, his act of opening a losing door proves to you that he knows where the prize is.

So, knowing human nature, you'd want to hold onto your first choice.

What that scenario "gets wrong" from the MontyHallProblem is that while the host does know where the prize is, the host is not given any choice about revealing a non-prize door. That's the fact that breaks the "common everyday" kind of reasoning used in this NotTheMontyHallProblem.

The solution to the first problem on this page is wrong. The table analysis admits that cases 3 and 5 are eliminated. Doing so results in the remaining probabilities not adding up to one. Cases 4 and 6 have probabilities 1/3 * 1. This is because when the first contestant picks a non-prize, the second contestant *HAS NO CHOICE* but to pick the other non-prize. To not do so violates the conditions of the problem. Suggesting the second contestant can pick the prize is along the lines of adding another case where there is more or less than one prize.

After the first guess, the first contestant *KNOWS* that at least one door could be opened to reveal a non-prize. Whether that door is opened by a knowing host or an unknowing contestant is irrelevant, because doing so adds no new information.

The true NotTheMontyHallProblem would be the following:

Contestant 1 picks a door, it is opened to reveal a non-prize. The second contestant *then* picks a door from the remaining two. Now the second contestant has an even chance of winning. -- JohnVriezen (I now realize that the above is incorrect. -- JohnVriezen

The solution to the first problem is not wrong: I said the probabilities were as calculated in advance. They are correct. Once you have seen the second contestant has been unlucky, you then eliminate 3 and 5, and you are left with two equal probabilities.

In short, 1/3 of the time you will switch and win, 1/3 of the time you will switch and lose, and 1/3 of the time you won't get the chance to switch because the second contestant will have got the prize.

In other words, in the 2/3 of cases where the second contestant doesn't happen to get the prize, you have a 50/50 chance whether you switch or not.

The same applies if the second contestant is a host who doesn't actually know where the prize is, which the overly vague definition on the MontyHallProblem page doesn't rule out.

I stand corrected. -- JohnVriezen

A bit of explanation to explain the difference between the two cases:

In the case where the host (or whoever opens a door second) doesn'tknow where the prize is, there is a 1/3 chance that s/he will pick the door with the prize. If we label the three doors W, L1, and L2, the possible outcomes are:

``` Case You  2nd  Probability  Win if switch
A    W    L1     1/6           No
B    W    L2     1/6           No
C    L1   W      1/6           N/A
D    L1   L2     1/6           Yes
E    L2   W      1/6           N/A
F    L2   L1     1/6           Yes
```
Thus, 1/3 of the time you win if you switch, 1/3 of the time you lose if you switch, 1/3 of the time the other guy wins and it doesn't matter. Simple enough.

The error that people make, in assuming that this equivalent to the MontyHallProblem, is assuming that in the case where Monty knows where the winner is and avoids it, that this 1/3 is eliminated and is distributed equally among the other two choices, leaving 1/2 possibility of winning if you siwtch. Not so; any instance where Monty MIGHT have picked the winner is forced into Monty picking the other loser, leaving the winner behind the last door - in other words, the 1/3 probability that Monty picks the winner is COMPLETELY given to the case where the winner is the last one remaining, and you win if you switch. The table is as follows, changes from the above table are in bold.

``` Case You  2nd  Probability  Win if switch
A    W    L1     1/6           No
B    W    L2     1/6           No
C    L1   W      0             N/A
D    L1   L2     1/3           Yes
E    L2   W      0             N/A
F    L2   L1     1/3           Yes
```
Adding them up, gives 2/3 chance of winning if you switch, 1/3 chance if you don't.

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