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# A simple Monty Hall variant

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Posted · Report post

Seeing as how every 8th grader with a math counts trophy knows the optimal play to his game, ol' Monty decided to shake things up.

He added two more doors, for a total of 5 instead of 3, behind which 4 held a goat, and 1 held a brand spanking new electric car (hey, gotta keep up with the times, eh?).

In the first round, the player chooses two doors instead of one. Then Monty opens up one of the other three doors and reveals a goat. Then the player may choose one of his two original doors or switch to one of the unchosen doors. Monty then releases one more goat (gotta give the first goat some company or else it'll eat Monty's brand new trousers), and the player is then given a final choice of the remaining three doors.

1) The second door Monty opens is random?

2) He is purposely playing against you, trying to keep that new car for himself?

And 3) If you weren't sure, how much would you have to suspect dear old Monty's honorable intentions to make either play?

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Call the doors A, B, C, D, E. I choose door A and B first.The prior probability of the goat being behind A or B is 2/5. Monty presumably chooses to show us a goat the first time - that's the way it works in the original problem - so we gain no new information about A or B. Let's suppose Monty chooses door E. There's still a 3/5 chance the goat's behind C or D.

Now it gets fun.

Let's assume I switch to door C. My odds of victory are currently 3/10 given no more info.

Then, Monty randomly chooses a door with a goat of the three remaining doors. If he chooses door A, there's a 3/10 chance of the car being behind C, 3/10 for D, and 2/5 for B. If he chooses door B, there's a similar result. If he chooses door D, there is a 3/5 chance of the car being behind door C, a 1/5 chance of the car being behind door A, and a 1/5 chance of the car being behind door B.

Let's suppose now that I choose door A instead of switching to C. My odds of victory are currently 2/10 given no more info.

Then, Monty chooses at random from the three remaining doors. If he chooses door B, there's a 2/5 chance of the car being behind door A and a 3/5 chance of it being behind doors C or D. If he chooses door C, there's a 3/5 chance of the car being behind door D and a 2/5 chance of it being behind door A or B. If he chooses door D, similar logic.

Looking at this, if I switch to C the first time, there are two cases where I have a 2/5 chance, and one case where it gives me a 3/5 chance. On the other hand, if I stick to A,, there are two cases where I get a 3/5 chance, and one case where I get a 2/5 chance.

So my play if Monty's randomly picking doors with goats behind them is:

Choose a door.

Don't switch.

If he opens the other door of my original pair, don't switch. If he opens one of the other two doors, switch.

If he's playing nonrandomly, I think this might degenerate into a game of "Is the poison in the glass of wine in front of me?"

Let's suppose I choose A + B, and don't switch, and the car's behind C.

If he opens door D, I'll figure that there's a 3/5 chance of the car being behind C (all other things being equal), so of course I'll switch.

But I know he knows this. He would never actually choose door D if the car were behind door C, because then I would switch to C. So the fact he chooses door D is evidence that I should not choose door C. In fact, if he assumes I'm gonna switch, he's given me hard evidence that the car is behind A or B!

But he knows I know he knows this...

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Posted · Report post

Since everything starts equal, it doesn't matter which two doors are picked first or which door is shown to have a goat. Of doors A, B, C, D, and E, assume doors A and B were picked and door E revealed a goat. From here (if I understand the problem correctly) you can pick any one of the remaining 4 doors. Currently, doors A and B each have 1/5 chance of having the car and doors C and D each have 3/10 chance.

There are 4 end results (well, 12, but they mirror the first 4) expanded to common denominators:

A is picked, B has a goat-

A = 4/20

C = 8/20

D = 8/20

A is picked, D has a goat-

A = 4/20

B = 7/20

C = 9/20

C is picked, B has a goat-

A = 6/20

C = 6/20

D = 8/20

C is picked, D has a goat-

A = 7/20

B = 7/20

C = 6/20

At best:

Repicking one of the original 2 doors gives you 9/20 probability of finding the car.

"Switching" to one of the previously unpicked doors gives you 8/20 probability of finding the car.

At worst:

Repicking one of the original 2 doors gives you a 50/50 shot at having 8/20 probability of finding the car.

"Switching" to one of the previously unpicked doors gives you a 50/50 shot at having 7/20 probability of finding the car.

If dear old Monty is choosing goats at random, the best plan would be to keep one of the two doors that you originally picked.

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Posted · Report post

On track with the play, but no one's gotten the complete correct reasoning yet. There's a consideration that makes the fractions somewhat uglier .

As for the WIFOM, it is a 30 minute show (well, 23 plus commercials), and if you don't make a play, you have 0 chance of getting the car. That said, okay, this problem may not be as simple as I originally thought.

I have realized that Monty's best (and sanest) play is probably to implement a probabilistic strategy, i.e. he pulls up a random number generator on his trusty smartphone and generates a number b/w 1 and 10, and if it's equal to or above X, he does one thing, and if it's less than he does another...he chooses X such that his expectation value or expected return is the same no matter which door you pick in the last round.

However, for the sake of there being a solution to 2) and 3) (which will also force you to calculate the consideration mentioned above), let's say that Monty doesn't know exactly which door the car is in either, i.e. he is told which door to open for the first and he picks a door he wants to open for the second over his bluetooth and the producers tell him to do it or to choose another door.

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Posted (edited) · Report post

Seeing as how every 8th grader with a math counts trophy knows the optimal play to his game, ol' Monty decided to shake things up.

He added two more doors, for a total of 5 instead of 3, behind which 4 held a goat, and 1 held a brand spanking new electric car (hey, gotta keep up with the times, eh?).

In the first round, the player chooses two doors instead of one. Then Monty opens up one of the other three doors and reveals a goat. Then the player may choose one of his two original doors or switch to one of the unchosen doors. Monty then releases one more goat (gotta give the first goat some company or else it'll eat Monty's brand new trousers), and the player is then given a final choice of the remaining three doors.

1) The second door Monty opens is random?

2) He is purposely playing against you, trying to keep that new car for himself?

And 3) If you weren't sure, how much would you have to suspect dear old Monty's honorable intentions to make either play?

Here's a play

The following play applies equally well to question 1) and question 2). I assume that when Monty opens a door, he will not open a door that is currently chosen by the player, and that he would not open a door that contains a car.

The algorithm for the player is as follows

1) After Monty opens the first door, randomly (with equal probability) choose 1 door among the remaining 4 doors.

2) After Monty opens the second door, we would have 3 doors left. Let's label those doors (R1, R2, C), where R1 and R2 are two of the remaining doors, and C is the currently chosen door. For the final choice, randomly pick between R1 and R2.

The chance of getting a car is 3/8, regardless of whatever strategy Monty uses.

Edited by bushindo
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Posted · Report post

Seeing as how every 8th grader with a math counts trophy knows the optimal play to his game, ol' Monty decided to shake things up.

He added two more doors, for a total of 5 instead of 3, behind which 4 held a goat, and 1 held a brand spanking new electric car (hey, gotta keep up with the times, eh?).

In the first round, the player chooses two doors instead of one. Then Monty opens up one of the other three doors and reveals a goat. Then the player may choose one of his two original doors or switch to one of the unchosen doors. Monty then releases one more goat (gotta give the first goat some company or else it'll eat Monty's brand new trousers), and the player is then given a final choice of the remaining three doors.

1) The second door Monty opens is random?

2) He is purposely playing against you, trying to keep that new car for himself?

And 3) If you weren't sure, how much would you have to suspect dear old Monty's honorable intentions to make either play?

Seeing as how every 8th grader with a math counts trophy knows the optimal play to his game, ol' Monty decided to shake things up.

He added two more doors, for a total of 5 instead of 3, behind which 4 held a goat, and 1 held a brand spanking new electric car (hey, gotta keep up with the times, eh?).

In the first round, the player chooses two doors instead of one. Then Monty opens up one of the other three doors and reveals a goat. Then the player may choose one of his two original doors or switch to one of the unchosen doors. Monty then releases one more goat (gotta give the first goat some company or else it'll eat Monty's brand new trousers), and the player is then given a final choice of the remaining three doors.

1) The second door Monty opens is random?

2) He is purposely playing against you, trying to keep that new car for himself?

And 3) If you weren't sure, how much would you have to suspect dear old Monty's honorable intentions to make either play?

Case A: Player selects 2 doors, one of which has a car; P[A] = 1 - (4 choose 2)/ (5 choose 2) = 2/5

Case B: Player's initial selection does not include car: 1 - P[A] = 3/5

Monty opens goat door.

In case A, one of the player's doors has a car, and both of Monty's remaining doors have goats.

In case B, both of player's doors have goats, and one of Monty's remaining doors have a car.

Case A.Stay: We are in case A, and player stays i.e. picks one of original 2 selected doors; With 1/2 conditional probability, will select car, conditional on being in case A. The net probability for pointing at right door here is 2/5*1/2 = 1/5

A.Stay.A = Pointing at right door (P = 1/5)

A.Stay.B = Pointing at wrong door (P = 1/5)

Case A.Switch: We are in case A, and player switches to one of Monty's doors; Player switches to goat with conditional probability 1, car with conditional probability 0. Net probability for pointing at right door here is 2/5*0 = 0;

A.Switch.A = Pointing at right door (P = 0)

A.Switch.B = Pointing at wrong door (P = 2/5)

Case B.Stay: We are in case B, and player stays; Player pointing at goat with condtional probability 1. Net probability for pointing at right door is 3/5*0 = 0.

B.Stay.A = Pointing at right door (P = 0);

B.Stay.B = Pointing at wrong door (P = 3/5)

Case B.Switch: We are in case B and player switches. Player points at right door with conditional probability 1/2; Net probability is 3/5*1/2 = 3/10

B.Switch.A = Pointing at right door (P = 3/10)

B.Switch.B = Pointing at wrong door (P = 3/10)

Monty is not allowed to open the door the player is pointing at, or the car door.

If player is pointing at the car, Monty may open any door the player is not pointing at.

If player is pointing at a goat, Monty, may not open the player's door, nor the car's door.

At this point, if player move was stay, player is pointing at right door with Prob 1/5

If player move was switch, then, player is pointing at right door with prob 3/10

If player is pointing at right door, Monty may open any of the 3 other doors. (Depending on strategy choice, we can make the probability of this case 1/5 or 3/10).

If player is pointing at wrong door, Monty may only open 2 of the other doors (since one of the other 3 is as car). (Depending on strategy choice, we can make the probability of this case 4/5 or 7/10).

Most likely to win if player is wrong at this point, and Monty gets rid of another goat.

If wrong, and Monty get's rid of another goat, then switch to one of 2 other doors, one of which is the car.

To maximize chance of being wrong at this point, should stay at first opportunity(P[wrong] in that case is 4/5), then switch at 2nd choice (P[win|was wrong] = 1/2).

With this strategy the probability of winning is 4/5*1/2 = 2/5.

To summarize, strategy is pick 2 of the 5 doors at random. After Monty opens a goat door, choose randomly of the 2 doors you initially picked.

After Monty opens another goat door, choose randomly of the other two that remain (not the one you were just pointing at). You will win 2/5 times regardless of how Monty picks doors and whatever his motives are.

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Posted · Report post

It's not equally probable that Monty picks all of the remaining doors in the second round as the goat door, even assuming he's attempting to be fair and impartial. If he's being fair, he picks one of the remaining doors which has a goat with equal opportunity to open. You have received information from the first round which should be taken into account.

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Posted (edited) · Report post

It's not equally probable that Monty picks all of the remaining doors in the second round as the goat door, even assuming he's attempting to be fair and impartial. If he's being fair, he picks one of the remaining doors which has a goat with equal opportunity to open. You have received information from the first round which should be taken into account.

I'm not following.

I think that is only true if we assume that Monty is making certain assumptions about our strategy, and incorporate that into our strategy.

My strategy does not make that assumption, and I think provides an accurate winning chance.

Does your strategy result in a winning chance greater than 2/5?

I think I do incorporate information from each round, whenever we are in scenarios where Monty's choice modifies probabilities of remaining doors in a meaningful way.

If my selection of 2 doors initially did contain the car, then Monty's choice in first round does not impact further probability calculations.

However, if my selection of 2 doors did not initially contain the car, then Monty's choice in first round increases the chance of one of the two doors in Monty's initial group of 2 of having a car. Previously, given my knowledge, each of those 3 doors had prob 1/5. Now that Monty removed a goat, each of the remaining two doors has prob 3/10. This is info from Monty which I gladly take.

Next I pick one door of the remaining four, and Monty removes another goat from the 3 I didn't pick.

If the car is not behind the door I picked, then Monty's choice communicates useful info, otherwise it does not.

There is no way of knowing whether it does or not at this point, but I'm working around that fact by doing an exhaustive analysis of all cases.

The strategy is to make choices which make it more likely that Monty communicates useful info at each step, and then to act on what door (or set of doors) is most likely to have the car at the final step.

Quote:

"It's not equally probable that Monty picks all of the remaining doors in the second round as the goat door, even assuming he's attempting to be fair and impartial."

Actually, I don't think my strategy even assumes that he selects equiprobably.

It only makes assumptions about what he is allowed to select, and my strategy proceeds forward with whatever selection he makes, regardless of whether he has tendencies to select certain doors more likely due to his belief in my strategy.

For example, if he always chose the lowest numbered door of those he were allowed to select, then my strategy wouldn't change.

etc.etc.

Perhaps I would agree with your answer if I saw it, but right now, I think I'm sticking with mine.

Edited by mmiguel
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Case A: Player selects 2 doors, one of which has a car; P[A] = 1 - (4 choose 2)/ (5 choose 2) = 2/5

Case B: Player's initial selection does not include car: 1 - P[A] = 3/5

Monty opens goat door.

In case A, one of the player's doors has a car, and both of Monty's remaining doors have goats.

In case B, both of player's doors have goats, and one of Monty's remaining doors have a car.

Case A.Stay: We are in case A, and player stays i.e. picks one of original 2 selected doors; With 1/2 conditional probability, will select car, conditional on being in case A. The net probability for pointing at right door here is 2/5*1/2 = 1/5

A.Stay.A = Pointing at right door (P = 1/5)

A.Stay.B = Pointing at wrong door (P = 1/5)

Case A.Switch: We are in case A, and player switches to one of Monty's doors; Player switches to goat with conditional probability 1, car with conditional probability 0. Net probability for pointing at right door here is 2/5*0 = 0;

A.Switch.A = Pointing at right door (P = 0)

A.Switch.B = Pointing at wrong door (P = 2/5)

Case B.Stay: We are in case B, and player stays; Player pointing at goat with condtional probability 1. Net probability for pointing at right door is 3/5*0 = 0.

B.Stay.A = Pointing at right door (P = 0);

B.Stay.B = Pointing at wrong door (P = 3/5)

Case B.Switch: We are in case B and player switches. Player points at right door with conditional probability 1/2; Net probability is 3/5*1/2 = 3/10

B.Switch.A = Pointing at right door (P = 3/10)

B.Switch.B = Pointing at wrong door (P = 3/10)

Monty is not allowed to open the door the player is pointing at, or the car door.

If player is pointing at the car, Monty may open any door the player is not pointing at.

If player is pointing at a goat, Monty, may not open the player's door, nor the car's door.

At this point, if player move was stay, player is pointing at right door with Prob 1/5

If player move was switch, then, player is pointing at right door with prob 3/10

If player is pointing at right door, Monty may open any of the 3 other doors. (Depending on strategy choice, we can make the probability of this case 1/5 or 3/10).

If player is pointing at wrong door, Monty may only open 2 of the other doors (since one of the other 3 is as car). (Depending on strategy choice, we can make the probability of this case 4/5 or 7/10).

Most likely to win if player is wrong at this point, and Monty gets rid of another goat.

If wrong, and Monty get's rid of another goat, then switch to one of 2 other doors, one of which is the car.

To maximize chance of being wrong at this point, should stay at first opportunity(P[wrong] in that case is 4/5), then switch at 2nd choice (P[win|was wrong] = 1/2).

With this strategy the probability of winning is 4/5*1/2 = 2/5.

To summarize, strategy is pick 2 of the 5 doors at random. After Monty opens a goat door, choose randomly of the 2 doors you initially picked.

After Monty opens another goat door, choose randomly of the other two that remain (not the one you were just pointing at). You will win 2/5 times regardless of how Monty picks doors and whatever his motives are.

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Posted · Report post

Let's say Monty's playing fairly and you stay with one of your doors from the first round. If Monty picks one of the two unpicked doors (7/11 chance) there's a 3/7 chance the car's behind the other unpicked door, and a 2/7 chance it's behind your door or the other door from the first round. Say he opens the other door from the first round (4/11) then there's a 3/8 chance it's behind either of the unpicked doors and a 2/8 chance it's behind your door. So at best a 3/7 chance of winning, at worst 3/8.

This is the same if you switch in the second round except you have a 4/11 shot at the 3/7 chance of winning and a 7/11 shot at the 3/8, so it's still the best play to stay with one of your doors.

You would have the upper hand if you knew Monty always tries to go for the other picked door from the first round, not sure if there's any way Monty can intentionally make the play harder though.

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