Here's my version:

Suppose the host is particularly bad at this game, and whenever he is faced with any kind of choice, he dithers (much to his wife's exasperation, we can imagine). Everyone notices this except for the host himself.

Now imagine a game in progress. The contestant has just selected door A. Now, there is a 1 in 3 chance that he has selected the winning door, and the host will then have to choose whether to open door B or C, and will dither. But there is a 2 in 3 chance that the contestant has selected the wrong door; in that case, the host has no choice (he must open the one door not chosen and not concealing the prize), and so he doesn't dither.

If the host dithers, the contestant knows that he has already selected the correct door and doesn't switch -- he wins the prize. If the host doesn't dither, the contestant knows he must switch -- and so wins the prize. That way, he wins every single time.

So, what are the chances he will switch? He switches when the host does not dither, but unhesitatingly opens a door. That happens in 2 out of every 3 cases, on average.

And sooner or later, the host is fired.

It's a game theory problem, in which the state of knowledge of the contestant has to be taken into account, including his assessment of the motivations and likely actions of the host.

The question put is not 'what should you do, knowing all this background', but 'what should the contestant do, unexpectedly presented with the second choice?'

The first question is a no-brainer. The second is interesting. Like should you draw to an inside straight, really - or, mor accurately, how should you respond to a confident raiser who has just drawn one to four cards.

This argument is similar to those in which the motives or integrity of the gameshow host are questioned. Except those arguments are made by addressing the RULES by which this game could be generally played in a REPEATED fashion. What I'm saying is that the problem as originally posed is a single instance, a specific subset of the possibilities, regardless of motives or rules.

I don't know of any such discussion, but I think I can clarify it here. In order for the problem to be a proper probability problem, you have to have a well defined set of possiblities, from which the various outcomes are selected by chance. By assuming that Monty Hall can open a door neither contestant chose, you are assuming that that door did not contain the big prize.

Therefore, your analysis applies only to the set of cases in which the unselected door definitely did not contain the prize. Among these cases, either contestant is equally likely to have picked the prize, so switching will not help either one. You can't take the conclusion of the usual analysis over to this new case, because the set of possible outcomes has been reduced, and the probabilities change when you do this. For example, your chances of getting a full house in poker change if you play with a deck with no spades.

Because of this, we don't know in advance that there is an advantage in switching doors, and your analysis does not lead to a contradiction, but rather to a perfectly valid solution to a different problem, in which the answer is that there is no advantage in switching.

--Stuart Anderson

1. This message is part of a thread
2. https://www.cut-the-knot.org/hall.shtml

Well, yes, of course. I have read these, but I cannot find in them a discussion of this particular variation of the problem, which is what Latecomer was looking for.

By the way, it has always seemed to me that the clearest way tp explain the original problem is to note that when you make your choice, 1/3 of the time you have the prize and a 2/3 of the time it is behind one of the other doors. If Monty Hall always opens one of them to show a goat, you now have new information, which lets you pick the right door in every one of those 2/3 of the cases. This is more or less the standard explanation, I'm just mentioning that it is the one that appeals to me the most.

As a sidelight, for people whose mathematical intuition works better with dollar signs in front of the numbers, this whole Monty Hall problem provides a simple example of why insider trading on the stock market is illegal. Suppose there are two investors X and Y considering buying stock in three companies A, B, and C. A, B, and C are in direct competition, and only one will be profitable. In an effort to avoid prosecution, X makes a deal with Monty Hall (who has insider information about all 3 companies). The deal is that after X has invested in one company, Monty Hall will give him a tip about which of the other companies is not profitable. After X invests in A, Monty Hall tells X that company B will not be profitable. X then sells his stock in A and buys C. Investor Y does not receive such a tip.

When they are brought before the Securities Exchange Commission, Monty Hall argues that he hasn't given X any information about the company X is currently invested in, nor has he told X which company is profitable. Now the question is, should this be illegal? Does the tip give X an unfair advantage?

Yes it does. From an investment point of view, the additional information allows X to avoid some risk, namely the risk of investing in B. If X acts on the tip, X can get the same potential profit at lower risk, so X's chances of profit go up. However, if X does not act on the tip, he has the same chance of making a profit as Y does. Therefore X should act on the tip. However, the only possible action is to sell A and buy C, since buying B would be stupid. Therefore, switching from A to C must increase the chance of profit.

--Stuart Anderson

It'seems to me that my suggested variation is a subset of the possibilities allowed by the original definition of the problem, rather than a completely different problem. From the point of view of contestant 1, who never knows about contestant 2, I think the setup assumptions can stay the same. And from his point of view, all he knows is that he's picked door A, and that there's a goat behind door C. And yet in this admittedly limited subcategory of the original definition, without him learning anything more, it'seems that another person's discovery of the same information trips up the conclusion that he will always double his chances of winning by switching.

As long as the original problem's definition can include the variation, and the variation leads to the possible inapplicability of the general analysis, I'm struggling . . .

I'm going to pose another variation a little farther down this chain.

"Contestant 1 chooses door A. The monitor in his isolation booth is switched off.

"The monitor and headphones in contestant 2's booth are turned on, and he chooses door B.

"Monty then opens door C for each in turn, revealing a goat. He offers each contestant in turn the chance to switch doors."

The choices of the two contestants can't be independent if Monty is still able to open door C with a goat behind.

If the players choose independently then there are 9 possible combinations (AA, AB, AC, BA,…). Two of these (BC and CB) do not allow Monty to open a door that is both losing and unchosen.

We need to define what our host is going to do. Consider two scenarios:

1. If he can open a losing door, then he always does, and then offers the contestants the opportunity to change their choice. We will ignore the 2 out of 9 cases when he cannot open a losing door.

2. He opens a losing door iff the contestants have chosen different doors, and he can open an unchosen door. We will ignore the 2 out of 9 cases when he cannot open a losing door and the 3 out of 9 cases where the contestants chose the same door.

In the first scenario, opting to switch wins on 4/7 occasions, and remains the winning strategy.

The second scenario, which is the one presented above, gives a 1/2 probability of winning whether you switch or not.

But thinking of that, I was thinking of the way Let's Make A Deal actually used to work, without any of the isolation booths and such that I mixed into my variation. So let me try this one on you:

Suppose contestants 1 and 2 are together in the audience. Monty offers 1 first choice, he picks door A. Monty offers 2 the next choice, he picks door B. Monty then opens door C, revealing a goat.

He then offers 1 and 2 the chance to switch doors. I don't think it matters whether it's a zero-sum, only one winner, both must agree to switch situation, or whether it's an either can switch, either can stay, both-can-win situation.

The standard analysis is that "my original pick had a 1/3 chance, and the other two doors had a 2/3 chance, which I can now have in a single door," right?

Contestant 1 thinks the 2/3 chance was contained in doors B and C.

Contestant 2 thinks the 2/3 chance was contained in doors A and C.

Both have learned that the 2/3 chance excludes door C, so both should switch, right? But that can't be right, right?

I realize that I've changed the game, but I don't think the changes I've made in either case have changed any of the elements that seem to be underlying the standard answer to the original Monty Hall problem as posed -- the original problem says that your initial choice had a 1/3 chance, that the remaining two doors had a 2/3 chance, and that eliminating one of the other doors gets you that same 2/3 chance all of the time. Don't those principles apply to my suggested variations? I don't think I've added any additional revelations, just different views of the same problem.

What do you think? I realize that this was a hot topic more than a decade ago, but I'm slow . . . hence my screen name!

It is not what I said. I said that, in so far as the experiments are not independent you cant' apply to each of them in separation the theory of an independent experiment. At least, this is what I meant.

If you require a stipulation that "one of the contestants has
chosen the prize," the experiments again are not independent.

The only way you may meaningfully consider two contenstants unaware of each other and each other's choices is to allow as much arbitrariness in one's choice as in the other's.

I was wondering, though, why adding a second contestant needed that contestant to be independent or unaware of the first contestant in order to illustrate my point about my problem with the initial Monty Hall solution. And on reflection, I don't think it matters whether contestant 2 knows about contestant 1, or can or cannot choose the same door, to demonstrate where I'm trying to go. Which is what I was getting at in restating my variation with a known and knowing contestant 2.

I think I'm correct in believing that one premise of the original solution is that the initial guess has a 1/3 chance of success. It'seems to me that this is a baseline premise of the standard solution for contestant 1 whether or not there is a second contestant, and whether or not the second contestant is known.

Similarly, contestant 2, whatever door he picks, would have the same 1/3 chance in being correct, according to the analysis that's at the foundation of the accepted answer to the original problem. We can set the problem so that he has the same number of choices, or we can set it'so he knows of contestant 1's choice and is locked out of that one. That doesn't matter, does it? Isn't he, according to the accepted analysis, faced with the same 1/3 probability for whatever choice he makes?

Now Monty opens door C, revealing a goat.

According to the accepted analysis, "the doors I didn't pick" held a 2/3 probability of being the winning door, correct? And that 2/3 probability is now contained in "the doors I didn't pick, excluding C," right?

What I'm not grasping is why a second contestant can't also say, "the doors I didn't pick" had a 2/3 chance of success, and why, if the second contestant didn't pick C and Monty opens it up to reveal a goat, he's not also covered by the standard analysis: my pick was 1/3, the doors I didn't pick were 2/3, that 2/3 is now down to one door, so I'm twice as well off by switching to that door.

The way I've seen the statement of the initial principle stated doesn't suggest that it depends on anything other than the idea that your initial guess was a 1/3 shot, and that the remainder was a 2/3 shot, whose payoff potential is now narrowed to one door. On principle, the way I've been reading it, nothing prohibits its application to a second contestant.

And yet in the instance where there is a second contestant, and he does pick a different door, and there is a zonk door for Monty to open, the application doesn't work -- I don't think. Since I think I'm understanding the premise of the original problem, and understanding the principle of the solution, applying the principle to a subset of the original without changing the facts of the original (haven't varied what contestant 1 knows, when he knows it, or where the prize is), when I get to that subset and the principle doesn't work, I'm thinking there's a problem with the principle.

Contestant 2's appearance, or his guess, whether he tells it to contestant 1 or not, shouldn't affect the right answer to a probability question, should it?

You are incorrect since it appears that you don't attach a equal probability of the car being in one of the three boxes.

>I would appreciate any and all opinions. Thankyou
Your welcome. Note that similar logic applies if the contestant opens the middle or right door instead of the left door. Hope this clarifies things.

That is part of my problem. How can it be so? If one box has been opened and shown to contain a goat, how can it contain both a goat and a car?

"You are incorrect since it appears that you don't attach a equal probability of the car being in one of the three boxes."

I don't, not once a box is open. Again I say it is impossible for a box to contain both a goat and a car. Isn't it? Remember that one box is open and shown to contain a goat. Therefore the car must be in one of the other two, must it not? I don't get this.

Thanks for your reply.

or box M

>box and show you the goat. This leaves the M box unopened.
>Similarly if I know that we have arrangement 2. If I know
>instead that we have arrangement 3, then I can open the M
>box and show you the goat.
>
>So, after I have opened a box containing a goat, the three
>possible arrangements are as follows:

>1: L has prize : M has a goat : R is open
>2: L has a goat : M has prize : R is open
>3: L has a goat : M is open : R has prize

I would say there is another possibility

1': L has prize : R has a goat : M is open

>So, remembering that these three possible arrangements are
>equally likely from your perspective, what should you do?
>(i.e.switch or stick)

(1 1') are equiprobable with 2 and 3.

>I read it and thought the explanation was flawed. The poster
>(a well known UK academic, by the way, so he should know)

Marilyn would disagree. A good many academicians scolded her after she published her solution.

>maintains that it is correct.

He is almost, with the above caveat.

>My reasoning is this: Once the contestant picks a box and
>the host opens one (containing a goat) surely that then
>becomes an individual game completely separate from the
>other two arrangements.

It does not. This has been chewed over and regurgitate so many times that I wonder whether your question is in good faith.

Please read this thread, or

https://www.cut-the-knot.org/hall.shtml

or just make a search on the web. Your quandary has been resolved many times over.

Perhaps I could try to explain my point by using a three horse race analogy?

Three horses "A", "B" and "C". The race starts. "A" falls down dead and is therefore out of the race (the open box containing a goat,it obviously cannot contain the car)

As "A" to all intents and purposes is last (dead) that then surely must give us just TWO possible outcomes to the race. (1)"B" wins, "C" is second, "A" is last. Or (2) "C" wins, "B" is second, "A" is last. Is there a third?

I hope you can now see that my problem is not with the Monty Hall paradox as such but wih the wording of the explanation I was given which says: "after" I have opened a box containing a goat. That then (in that game) is surely unalterable, like the dead horse.

I am not a mathematician so please bear with me.

No, there are just two of them.

The problem is the analogy is not accurate. Instead consider this: assume under the saddle of one of the three hourses a chance concealed a $1,000 bill. You jump on one of them and are about to start looking under the saddle while another one loses its saddle revealing nothing beneath it. It drops dead if you wish. The other two hold steady.

Or you can think of a situation where it is known up front that two of the horses are going to drop dead and not finish the race. You jump on one of them. On the first loop one of the other hourses drops dead. You are given a chance to change horses. Should you?

>I hope you can now see that my problem is not with the Monty
>Hall paradox as such but wih the wording of the explanation
>I was given which says: "after" I have opened a box
>containing a goat. That then (in that game) is surely
>unalterable, like the dead horse.

I did not use the word "after" in the two scenarios. However, each of them is clearly sequential. Horses start droping dead after you chose one of them.

Your scenario does not model the Monty Hall problem. What is the horse metaphor? A door? A prize?

If you say "door", I reply "Is it a door that you are after or the prize?"

If you say "prize", I reply "How do you know which horse has died? Could not it be the prize?"

Neither really. A more accurate comparison would be the box itself and it's contents. The car/prize is represented by winning of the race.

So horse "A" (the dead one) would be = The open box containing the goat.

In (1)"B" would be = the box containing the car/prize.
"C" would be = the box containing the second goat.
"A" would be = the open box containing the other goat.

In (2)"C" would be = the box containing the car/prize.
"B" would be = the box containing the second goat.
"A" would be = the open box containing the other goat.

"The problem is the analogy is not accurate."

Any better?

Suppose I am the contestant. Three boxes Left, Middle and Right. Let's say I pick L. Monty opens M to show a goat. That means the car/prize is in either L or R. It cannot be in M. That also means there are two possibilities.

M contains a goat.
L contains a goat.
R contains the car.
or
M contains a goat.
R contains a goat.
L contains the car.

Now my man says "So, after I have opened a box containing a goat"
and
"So, remembering that these three possible arrangements are equally likely from your perspective, what should you do? (i.e.switch or stick)"
How can that be possible? For it to be so, the third arrangement would have to be:

M contains the car/prize.
R contains a goat.
L contains a goat.

As I already said that is not possible because we know that M contains a goat. Monty has showed us it.

Unless you know different?

>Now my man says "So, after I have opened a box containing a
>goat" and "So, remembering that these three possible
>arrangements are equally likely from your perspective, what
>should you do? (i.e.switch or stick)"

I do not see three possibilities, just two. But they are not equally likely. They occur in ratio 2:1. This is because, before you made your selection of L, the third possibility was present. If you went with it, Monty would not open M, but R instead. This possibility has been implicitly folded into your #1.

Try to switch a view point. Assume, after Monty opens a door, you are not required to do anything but just compute the probabilities. Surely the probability of having the prize in L is 1/3, whatever Monty does afterwards. So the probability of not having a prize in L is 2/3, whatever Monty does or does not do.

Thankyou. That is what I have been saying all along. Both here and at the other forum.

"But they are not equally likely."

I know they are not. I have never said they were. Either here or at the other forum.

"Surely the probability of having the prize in L is 1/3, whatever Monty does afterwards. So the probability of not having a prize in L is 2/3, whatever Monty does or does not do."

I agree. I have never argued against it. Either here or at the other forum.

So, would you now agree with me that my U.K. academic's explanation, where he says that after a box is opened and shown to contain a goat, there are three possible arrangements, is incorrect?

Do not know about other forums, but this is not what you were saying on this one.

>So, would you now agree with me that my U.K. academic's
>explanation, where he says that after a box is opened and
>shown to contain a goat, there are three possible
>arrangements, is incorrect?

No, I just checked your original post at this forum. There is not too subtle a difference between what you posted originally and what you said lately. You also changed roles. At the beginning "I" meant Monty; later you converted it into the contestant.

The wording of the fellow you can't agree with may be unfortunate, but the reasoning is correct. After a door is opened by Monty there are indeed three equally likely possibilities:

1: L has prize : M has a goat : R is open
2: L has a goat : M has prize : R is open
3: L has a goat : M is open : R has prize

This was the original statement. Later, you formulated it thus

Suppose I am the contestant. Three boxes Left, Middle and Right. Let's say I pick L. Monty opens M to show a goat. That means the car/prize is in either L or R. It cannot be in M. That also means there are two possibilities.

M contains a goat.
L contains a goat.
R contains the car.

or

M contains a goat.
R contains a goat.
L contains the car.

You see "Monty opens a door" and "Monty opens door M" are two different things. (Note that, for the sake of juxtaposition, it would have been less confusing had you said "Monty opens R to show a goat," not "M".)

That is not so. The act of opening the door tells me, you, Monty, whoever that that one contains a goat. Since we know that Monty always opens a door with a goat then we also know that that leaves only two possibilities after the door has been opened. It matters not which one it is.

"You see "Monty opens a door" and "Monty opens door M" are two different things."

I know they are but as I said above it doesn't matter. The end result is the same.

"(Note that, for the sake of juxtaposition, it would have been less confusing had you said "Monty opens R to show a goat," not "M".)"

OK we/I can do that but it doesn't alter the fact that there will still only be two possibilities.

I am begining to think you and my man are one and the same. :)

You should read this as "With what me know after Monty opened a door there may be three equally likely possibilities." The possibilities do not reflect the physical state of affairs which is "one door open, two doors closed," but are those that might have led to that physical state.

As I said before and which you prefer to ignore, the wording - the usage of the word after - may be unfortunate, but this does not change the author's intention, viz.,

Before Monty opened a door, we had such and such possibilities. The fact of Monty's opening a door made some of them unrealized in principle, so that after he opened a door only such and such possibilities remain.

This is what the fellow wanted to say.

This is what the fellow wanted to say."

Finally, we seem to have arrived at what seemed to me to be the unavoidable and obvious conclusion. Albeit by a rather tortuous route. At times I felt I was flogging my own dead horse.

Thankyou.

Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true.
Bertrand Russell (1872 - 1970), Mysticism and Logic (1917) ch. 4

Just a joke, bud.

You are welcome.

>Mathematics may be defined as the subject in which we never
>know what we are talking about, nor whether what we are
>saying is true.

But always strive to understand each other, right?