# Monty Hall vs. Monty Fall

1. Jan 1, 2013

### Jiggy-Ninja

Standard Monty Hall Problem:
Contrary to what you may be thinking this is the easy one for me to understand. Once it was explained, the 2/3 odds make perfect sense.

The trouble comes with the related "Monty Fall" problem:
The probabilities quoted for this are 50% win chance whether you stick or switch, rather than the 2/3 win chance for switching with the original Monty Hall problem. I ran 10,000 simulations of this game with an Excel spreadsheet. The prize was put behind a random door, the contestant picked a random door with even probability, the host picked a door from the other two at random with even probability, then the contestant randomly decided to switch or not with even probability. Games were declared either Win, Lose, or Void (if the host picked the prize door). On games that were not declared Void, switching had a 50% win record.

As a check, I ran the exact same simulation with the host pick changed to reflect the original Monty Hall problem, and it resulted in the expected 2/3 chance of winning with switching.

How do I make sense of this? Why should the host's intentions matter when the mechanics of the situation (Contestant chooses door, host ends up revealing goat) are identical?

One thing I noticed is that all situation that end up Void in Monty Fall would end up as wins for switching in Monty Hall, as the host's pick would be constrained by the presence of the prize in the choose-able doors, which may lead to those situations being "double-counted" in the Monty Fall scenario as opposed to Monty Fall.

However, I still can't get past the fact that that explanation is too Deepak Chopra-ish. The power of human intention to affect the laws of probability, and nonsense like that. I still don't get why two situations with the same objective facts (Contestant chooses door, host reveals goat) can retroactively change probabilities if you know the host's intention.

Can anybody help explain that?

2. Jan 2, 2013

### Mentallic

That's exactly it. The Monty Fall scenario doesn't give better odds for switching because the host doesn't know where it is. The reason for this is because all we can conclude from him opening a door at random is that he didn't land the 1/3 chance of revealing the car, so when there are two doors left, they are indeed both equally likely of having the car.

3. Dec 24, 2013

### puzzlenut

You are right! Whenever the switch is offered the contestant doubles his chances of winning by switching. The difference is that if Monty makes a mistake the switch is not offered. This will happen 1/3 of the time.

A fallible Monty does reduce the value of the game. Normally it is worth 2/3 times the value of the prize, but if the switch is offered only 2/3 of the time because of Monty's error it is worth only 4/9. For the player who never switches it is worth 1/3 whether Monty is fallible or not. The player who switches whenever he can still comes out ahead.

I think the "Monty Fall" theorists merely have discovered a new way of promulgating the confusion about this game.

Last edited: Dec 24, 2013
4. Dec 24, 2013

### D H

Staff Emeritus
There's yet another variant of the Monty Hall problem, the Monty Crawl problem. Like the original Monty Hall problem, Monty has to open a door, but if he has a choice of doors he always opens the lower numbered door. (Monty is tired, so he kinda crawls to the door he's going to open rather than walking crisply to it. Hence the name "Monty Crawl".) With this variant, it doesn't matter if you stay or switch if Monty shows the lower numbered door of the two unopened doors. Your odds of winning are 50/50 either way. But you had better switch if Monty shows the higher numbered door. Now the odds have changed to zero if you stay, 100% if you switch.

The mechanics aren't identical. In the original Monty Hall problem the host has little choice. He has to show you a door per the rules of the game, and he's not going to show the door that hides the car. The only time Monty has any volition is on the one out of three times that the guest has chosen the winning door.

That Monty is required to open a door gives new information, new information that the players can use to their advantage. In the Monty Fall problem, that Monty happened to show a door that hides a goat is also new information, but there is no advantage in switching with this information. In the Monty Crawl version of the problem, the quality of the information depends on which door the tired Monty opens.

5. Dec 25, 2013

### puzzlenut

You report that your simulation gave a 50% win rate for the Monty Fall scenario, but there
are some other values very close to 50% that might equally well be supported by your results.

You do not report having done the simulation with a "never switch" strategy. This is the
best way to decide the question. Do it now and if your result is close to my predicted
result of 1/3 you are in an excellent position to drive the Monty Fall holdouts back into
the woodwork.

Their arguments generally are based on something like the following:

A door you did not choose is opened randomly and discloses a goat? What is the chance that

A = initial choice is correct
B = initial choice is incorrect
G = The opened door shows a goat

Pr(A|G) = Pr(G|A)Pr(A)/Pr(G|AvB)

Pr(G|A)Pr(A) = (1)(1/3) = 1/3

Pr(G|AvB) = 2/3, therefore Pr(A|G) = 1/2 and there is no point in switching.

I do not believe that this conclusion is reliable but at present cannot say why. I can only
suggest alternatives.

In some cases it is possible to do a complete simulation by cycling through all possible

outcomes.

I. Choose A, no-switch strategy

Pr is 1/3 no matter what Monty does.

II. Choose A, Monty never opens prize door, Switch strategy

1. Car is behind A, Monty opens B or C, switch - lose
2. Car is behind B, Monty opens C, switch to B - win
3. Car is behind C, Monty opens B, switch to C - win

Contestant wins two times out of three

III. Choose A, Monty may open prize door at random, Switch strategy

1. Car is behind A, Monty opens B or C, switch - lose
2. Car is behind B, Monty opens B, no switch possible - lose
3. Car is behind B, Monty opens C, switch to B - win
4. Car is behind C, Monty opens B, switch to C - win
5. Car is behind C, Monty opens C, no switch possible - lose

Here the score is 2/5, which is better than 1/3. It still pays to switch.

II.1 and III.1 are both counted as single cases because it is immaterial which door is
opened.

6. Dec 25, 2013

### mathman

Your description of III is correct, but your conclusion is misleading. The question is whether or not to switch. Since 2 and 5 are ruled out (no choice allowed), that leaves 1, 3, or 4, where switching is better 2/3 of the time.

7. Dec 25, 2013

### Staff: Mentor

It is useful to count III.1 as two separate cases, otherwise you have cases with different probability (as all cases "Car is behind A" added have to have 1/3 total probability).

In III, the question is "what should you do once Monty opens an empty door?" - and then switching does not change the probability as you know (2) and (5) did not happen.

In total, in III:
1/3 you picked the right door and Monty opens an empty door (switching bad)
1/3 you picked a wrong door and Monty opens an empty door (switching good)
1/3 you picked a wrong door and Monty opens the right door (switching not possible, or trivial)

With Monty Fall, switching is pointless.

8. Dec 26, 2013

### D H

Staff Emeritus
puzzlenut, you did two things wrong in your scenario III. One is that your events 1 to 5 are not equiprobable, but you treated them as if they were. The other is that scenario III is not Monty Fall. In the Monty Fall scenario, Monty slips on a banana peel and accidentally opens a door behind which there happens to be a goat. Your events 2 and 5 are not a part of the sample space.

mathman, you got the sample space correct, but you repeated puzzle nut's first error. Your three events are not equiprobable.

Sans the qualifier that the door Monty opens happens to hide a goat, there are six equiprobable events:
1. Car is behind door A, Monty accidentally opens door B.
2. Car is behind door A, Monty accidentally opens door C.
3. [strike]Car is behind door B, Monty accidentally opens door B.[/strike]
4. Car is behind door B, Monty accidentally opens door C.
5. Car is behind door C, Monty accidentally opens door B.
6. [strike]Car is behind door C, Monty accidentally opens door C.[/strike]
I marked the cases excluded from the Monty Fall sample space with a strike-through. The remaining four possible events are still equiprobable, but with a posterior probability of 1/4 rather than the a priori probability of 1/6. Staying wins on events 1 and 2, loses on the other two. With switching, it's the other way around. Both strategies become a 50/50 proposition in the Monty Fall scenario.

Last edited: Dec 26, 2013
9. Dec 26, 2013

### puzzlenut

Today I think that nearly everyone will agree that the contestant who picks one of the three
doors and sticks with his decision has a 1/3 probability of being right, but it was a
struggle getting there. The principal holdouts are some of the Monty Fall theorists who think
that something that Monty does other than revealing the winning door can affect this probability.

Most now also are willing to accept that if Monty reveals a losing door not chosen by the
contestant and the contestant switches to the remaining door he has a 2/3 probability of
being right. This is the standard problem and the standard solution.

The problem now before us is what is the probability of the contestant winning if he always
accepts a switch, if offered, but Monty is as ignorant as he is and is just as likely to
select a winning door as a losing one. A simulation has suggested 50% and I have given
arguments suggesting 4/9 and 2/5. I now think that these previous estimates are incorrect.
I would delete these previous solutions if they had not been commented on. The logic is
completely different, so any objections to the previous solutions do not necessarily apply here.

Here is my current thinking on the subject. I shall speak as though both the contestant and Monty
are trying to guess which door hides the prize.

In the ignorant Monty scenario The contestant will win only if both he and Monty choose incorrectly.
If he chooses correctly he will lose because he accepts the switch and if Monty chooses correctly
he will lose because he chose incorrectly and cannot switch. If both choose incorrectly the switch
will give him the winning door.

For each of the three doors he can choose, Monty can choose two, for a total of 6 ways he
and Monty can choose two doors. There are two ways that he and Monty both can choose
incorrect doors, so the probability that he and Monty will both choose incorrectly is 1/3.
The contestant who always switches when he has a chance has a probability of winning of 1/3,
the same as that of the player who never switches. This contradicts the Monty Fall doctrine
that either player has a probability of winning of 1/2.

Despite the fact that there appears to be no advantage in switching, a player should
switch anyway. Here is why.

While neither strategy has an advantage over the other if Monty's error rate is exactly 1/3, in practice
it is only an average. In any particular series of games he will open the wrong door either more or less
than 1/3 of the time. If he opens the wrong door less than 1/3 of the time it will benefit the switch player
by giving him more opportunities to switch. If he opens the wrong door more than 1/3 of the time it will not
hurt the switch player because his expected return is still 1/3. The no-switch player is unaffected in either case.

A player should always switch on the chance that Monty will make fewer than the average
number of mistakes.

Last edited: Dec 26, 2013
10. Dec 26, 2013

### D H

Staff Emeritus
puzzlenut, you are still ignoring the fact that Monty did not open the winning door. You have the wrong sample space for the Monty Fall problem.

11. Dec 26, 2013

### mathman

Attn DH: You're right. I stand corrected.

12. Dec 26, 2013

### puzzlenut

Correction:

If Monty opens the prize door with greater than the expected frequency it will hurt the
switch player. Here are three variations of Monty's behavior:

I. Monty never opens a prize door:

Switch player wins 2/3 of the time
non-switch player wins 1/3 of the time

II. Monty opens a door at random:

Switch player wins 1/3 of the time
non-switch player wins 1/3 of the time

III. Monty always opens a prize door when it has not been selected by the contestant:

Switch player always loses.
non-switch player wins 1/3 of the time.

13. Dec 27, 2013

### Staff: Mentor

Right.

14. Dec 27, 2013

### D H

Staff Emeritus
I disagree, mfb. For one thing, not one of puzzle nut's scenarios is on-topic. They don't address Monty Fall, which is the main topic of this thread. Secondly, scenario I is arguably incorrect. The answer subtly depends on how Monty picks a door. There's no advantage to switching versus not switching if Monty doesn't know where the car is and just happens to open a door that hides a goat. The probability is 1/2 rather than 1/3 in this case. (This is the Monty Fall scenario.) There's also no advantage to switching versus not switching if Monty opens the lower numbered unopened door in the case that Monty does know where the car is and always open the lower numbered door that hides a goat.

15. Dec 27, 2013

### PeroK

For what it's worth, I agree with mfb. Puzzlenut has cracked this problem and shown an understanding of the variations. It's clear that in scenario I MH must know where the car is if he never reveals it.

16. Dec 27, 2013

### D H

Staff Emeritus
Once again, I disagree. The title of this thread is "Monty Hall vs. Monty Fall". The probabilities in the Monty Fall problem are 50/50, whether one stays or switches. puzzle nut has never acknowledged this.

More to the point, nobody has answered the OP's key question. So let's get back to that question.

The true probabilities don't change. Say the car was put behind door #3. The probability that it is behind doors #1, 2, and 3 are not 1/3, 1/3, and 1/3. The true probabilities are 0, 0, and 1, respectively. Let's Make a Deal is a question about the everyday world, not the quantum world (where the car can be behind each door, simultaneously). The car has been placed behind one specific door, door #3 in this case, with a probability of 1.

There's one minor problem here: The people selected to play the game don't know that the car is hiding behind door #3 this time. They have to guess. The act of assigning equal probabilities of 1/3 to each door is done in the face of unknown probabilities, based on the principle of indifference (Bayesianist POV) or based on thousands of observations of where Monty's staff hides the car (frequentist POV). With either interpretation, this initial equiprobable assignment is but a rough guess at the true probabilities.

Both the Bayesianist and frequentist points of view provide tools to update that initial estimate based on new information. It's our interpretations of that new information that result in different estimates of the probabilities, not Monty's intentions. Note well: No matter how we interpret that new evidence, our updated probabilities are still estimates. We don't know the true probabilities until Monty opens the door that hides the car (or opens all the doors that hide goats). We can only learn the true probabilities after the fact.

17. Dec 28, 2013

### pwsnafu

This is not Monty Fall.

This is not Monty Fall.

This is not Monty Fall.

In terms of general player strategies:
In order for the Nash equilibrium to coincide with switch strategy the player must always be allowed to switch, and the host must always reveal a door. That's the criteria. If the host has the option to not offer a switch, we have "Monty from Hell" (switch always loses). If the host doesn't need to reveal, then not switching is optimal at 1/3.

18. Dec 28, 2013

### PeroK

Consider an alternative problem. There are three boxes, only one has a prize and you're going to be given one of them.

In the first case, a box is chosen at random and you know that your chance of having the prize in 1/3.

In the second case, your friend chooses the box (and he knows which one has the prize), so your chance of having the prize is 1 (as long as you can trust your friend!).

So, to an unwitting observer, these are the same objective facts. But, they are not the same event if you know the intentions of the participants.

The Monty Hall problem is not that different. He knows where the car is. So, by deliberately showing you where it's not, he is helping you.

Consider, finally, a variation on Monty Hall. This time there is 1 car but 100 doors.

You choose a door. Monty then opens 98 doors, none of which has the car. Well, he's practically telling you where the car is! Okay, there's a 1/100 chance that you chose the "right" door. But, apart from that 1/100, the car must be the behind the remaining door. And, if you switch you have a 99/100 chance of getting he car.

If Monty repeatedly fell against 98 doors, then almost always (98/100) he would reveal the car and spoil the game. It would be a fluke if he didn't. But, in this case, when he accidently opens 98 doors without revealing the car, there's then an even chance that the car is behind the remaining 2.

19. Dec 28, 2013

### Staff: Mentor

"Monty never opens a prize door" includes that Monty has to know where the prize is (otherwise he would open the prize door with some probability >0).

Right, that is a valid option in scenario I, but I think puzzlenut meant the regular Monty Hall scenario here.

I disagree, D H explained the reasons in post 16.

Funny situation:
:D

20. Oct 16, 2016