Two-Player Monty

Here is an interesting variation on the Monty Hall problem. For now I will simply present it cold, without indicating the context of where I saw it. Feel free to leave your proposed solutions in the comments. Everything from vague intuitions to hard-core Bayesian analysis is welcome.

Adam and Barney are contestants on a game show hosted by Monty Hall. Each player knows that the other one exists. They are confronted with three identical doors. One of the doors conceals a car, while the other two conceal goats. Both players select one of the doors, but neither player knows which door was chosen by the other player. After making their selections, Monty now opens a door according to the following rule: If Adam and Barney have selected the same door then Monty opens one of the remaining two doors, always opening one he knows to conceal a goat. (If he has more than one door to choose from, he chooses randomly from his available options). If Adam and Barney have selected different doors, then Monty opens the one remaining door, even if by doing so he reveals the car. After Monty opens a door, both players are given the chance to switch. If both players land on the door with the prize, then they both recieve the car.

Place yourself in the role of Adam. You have chosen door one. You now see Monty open door 3 and it turns out to be empty. Should you switch?

To help you parse this, let me mention a few possible scenarios.

The only way Monty will open the door with the car is if Adam and Barney select different doors, and neither of their doors conceals the car. (For simplicity, we will assume that both players win if Monty reveals the car.

Suppose Adam and Barney both choose door one. If the prize is behind door two, then Monty will be forced to open door three. If the prize is behind door one, then Monty will be able to open either door twoor door three, and he will make his choice randomly.

If Adam choose door one and Barney chooses door three, then Monty will open door two regardless of what is behind it.

Keep in mind that you do not know which door Barney chose. So when you see Monty open the empty door three, it might be the case that Barney chose door two, thereby forcing Monty to open door three, whereupon it was discovered by chance that door three was empty. Or it might be that Barney chose door one, and Monty opened door three because he knew that it was empty. There are other scenarios, of course.

Okay! Have at it. I’m going to go grade finals for the next forty-eight hours. I’ll expect answers when I return.

Comments

#1 Brian
December 12, 2007

By “empty”, I’m assuming you mean it had a goat behind it.

It would seem to me that it would still make sense to switch.
#2 Jason Rosenhouse
December 12, 2007

Yes, “empty” means that the door concealed a goat.

If you’re answer is right, it has an interesting consequence. Suppose Adam and Barney chose different doors. Both have now seen Monty open the empty door three. Are you saying that both of them should prefer switching?
#3 Trey Jackson
December 12, 2007

I think I worked out the table properly. If you see an empty/goat door there are 7 possibilities, and switching would end up with a win 4 of those times. So I say switch.
#4 Jeff the Math Illiterate
December 12, 2007

I would stay with my original choice, because if I had it right to begin with and waffled, I’d hate myself. I’d rather be right or wrong the whole way than have it in my grasp (even though I wouldn’t have known it at the time) and lose it by my own indecision.

Then if I lost, I’d accuse Monty Hall of being in cahoots with an Indian airline, clearly having sold the goat for sacrifice while I was deciding, and then trying to claim that “empty” is the same as “goat”. In the ensuing confusion I’d throw one of the hot model hostesses into the car, jump in, and drive away, making game show history.
#5 Colst
December 12, 2007

“If you’re answer is right, it has an interesting consequence….”

But that only becomes interesting if you know that they have not picked the same door.

playing from Adam’s point of view…

There are 9 possible arrangements for the first choice, but assuming a goat is behind the opened door, only 7 of them could have actually happened.

2 possibilities involve both picking the wrong door. In these cases, switching would win.

2 possibilities involve Adam picking the wrong door and Barney picking the right door. In these cases, switching would also win for for Adam.

2 possibilities involve Barney picking the wrong door and Adam picking the right door. In these cases, switching would lose for Adam.

1 possibility involves both of you picking the right door. In this case, switching would lose for Adam.

So, we have 4 possibilities where switching wins, and 3 where switching loses. Either Trey did the table right, or we both screwed up and got the same result.
#6 Jason Rosenhouse
December 12, 2007

But it’s certainly possible that Adam and Barney selected different doors, and if both of them reason the way you have described then both of them will think they are improving their chances of winning by switching to the other one’s door. But they can’t both be right. Can they?

Just to be clear, I’m not saying that your answer is wrong, just that it has this potentially paradoxical consequence.
#7 Dave M
December 12, 2007

I got the same answer as Trey and Colst, and I at least will embrace that paradoxical consequence. I don’t see why they both wouldn’t improve their *chances* of winning even when one guy ends up switching from car to goat.

Of course Monty’s burned me here before (in one of the variant versions)…
#8 DominicH
December 12, 2007

If I have it correctly, you have to account for the likelihood of each player/host combination and not all are equally likely. This is because the host is constrained in his choice such that he MUST choose a goat if he can.
There are 9 possible arrangements of player choice for each of the 3 arrangements of goats and cars so each player/goat combination has a 1/27 probability.
Factoring on the hosts involvement: in most cases (24 of 27) there is either only one door or one goat-hiding door remaining so Monty has to choose it giving a 1/27 chance for each of those occurrences. The remaining occasions where Monty can chose between goats have a 1/54 chance of occuring.
Examining the specific case here, there are 5 chances of Adam selecting door 1 and Monty picking door 3. In three of these Adam gets the goat at a 1/27 probability but in the other 2 where he gets the car, 1 is a 1/27, the other a 1/54 chance. Doing the maths we get a 1/3 chance that Adam has picked right. He will therefore double his chances of winning by switching (I didn’t believe it either).
In the final analysis it makes absolutely no statistical difference whether there are 2 players or one. In each case the player doubles his chance of winning by switching.
#9 Andrew Wade
December 12, 2007

Yeah I’m getting 3/7th for staying pat, and 4/7ths for switching too.
My probability calculations looked like:

1/3 = P(both picked same door)->
1/3 = P(car when staying pat)
2/3 = P(car when switching) (classic Monty Hall)

2/9 they picked different doors, both goats ->
0 = P(car when staying pat)
1 = P(car when switching to Monty’s door)

4/9 they picked different doors, one of which has a car ->
1/2 = P(car when staying pat)
1/2 = P(car when switching)

(Where -> indicates the following probabilities are conditional.)

This works out to a probability of 1/3 of obtaining the car when staying pat unconditionally, which is as it should be.

I then arrived at the conditional probability
P(car when staying pat | not (they picked different doors, both goats)) = 3/7

The tricky bit is identifying the condition “not (they picked different doors both goats)” with “Monty reveals a goat”. But the latter condition occurs if and only if the former does.
#10 Dave M
December 12, 2007

The tricky bit is identifying the condition “not (they picked different doors both goats)” with “Monty reveals a goat”. But the latter condition occurs if and only if the former does.

That’s right, “they picked different doors both goats” = “Monty reveals a car.”
#11 DominicH
December 12, 2007

Oops, I missed out the case where Monty has picked the car. Thats what I get for doing probabilities at 2am!
So, we actually get 2/9 likelihood that Monty opens a door with a car, 3/9 likelihood that Adam has picked correctly, 4/9 that he would be better off switching. He should therefore switch on the 4:3 odds.
This also defines a difference to the 1 player case where Monty can always pick a goatly door and the player doubles his win probability by switching.
#12 qetzal
December 12, 2007

But it’s certainly possible that Adam and Barney selected different doors, and if both of them reason the way you have described then both of them will think they are improving their chances of winning by switching to the other one’s door. But they can’t both be right. Can they?

In that case, they wouldn’t both be right, but they wouldn’t be wrong either. IF they picked different doors, and then Monty revealed a goat, then switching doesn’t change their chances at all.

But they (and we) don’t know that they picked different doors. They know that either they both picked the same door, or they picked different doors and one of them picked correctly. The first case is equivalent to the classic Monty Hall situation, so switching is better. Staying or switching makes no difference in the second case, so switching is best overall.
#13 Don Cates
December 12, 2007

In the original MH problem his pick gives you no additional info about your original pick. That win probability remains 1/3 so switching gives you a 2/3 chance of winning.

In this scenario his pick does give you additional info about your original pick. There are 9 combos in 3 of which you have picked a winner, 1/3 chance of winning. In 2 of the 9 you both are immediate winners and are eliminated from the possible original choices (which were losers). So your original probability of having picked a winner is now 3/7 and switching will change that to 4/7, so switch.

The apparent paradox isn’t because the situations are not symmetrical. In 4 of the 7 possibilities there is a symmetry, switching exchanges a win for a loss and vise/versa. But in 2 cases you both switch to a win and in the other case you both switch to a loss.
#14 Divalent
December 12, 2007

I calculate that it makes no difference.

I worked it this way (I’m Adam):

There are nine possible combinations of where the car is and what door Barney chose: The car can be behind either door 1,2 or 3, and for each Barney could have selected 1,2, or 3 (I chose door 1).

However, 3 of those nine are eliminated for this particular play of the game because Door 3 was shown not to have a car.

A further two more are not possible because Monty opened #3, so Barney did not pick door 3.

So that leaves only 4 possible combinations: the car behind door #1 and Barney chose either 1 or 2, (two possiblities) or the car behind door 2, and Barney chose either 1 or 2.

So it’s a 50-50 shot: you win half the time by holding, half the time by switching.

In table form, it looks like this (b=barney’s door, c=car door):

1cb – 2. – 3. Win by HOLDING
1c. – 2b – 3. Win by HOLDING
1c. – 2. – 3b … Not possible, door 3 was opened

1b – 2c. – 3. Win by SWITCHING
1. – 2cb – 3. Win by SWITCHING
1. – 2c. – 3b … Not possible, door 3 was opened

1b – 2. – 3c … Not possible (3 had goat)
1. – 2b – 3c … Not possible (3 had goat)
1. – 2. – 3cb … Not possible (3 had goat)
#15 Duae Quartunciae
December 12, 2007

If they both pick the same door, and that door has a goat, then Monty will reveal the other goat. If both have read these comments then both will select the car on switching.

It’s not clear what happens in this case. Do they share the car? In that case, they get a value of 0.5 of a car. The value of the game becomes 3/7 for both players. 2 times out of 7 player A gets the car, 2 times out of 7 player B gets the car, and 2 times out of 7 they share the car. Monty keeps the car 1 time in 7.

The value of the game before any doors are opened is 1/3 of a car. 2 in 9 times Monty has to reveal the car, and hence he keeps it.

OK… another puzzle. Assume one player always switches, and the other never switches. What is the expected value of the game for the two players? If they both end up selecting the car, they have a value of 0.5 of a car.
#16 Duae Quartunciae
December 12, 2007

Postscript. The error in Divalent’s analysis is assuming equal probabilities for the four cases. This is the usual problem with getting these kinds of puzzle wrong.

1cb – 2. – 3. Monty had a free choice between 2 and 3.
1c. – 2b – 3. Monty was forced to open 3.
1b – 2c. – 3. Monty was forced to open 3.
1. – 2cb – 3. Monty was forced to open 3.

Assuming (as provided) a 50/50 choice when Monty gets to pick between doors, you actually have the first case half as often as the other three. Divalent’s analysis only works if you know in advance that Monty always prefers to show door three rather than two, given a choice. But that is not the case.
#17 Blaise Pascal
December 13, 2007

OK, I’d have to be confident with my answer if I’m willing to give it using this nom-de-net for a probability problem…

The way I see it, there are 27 (3^3) possibilities of Adam’s choice(A), Barney’s choice(B), and where the car is(C). I suppose I could narrow it down by fixing one of A, B, or C and thus really only have 9 possibilities to consider, but I’ll go with the whole bundle.

Of the 27 possibilities, in 6 of them (ABC in {123,231,312,321,213,132} have a car revealed by Monty. 9 of them (ABC in {111, 121,131,212,222,232,313,323,333}) have Adam initially chooses the car, and thus does not benefit from a switch. The remaining 12 options (ABC in {112,113,122,133,211,221,223,233,311,322,331,332}) Adam fails to choose a car initially, and thus benefits from a switch.

So of the 21 times out of 27 that Adam sees Monty reveal a goat, he benefits from switching 12 times and loses 9 times. He should switch.

There is no conflict with the idea of Barney also preferring a “switch” strategy. Out of the 21 possibilities where Monty reveals a goat, Adam and Barney have chosen different options 12 of them, and one will win and one will lose (Adam 6 times, Barney the other 6). However, they will have chosen the same option the other 9 times. In 3 of those 9, they will have chosen the car, and switching is bad, but in the other 6 cases, switching will garner a car.

So to recap, if Adam adopts a switch-always policy, he’ll win 4/9ths of the time, lose 3/9ths of the time, and in 2/9ths of the time, the strategy will be moot. 2/9ths of the time, both Adam and Barney will both benefit from switching, 1/9th of the time, both Adam and Barney will both lose from switching, and in the remaining 4/9ths of the time that the strategy isn’t mooted, only one benefits from switching.
#18 Kay Aull
December 13, 2007

7/12 (~58%) in favor of switching?

Barney has picked either door #1 or #2. 1/2 chance each.

If both of you are on #1, you have a 2/3 chance of goat. If Barney is on #2, and there’s a goat in #3, then there is a 1/2 chance of goat for each of you.

Thus, half the time, you win 2/3 by switching, and the other half you break even. (1/2)*(2/3 + 1/2) = 7/12.

But don’t mind me, I’m studiously avoiding a final paper right now…
#19 Chris Noble
December 13, 2007

I get initial odds of being correct 1/3 (6/18).
Odds of being correct if they switch 11/18.

Definitely switch.

Of the 9 possible initial guesses from A and B.
7/9 would give monty opening a door with a goat. That leaves the car and a goat. This would give 1/2 odds for a switch.

2/9 would give Monty opening a door with a car. Obviously switch to the car. This would give 1/1 odds for a switch.

7/9 * 1/2 + 2/9 * 1 = 11/18.
#20 Chris Noble
December 13, 2007

Just to be clear, I’m not saying that your answer is wrong, just that it has this potentially paradoxical consequence.

There isn’t a paradox.

It’s a not a zero sum game. Both of them can increase their odds from 1/3 to 11/18. If they repeated the game 180 times then both would increase the number of cars that they each win from ~60 to ~110. You could set up a program with a random number generator.

Even in the sub-scenario where Monty reveals a goat then they both increase their odds from 1/3 to 1/2
#21 Chris Noble
December 13, 2007

Damn!

Now I’ve gone though the 9 initial possibilities and calculated the results.

The odds with the switching strategy rise from 3/9 to 4/9.

Trey was correct the first time.

With the switching strategy

2 times out of 9 Adam will win the car.
2 times out of 9 Barney will win the car.
2 times out of 9 they’ll both win the car.
3 times out of 9 neither of them win the car.

Still no paradox.

With the no switch strategy.

2 times out of 9 Adam will win the car.
2 times out of 9 Barney will win the car.
1 times out of 9 they’ll both win the car.
4 times out of 9 neither of them win the car.
#22 Chris Noble
December 13, 2007

The other way to see through the apparent paradox.

After Monty has opened a door with a goat there are 7 possibilities left.

In 4 of these Adam and Barney have chosen different doors. If both of them swap then the car will just swap from one to the other. It is zero sum. If one of them gains the other loses.

In the 3 remaining possibilities they’ve both chosen the same door. In one of these they both originally chose the car and when they switch they both lose. In two cases they both picked the same door with a goat and they swap then they both win a car. The switching strategy is a postive for both of them. 1 times out three they lose together. 2 times out of 3 they win together.
#23 Paul King
December 13, 2007

While calculating the details proves the point it really is a simple problem.

If the car is revealed than there’s nothing more to do, so we can ignore that case.

If both chose the same door you have the regular Monty Hall problem – so both should switch.

If they chose different doors the odds are still 50/50 so there’s no reason to prefer switching or staying.

So the correct choice is to switch, because if they chose the same door they have a better chance of winning and if they chose different doors they will do no worse.

There is no paradox in the fact that the rule prescribes the same behaviour for each contestant. That is because the information available to the two is – from their perspective – exactly the same. So the rule should tell both to stay or both to switch. Only be creating an asymmetry in the information available to the two contestants should we expect them to decide differently.
#24 Samantha Vimes
December 13, 2007

I’m sorry to disagree with the rest of you, but the fact is there are just two doors at the end of the puzzle.
And like flipping a coin, when you’ve only got 2 possible outcomes, it’s 50/50.
I say it doesn’t actually matter if you switch doors or not. The revealing of one goat does not alter what is behind the other doors.

By the way, if I do pick the goat do I get to keep it? I’ve long wanted a goat.
#25 daenku32
December 13, 2007

Switch! switch! switch!!!

Barney is just a distraction.
#26 BrightonRocks
December 13, 2007

I think that by switching you increase the odds of winning form 4/9 to 5/9.

There is a 1/3 chance that Barney picked the same door as me and in this case the situation is identical to the original ‘single player’ Monty Hall problem so if I stick my chances of winning are 1/3 and if I switch my chances are 2/3.

There is a 2/3 chance that Barney picked a different door to me, in this case Monty Hall has no option of what door to pick so by opening it and showing that it contains a goat this gives me extra information (that I do not have in the original Monty Hall situation because Monty gets to pick the door). I now know that the car is behind one of two doors, mine or Barneys and I have a 1/2 chance of winning if I stick and a 1/2 chance of winning if I swap.

Therefore, by combining the odds from the two situations:

The odds of winning if I stick are: (1/3 * 1/3) + (2/3 * 1/2) = 4/9

The odds of winning if I swap are: (2/3 * 1/3) + (2/3 * 1/2) = 5/9
#27 MartinM
December 13, 2007

I’m sorry to disagree with the rest of you, but the fact is there are just two doors at the end of the puzzle.
And like flipping a coin, when you’ve only got 2 possible outcomes, it’s 50/50.

But just because there are two possible outcomes doesn’t mean they’re equally likely. The Sun may come up tomorrow, or not. Two options. I wouldn’t decide which way to bet based on a coin flip, though.

Try looking at it this way. There’s a 2/3 chance that the players will select different doors; in that case, Monty has no choice but to open the third door. There’s a 1/3 chance that this door will be the one with the car, for a total of 2/9 probability that you lose outright, without being given the option to switch.

That leaves 7 out of every 9 trials in which you get to choose. If there’s really no benefit to switching, you may as well stick with your original choice every time, and win half, for a winning rate of 7 out of 18. Of course, the only way your original choice can win is if you chose correctly the first time. If you can consistently choose correctly from three indistinguishable options 7/18 of the time, there’s a man in Florida with a million dollars for you.
#28 Dunks
December 13, 2007

The think the reason it is better to switch is because Adam and Barney can count on picking the same door 1/3 of the time, thus forcing the host’s door opening behaviour to mimmic that of the single contestant scenario.

When Adam and Barney pick different doors (2/3 of the time) there is no net advantage or disadvantage in switching because no extra information can be garnered from the host’s door opening behaviour (since he is constrained to pick the remaining door). It’s pure chance unless the host reveals the car.

I don’t think it’s a paradox because the switching-advantage-bias originates from both contestants knowing if they play the game multiple times there is a chance they will pick the same door (and initiate host behaviour which favours switching). Since they don’t actually know when they have picked the same door it is in their interest to ALWAYS switch since there is no net disadvantage in doing so when they pick different doors (win 50%, lose 50%) but there is a net gain over all games played.

Am I missing something?
#29 JBL
December 13, 2007

I think Paul King has the clearest explanation of why there is no paradox here (and the simplest argument in favor of switching, no computations involved). The people who said that “if they pick different doors, they both can increase their odds by switching” are wrong: if they have picked different doors, it’s back to 50-50 between the two doors they picked, so switching gives you even odds. The difference is that we (and they) don’t know a priori that they have picked different doors: there’s a nonzero chance that they’ve picked the same door. When they’ve picked the same door, we’re back to the standard Monty problem (as several people have pointed out) and both players do better by switching, so in the aggregate they’ll do better by switching (but not in every subcase, e.g. in the subcase that they picked different doors, or for that matter in the subcase where they both initially picked the car).
#30 Jason Rosenhouse
December 13, 2007

Thanks for all the comments. I’m still thinking about some of the more complicated ones. And I haven’t completely decided for myself what the correct answer ought to be.

Let me toss out one more thing to think about. After Monty opens door three, we know that Barney chose either door one or door two. Which of these is more likely?

Before Monty does anything we would judge these possibilities to be equally likely. But the event, “Monty opens door three and shows that it is empty” is more likely to occur when Barney has chosen door two, than when Barney has chosen door one.

That implies, via Bayes’ Theorem, that after seeing Monty open the empty door three I should now think it is more likely that Barney chose door two than that he chose door one. This complicates any attempt to solve the problem by a straightforward enumeration of cases.
#31 Rieux
December 13, 2007

If, instead of doors, goats and a car, the situation were modeled with playing cards–two black and one red–would that make it….

Never mind.
#32 One Brow
December 13, 2007

“But it’s certainly possible that Adam and Barney selected different doors, and if both of them reason the way you have described then both of them will think they are improving their chances of winning by switching to the other one’s door. But they can’t both be right. Can they?

Just to be clear, I’m not saying that your answer is wrong, just that it has this potentially paradoxical consequence.”

What’s so paradoxical? If they picked different doors, they either both improve by swicthing, both improve by staying, or the result of switching is exactly 50-50. Since Monty Hall only gets to pick a door when both players and the car are all in the same location, I find the 50-50 most unlikely.
#33 Ned Rosen
December 13, 2007

Having chosen door 1, there are 9 = 3×3 equally likely possibilities for (B,C) where B = Barney choice and C = location of car. Draw and tree and then:

2 of these 9, namely (2,3) and (3,2) will reveal the car.
3 of these 9 , namely (3,3), (3,1) and (1,3) will force a goat reveal of door 2.
3 of these 9, namely (1,2), (2,1), and (2,2) will force a goat reveal of door 3.
1 of these 9, namely (1,1), will reveal a goat, 1/2 time at 2 and 1/2 time at 3.

So given that a goat was revealed behind door 3, there are four possibilities (1,2), (2,1), (2,2) of equal frequency, and a fourth possibility [(1,1) with 3 revealed] which has half the frequency of the other three. So:

i) Switching wins 4/7 of the time {(2,2), (1,2)} and staying wins 3/7 of the time {(1,1), (2,1)

ii) Barney chose door 2 with probability 4/7 {(2,2), (2,1)}. And door 1, 3/7 {(1,2), (1,1) }

That’s my guess for now.
#34 Divalent
December 13, 2007

It makes no difference. Definitive proof from a Monty Carlo simulation of the Monty Hall variant. (I think this is refered to as a Full Monty analysis)

I set up a spread sheet to randomly decide where the car was and what door barney chose. I then had Monty open a door based on the rules stated above.

10,000 simulations at a time (re-run each time I hit the recalculation button) show the proof: it makes no difference.

You have a 1/3 chance of winning a car with either strategy. There is no difference between them.

(Note: you have no information that would indicate any difference between doors 2 and 3, so in terms of the information at hand, nothing can be inferred from the fact that, in the above example by Jason, Monty opens #3 rather than #2)

If you’re interested in doing this, here’s the spread sheet stucture:

Col A and B are random numbers from 1 to 3. Col A represent barneys choice and Col B is the location of the car [The exact function I used is INT(RAND()*3)+1.] Col C is Monty’s choice: either the remaining door if Barney and I chose different doors; the non-car door if we chose the same door that didn’t have the car, and a random door if we both chose door 1 and that’s where the car was.

Col D is 1 if Monty showed a goat, or 0 if he showed a car(i.e., if Col C = Col B, place a 0 here, otherwise place a 1)

The next col E is just an intermediate result: it is 1 if the car is behind Door 1, 0 if not. (if Col A = 1, then Col E if 1, otherwise its 0)

The next two colums reflect whether you win for Holding or switching. Col E = Col

Col F is the product of col E times Col D, and represents the result if you adopt a HOLD strategy: it’s “1″ if you won a car, 0 if you didn’t.

Col G is the product of col D times (1 – Col F), and represents the result if you adopt a SWITCH strategy: it’s “1″ if you won a car, 0 if you didn’t.

I copied these cell entries into 10,000 rows, then summed the total of Col F and Col G.
#35 BrightonRocks
December 13, 2007

Sorry, my initial solution was wrong because I had incorrectly assumed that the probability of Barney choosing the same door as me (Adam) was 1/3. This is the case before a goat is revealed but not once a goat has been revealed as all of the 6 now discounted possibilities (of the original 27) have Barney choosing a different door to me, thus my assumption was wrong.

Jason, I think that you can actually solve this by a straightforward enumeration of the remaining 21 states that are possible once a goat has been revealed. This is what Blaise Pascal did and I agree with him/her that the odds of winning are:

The odds of winning if you stick: 9/21

The odds of winning if you swap: 12/21
#36 One Brow
December 13, 2007

Also, put me in for the 3/7 stay, 4/7 switch category. There are four possible starts after Hall opens door 3 to goat, and in three of them his choice was forced, while in the fourth he got to pick. That’s a relative weight of 4 for the car being behind door 2 (switching) and 3 behind door 1 (staying), while in the original problem the relative weights were 2 to 1.
#37 MartinM
December 13, 2007

The next col E is just an intermediate result: it is 1 if the car is behind Door 1, 0 if not. (if Col A = 1, then Col E if 1, otherwise its 0)

You might want to check that; Column A is supposed to be Barney’s choice, not the car’s location. Should be (if Col B = 1, then Col E = 1, else 0).

Just tested my own simulation in Java over a million runs, but I won’t post the results until everyone’s happy with their answers
#38 qetzal
December 13, 2007

@ Divalent:

I did a spreadsheet as well, but mine shows a 4:3 advantage to switching. Details:

ColA, Adam’s Door: =1
ColB, Barney’s Door: =INT(RAND()*3+1)
ColC, Car’s Door: =INT(RAND()*3+1)
ColD, Monty’s Door: =IF(B1=A1,IF(C1=A1,INT(RAND()*2+2),5-C1),5-B1)
ColE, 0 if Monty shows the car, 1 if he shows the goat: =IF(D1=C1,0,1)
ColF, Adam’s payoff if he stays: =IF(E1=0,”",IF(C1=A1,1,0))
ColG, Adam’s payoff if he switches: =IF(E1=0,”",IF(C1=A1,0,1))

For the last two columns, note that if Monty shows the car (E1=0), the cell is left empty (“”). Otherwise, the cell value is 1 if Adam wins the car, 0 if he loses.

Now, if we copy columns A-G down to row 10,000, we can calculate Adam’s average chance of winning:

Adam always stays: =SUM(F1:F10000)/COUNT(F1:F10000)
Adam always switches: =SUM(G1:G10000)/COUNT(G1:G10000)

Note that the COUNT function counts how many cells contain numbers (i.e. 0 or 1), but ignores empty cells (representing cases where Monty already revealed the car).

@ JBL:

I agree that Paul King’s explanation was the clearest. I wish mine had been as well worded! (See comment at Dec 12, 10:09 PM.)
#39 qetzal
December 13, 2007

Jason:

Let me toss out one more thing to think about. After Monty opens door three, we know that Barney chose either door one or door two. Which of these is more likely?

Before Monty does anything we would judge these possibilities to be equally likely. But the event, �Monty opens door three and shows that it is empty� is more likely to occur when Barney has chosen door two, than when Barney has chosen door one.

That implies, via Bayes’ Theorem, that after seeing Monty open the empty door three I should now think it is more likely that Barney chose door two than that he chose door one. This complicates any attempt to solve the problem by a straightforward enumeration of cases.

Maybe the solution is to look at it slightly differently.

Before Monty opens a door, you know there’s a 2/3 chance that Barney’s door is not #1. That 2/3 chance is evenly distributed to doors #2 & #3.

After Monty opens door #3, there’s still a 2/3 chance that Barney’s door is not #1. But Monty gave you new information that lets you assign the full 2/3 chance to door #2.

Does that resolve the apparent complication?
#40 Chris Noble
December 13, 2007

I think Paul King has the clearest explanation of why there is no paradox here (and the simplest argument in favor of switching, no computations involved).

I agree. After starting by getting it completely wrong my posts a 3:39 AM and 4:15 AM say the same thing but less clearly.

The people who said that “if they pick different doors, they both can increase their odds by switching” are wrong: if they have picked different doors, it’s back to 50-50 between the two doors they picked, so switching gives you even odds.

In fact if Adam goes first then Barney’s best strategy is to pick the same door (if he has the information). Apart from tabulating the possibilities the obvious reason is that Monty will not be able to reveal the car according to the rules.

So the best strategy is to both pick the same door as long as they can both get a car each and not have to share it.
#41 BrightonRocks
December 13, 2007

Divalent, qetzal, nice idea to use a spreadsheet to test this.

I think that qetzal’s result of a 4:3 advantage of switching is correct and confirms the solution of enumerating the 21 possibilities of Adam’s choice, Barney’s choice and the car that remain from the initial 27 permutations (3*3*3) once Monty has revealed a goat. The six possible permutations where Adam’s choice, Barney’s choice and the car are all different can be discounted as no goat can be revealed. Doing this gives the chances of winning as:

chance of winning by sticking = 9/21 = 3/7
chance of winning by switching = 12/21 = 4/7

As can be seen the advantage of switching is to increase your chances of winning by 4/3.
#42 Chris Noble
December 13, 2007

I set up a spread sheet to randomly decide where the car was and what door barney chose. I then had Monty open a door based on the rules stated above.

I started to write some code to do a monte carlo calculation but in doing so it became obvious that it wasn’t necessary.

The initial possibilities are described by a 3×3 matrix for the choices of doors 1-3 for A and B. It doesn’t matter which door the car is behind so there isn’t any need to go to the 3x3x3 cube. Just take the car to be behind door 3.

The doors that Monty opens for the 9 original possibilities are then

(2) (3) (2)
(3) (1) (1)
(2) (1) (1/2)

In all cases the door that Monty opens is constrained except for the case where both A and B initially chose door 3 with the car then Monty can open door 1 or door 2.

In the switching strategy A’s choices are completely constrained.

After the switch they are then

(3) (1) (1)
(2) (3) (2)
(3) (3) (2/1)

There are now 4 out of 9 cases where A selects door 3 with the car.

Compare this to the original choice matrix where adam selects the door with the car 3 times out of 9.

(1) (2) (3)
(1) (2) (3)
(1) (2) (3)

The corresponding matrices for Barney can be obtained by transposing those for Adam.
#43 rmp
December 13, 2007

My apologies if this is redundant. Here is how I went about it.

Scenario 1: Monte picked a door with a goat because 2 doors were available and he chose the one with the goat. In this scenario it’s a standard Monte Hall and obviously you switch.

Scenario 2: Monte picked a door with a goat because it was the only door available. Since you don’t know which door Barney chose, there is no reason to treat this any different than Scenario 1.

Scenario 3: Monte picked a door with the car. Your screwed BUT, we have been told that this wasn’t the scenario.

Since both Scenario 1 and Scenario 2 indicate swapping doors, ……

my $0.02
#44 Ned Rosen
December 13, 2007

At the risk of being rerpetitive:

Given that Monty revealed a goat behind #3, the probability that Barney chose door 2 is 4/7 (assuming that a priori, both the car and Barney’s initial choice are randomly made from the three doors). This is because there are four possible combinations which can lead to a goat revealed behind door three. These are

Barney = 2 Car = 2
Barney = 2 Car = 1
Barney = 1 Car = 2
Barney = 1 Car = 1

The first three are equally likely but the fourth will happen only half as often as the others because, although (B=1, C=1) is equally likely to the other three, half the time that (B=1,C=1) happens Monty will open door 2. Thus the relative frequencies of the four events, given that Monty revealed a goat behind #3, are respectively 2, 2, 2, 1. The probability of (Barney = 2) is (2+2)/(2+2+2+1) = 4/7. The probability of the car being behind door 2 is (2+2)/7 = 4/7. The probability of both Barney and the car being #2 is 2/7. The probability of at least one of Barney and the car being #2 is 6/7. Etc.

In the language of conditional probability, given that Monty revealed a goat behind #3, the four events above have conditional probabilities 2/7, 2/7, 2/7, 1/7 respectively. Without the condition, the four events each have probability 1/9.

Anyway, you CAN enumerate the cases as long as you leave the “revealed door” to the end of the tree. Then the branch probabilities are just 1/3 for Barney’s choices, 1/3 for the car’s locations, and either 1 (Monty forced) or 1/2 (the case (1,1) above, where Monty flips a coin) for the revealed door. You get 10 outcomes, eight with probability 1/9 and two (B=1, C=1, Mg2) and (B=1, C=1, Mg3) with probability 1/18.
#45 divalent
December 14, 2007

Updated “Full Monty” analysis: There was an error in my spread sheet (as some probably figured out by their different results). Not the one pointed out by MartinM (that “error” was just a translation error in my post), but in my “Monty’s choice” column: I originally had him do a random pick from 2&3 when barney selected door 1 even if the car was in 2 or 3 (that’s what I get for rushing this before heading off to work).

When corrected, I get what most others have found: you win a car 1/3 of the times you play the game by holding, and 4/9th by switching (if you can). But 2/9th of the time you never make it to the situation Jason described, because on those outcomes Monty reveals a car, and you lose.

So the odds relative to the question posed is 3/7th by holding, and 4/7th by switching.
#46 Skye
December 14, 2007

My take was:
-If Adam and Barney had chosen different doors, 1/3rd of those times, Monty would be forced to reveal a car.
-Monty was able to reveal a goat, so its more likely that both Adam and Barney were on the same door (3/5ths of the times Monty can reveal a goat, they are on the same door, 2/5ths of the times Monty can reveal a goat, they are on different doors).
-If they are on the same door, the Monty Hall statistics from the original Monty Hall puzzle apply: they would both benefit from switching, because their current chances are 1/3rd whereas their chances with door #2 are 1/2.
-If they are on different doors, there is no benefit from switching- either door could contain a car (50/50).
-Combine the odds from both scenarios to get the odds for it being unknown if Adam and Barney have chosen the same door: 3/5ths (1/3 chance by staying vs 1/2 chance by switching), 2/5ths (1/2 chance by staying vs 1/2 chance with switching).
So I’d say the average chance is 2/5 staying vs 1/2 switching
#47 BrightonRocks
December 14, 2007

Skye,

Your approach is correct but your assignment of a 3/5 chance that Adam and Barney chose the same door when Monty revealed a goat is wrong. The chance of this is actually 3/7, with a corresponding 4/7 chance that they chose different doors.

You have also assigned a 1/2 chance of success when switching in the original Monty Hall puzzle (when both Adam and Barney are on the same door) but this should be 2/3.

If you change these figures in your calculation you come out with the probabilities of winning when sticking as 3/7 and the probability of winning when switching as 4/7 which, I think, is correct.
#48 ctw
December 14, 2007

Yet another (highly non-intuitive) approach:

Let C=door with car, B=door Barney picks, M=door Monty opens, G={door 3 has goat}. Assume Adam picks door 1. Then we want to calculate:

P{C=1|M=3,G} = P{C=1,M=3|G}/P{M=3|G}

Given that door 1 has the car and door 3 has a goat, Monty will open door 3 with probabilities 1/2 if B=1, 1 if B=2, 0 if B=3. Hence, the numerator is 1/3(1/3)(1/2+1+0)=1/6.

Given only that door 3 has a goat, Monty will open door 3 with probabilities 1/2 if both P=1 and B=1, 0 if B=3, and 1 otherwise. Hence, the denominator is 1/9(1(1/2)+2(0)+3(1))=7/18.

The answer is then (1/6)/(7/18)=3/7.

- Charles
#49 g
December 14, 2007

Yes, switch, with odds ratio 4/3.

Jason’s paradox isn’t a paradox; there’s no difficulty about two people in possession of different incomplete information both being correct to do things that can’t both be good. (Adam and Barney are picked at random from the population, and both know this. They both happen to be very — and equally — poor, but neither knows the other’s wealth. They’re asked “would you like to swap possessions with the other guy?”. Of course both say yes.)

Jason, you do Bayesian updates using your prior probabilities, not using posterior probabilities derived from another Bayesian update using the same information. So the fact that *given Monty’s choice of door 3* Adam no longer thinks Barney equally likely to have chosen each door doesn’t mean that he should do his calculation of whether to switch using unequal probabilities for Barney’s choice of door.
#50 WhatMeWorry
December 14, 2007

LOL why is this difficult? By switching, you are getting two doors for one. DO IT! At the very worse, you are even-money. At best, after you do the 4/3 of the 1/2 of the pi r squareds, you are somewhat better than even money. You CANNOT be LESS than even money, so just do it. Geesh.
#51 Skye
December 14, 2007

Thank you for the correction, Brighton:)
#52 Michael
December 18, 2007

Here’s my take:

There are 27 possible game boards (3 places for the car, 3 choices for player 1, and 3 choices for player 2… 3^3 = 27). Of those 27 boards, the break out as follows:

3 same door, car
6 same door, no car
6 different door, p1 has car
6 different door, p2 has car
6 different door, no car

The last 6 possibilities are not in consideration, due to the phrasing of the question, because this would result in Monty showing a car. This leaves only 21 possibilities. “same door, car” and “different door, p1 has car” will result in a loss if you switch, leaving only 12 possibilities. Of the remaining two, they guarantee a win… “same door, no car” results in Monty picking a goat and both players switching to the car. “different door, p2 has car” results in Monty showing a goat and p1 switching to p2′s door, which is a win. This means all 12 remaining scenarios are sure things.

12:21 are your odds, which is slightly better than half, so I’d switch.
#53 Michael
December 18, 2007

Rephrasing my last post in Bayesian theory:

H = picking the car
E = seeing a goat

If you pick the car, then you always see a goat:

P(E|H) = 1

The probability of picking a car is:

P(H) = 1/3

The probability of seeing a goat is twofold:

P(E|H)P(H) + P(E|!H)P(!H)

The probability of not picking the car is:

P(!H) = 2/3

The probability of seeing a goat if you didn’t pick the car is broken into three scenarios, cooresponding to which door your opponent picked:

Either you share a door, you both picked the wrong door, or your opponent landed on the car:

P(E|!H) = 1/3*1 + 1/3*0 + 1/3*1 = 2/3

Now the result using Bayes’ Theorem:

P(H|E)
= P(E|H)P(H)/P(E)
= P(E|H)P(H)/(P(E|H)P(H)+P(E|!H)P(!H))
= (1/3 * 1) / ((1/3 * 1) + (2/3 * 2/3))
= 1/3 / (1/3 + 4/9)
= 1/3 / (3/9 + 4/9)
= 1/3 / 7/9
= 1/3 * 9/7
= 9/21

There’s your 9:21 odds for having picked the car. So you have 12:21 odds for NOT picking the car and thus switching.
#54 Cecil
December 19, 2007

Thinking out loud (as it were)…

Starting from the beginning, the car is behind each door with equal probability, and after you pick door one Barney picks one of three doors with equal probability. There are thus nine cases each with probability 1/9 – C1B1, C1B2, C1B3, C2B1, C2B2, C2B3, C3B1, C3B2, C3B3.

Keep in mind this is we know at this point, before Monty makes his choice. Now lets consider what he does in each of those nine cases.

If C1B1, Monty shows a goat behind door 2 or door 3 with equal probability (since they both have goats) – probability 1/18 for each of cases C1B1M2-G, C1B1M3-G.

If C1B2, Monty shows a goat behind door 3 – probability 1/9 (case C1B2M3-G).

If C1B3, Monty shows a goat behind door 2 – probability 1/9 (case C1B3M2-G).

If C2B1, Monty shows a goat behind door 3 – probability 1/9 (case C2B1M3-G).

If C2B2, Monty shows a goat behind door 3 – probability 1/9 (case C2B2M3-G).

If C2B3, Monty shows a car behind door 2 – probability 1/9 (case C2B3M2-C).

If C3B1, Monty shows a goat behind door 2 – probability 1/9 (case C3B1M2-G).

If C3B2, Monty shows a car behind door 3 – probability 1/9 (case C3B2M3-C).

If C3B3, Monty shows a goat behind door 2 – probability 1/9 (case C3B3M2-G).

So there are 10 cases possible given only that you picked door 1.
1. C1B1M2-G – 1/18
2. C1B1M3-G – 1/18
3. C1B2M3-G – 1/9
4. C1B3M2-G – 1/9
5. C2B1M3-G – 1/9
6. C2B2M3-G – 1/9
7. C2B3M2-C – 1/9
8. C3B1M2-G – 1/9
9. C3B2M3-C – 1/9
10. C3B3M2-G – 1/9

Now, we also are given that Monty actually opened a goat behind door 3. This rules out cases 1, 4, 7, 8, 9, and 10, leaving:

2. C1B1M3-G – 1/18
3. C1B2M3-G – 1/9
5. C2B1M3-G – 1/9
6. C2B2M3-G – 1/9

These five cases have total probability 7/18. That is, the probability that Monty opened a goat behind door 3 given only that you picked door one is 7/18.

Only in cases 5 and 6 does it pay to switch. These cases together have relative probability (2 * 1/9) / (7/18) = 4/7.

So I guess my vote falls in the 4/7 camp. That is, switching is 4:3 in your favor.
#55 john
December 31, 2007

If, instead of doors, goats and a car, the situation were modeled with playing cards–two black and one red–would that make it….
#56 larry
January 9, 2008

Paul King’s explanation was the clearest.
#57 hit payla?
May 1, 2008

te?kk
#58 Father
July 20, 2008

That leaves 7 out of every 9 trials in which you get to choose. If there’s really no benefit to switching, you may as well stick with your original choice every time, and win half, for a winning rate of 7 out of 18.
#59 izmir oto kiralama
September 15, 2008

thanks you.
#60 temizlik ?irketleri
October 16, 2008

thanks.
#61 s�ve
October 17, 2008

thanks.
s�ve
boya
mantolama
kartonpiyer
yal?t?m
antibakteriyel boya
#62 s�ve
October 18, 2008

thanks.
#63 boya
October 18, 2008

thanks.
#64 mantolama
October 21, 2008

thanks.
#65 d?? cephe
December 16, 2008

Thanks a lot.
#66 Hosting
March 5, 2009

Thank you for the correction…:)

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