3 Door Logic Problem

[Mendeli]

Assuming he's not choosing the second coin only because the first was tails, yes.

then the odds should be the same as B because he hasn't given us enough information to rule out another outcome besides two tails.

He is ruling out the same outcomes as either A or C. He's just not telling us which. Why would what he chooses to tell us affect his odds of hittin another heads when throwing a coin? I can't accept that his choosing or not choosing to give us information would affect the physical trajectory of a tossed coin.

 

[Lamuella]

He is ruling out the same outcomes as either A or C. He's just not telling us which. Why would what he chooses to tell us affect his odds of hittin another heads when throwing a coin? I can't accept that his choosing or not choosing to give us information would affect the physical trajectory of a tossed coin.

The coins have already been tossed. The outcome is already decided. What is at question here is our incomplete knowledge of the outcome.

Of course information is the key here. If someone tells you truthfully that both their coins landed on heads, all options APART from heads are ruled out.

This isn't about the odds of what WILL happen (what a series of coins will come down as when tossed) but what HAS happened (what we know about the results of a series of coin tosses).

Besides, the man who says "In my situation, either Person A's or person C's comment applies." has given you substantially less information than either of them, in the same way that saying "my coin either came up heads or tails" gives you substantially less information than saying "my coin came up heads"

 

[tsg]

He is ruling out the same outcomes as either A or C. He's just not telling us which. Why would what he chooses to tell us affect his odds of hittin another heads when throwing a coin? I can't accept that his choosing or not choosing to give us information would affect the physical trajectory of a tossed coin.

It doesn't. It's the information that is given to us that allows us to determine what the possible outcomes were, and therefore the probability of each having happened.

If someone has tossed two coins and given you no information about it whatsoever, the only thing you can deduce is there is a 1/4 chance that he threw two heads, 1/4 of two tails, or 1/2 of a head and a tail. By telling us which coin landed heads up, we can rule out the possibilities where that coin landed tails up. By telling us only that at least one coin landed heads up, we can only rule out the possibility of both coins landing tails up. If he tells us they were both heads, we can rule out all possibilities where either coin landed tails up. Likewise, if he tells us that one is a head and the other a tail, we can rule out the possibilities of two heads or two tails.

We are simply calculating the probabilities of what happened based on what information we have. What he tells us doesn't affect the outcome. What he tells us affects how well we can predict what the outcome was.

 

[DevilsAdvocate]

Try it with cards version.

Each of Persons are given a deck of four cards. two black cards, two red ones in each deck.

Person A says his first card was black.

Person B says one of his cards was black

lets examine what there is in the remaining deck in every possible scenario:

Person A

hand deck chance
BB RR 50%
BR BR 50%



Person B

hand deck chance
BB RR 50%
BR BR 50%
RB RB 50%

wait... something's fishy.

There are more chances for BR and RB than there are for BB and RR. Here are the possibilities:

B1 B2 R1 R2
B1 B2 R2 R1
B1 R1 B2 R2
B1 R1 R2 B2
B1 R2 B2 R1
B1 R2 R1 B2
B2 B1 R1 R2
B2 B1 R2 R1
B2 R1 B1 R2
B2 R1 R2 B1
B2 R2 B1 R1
B2 R2 R1 B1
R1 B1 B2 R2
R1 B1 R2 B2
R1 B2 B1 R2
R1 B2 R2 B1
R1 R2 B2 B1
R1 R2 B1 B2
R2 B1 B2 R1
R2 B1 R1 B2
R2 B2 R1 B1
R2 B2 B1 R1
R2 R1 B2 B1
R2 R1 B1 B2

So we get 24 possibilities of the first two cards:

BB 4
BR 8
RB 8
RR 4

If the first card is black, we have BB (4) + BR (8) = 12 of 24 = 50%.

If any one of the cards is black, we get BB (4) + BR (8) + RB (8) = 20 of 24 = 83%.

It breaks down like this:

Draw Count Percent

Both Black 4 17%
First Black 12 50%
Scond Black 12 50%
No Black 4 17%

At least one Black 20 83%
Only one Black 16 67%

 

[CurtC]

Therefore, if you switch, the probability of ending up with the prize is 2/3

I skipped part of this thread, so sorry if this has already been brought up, but there is a subtlety that the "2/3" people usually overlook, and I myself did when I first heard this puzzle more than 15 years ago.

The 2/3 answer is correct only if Monty is forced to offer the choice to switch every time. If he gets to decide whether he wants to offer the switch, there isn't enough information to formally solve the question. For example, if his goal is to make you lose, he may offer the switch only when you've picked correctly on your first guess, in which case you will always lose by switching.

Or, he could be trying to make the show more exciting by trying to give away prizes, and offering the switch only when you've picked wrong initially. In that case, you'd win 100% of the time by switching.

But I never see the puzzle worded so that the host is constrained to always offering the switch, and with only as much information given, any answer can be justified based on your assumptions about Monty's motiviation for offering the switch.

 

[Ashles]

Okay, but assuming that Monty does not have to show you a door, would it still not apply that in any instance where he does show you a door, it still makes sense to swicth?

When he doesn't show you a door then your guess is obviously back to 1/3

 

[pgwenthold]

Okay, but assuming that Monty does not have to show you a door, would it still not apply that in any instance where he does show you a door, it still makes sense to swicth?

No. As CurtC noted, if, for example, he only offered you the chance to switch because your initial guess was right, then by switching you have a 0% of winning.

 

[CurtC]

Okay, but assuming that Monty does not have to show you a door, would it still not apply that in any instance where he does show you a door, it still makes sense to swicth?

No, not at all. Here's an example I've used before. It has the same exact mathematical characteristics as the Monty Hall puzzle:

You approach a street hustler offering a game of three-card monte. It's on the up-and-up, by which I mean that there really is the prize card on the table after he shuffles them, and you have a 1/3 chance of winning assuming that he shuffles them so quickly you can't keep up.

You make your pick. The street hustler then turns over a card you didn't pick, and gives you the option to switch to the other card. Should you?

I think it's obvious in this situation that you will lose 100% of the time if you switch, because the hustler will give you the option only if you picked the prize card on your first try.

This situation is completely equivalent technically to the Monty Hall problem, the only difference being what you assume the host's motivations are.

 

[Number Six]

As some others have said, whether you should switch depends on your knowledge of Monty's strategy for showing you a door in the first place. If you know beforehand that he's going to offer you a chance to switch, then switch. Or if you otherwise have some knowledge of what procedure he's following in determining whether to offer you a chance to switch, then you can use that info to help you decide whether to switch.

But absent any info of Monty's strategy, we can't say whether you should switch or not. Unless you're willing to make _assumptions_, there is no best decision for you to make.

 

[RandFan]

As some others have said, whether you should switch depends on your knowledge of Monty's strategy for showing you a door in the first place. If you know beforehand that he's going to offer you a chance to switch, then switch. Or if you otherwise have some knowledge of what procedure he's following in determining whether to offer you a chance to switch, then you can use that info to help you decide whether to switch.

Sorry, no.

But absent any info of Monty's strategy, we can't say whether you should switch or not. Unless you're willing to make _assumptions_, there is no best decision for you to make.

No.

I know it is difficult to understand but the answer is you switch. There have been many explanations in this thread why that it is so. It is simple logic. I don't mean to be patronizing. It was really difficult for me to understand but it really is very simple logic.

Get 3 cups and a friend and play the game for awhile. The odds will prove it to you. You can play the game here but you will have to trust the programmer.

You need to get Monty Hall out of your minds eye. Monty could only matter if he changed the location of the prize after you chose. That is the only way Monty can change the problem.

So, forget Monty, get a friend, a coin and 3 coffee cups. Have your friend position the coin under one of the cups while your back is turned. Pick a cup and have him show you the one cup that is empty. Make your choice.

Fact: Your choice will always be wrong 2/3 of the time. What do you think Will happen? What is said doesn't matter. Whether you change your mind really doesn't matter.

Your first choice will still always be wrong 2/3 of the time.

If your first choice will always be wrong 2/3 of the time then that means that the coin is in one of the other two cups.

Right?

So if I eliminate one of the cups then there is a 2/3 chance that the coin is under the cup that wasn't chosen.

It is that simple.

Keep working at it, you'll get it.

 

[Number Six]

Sorry, no.

No.

I know it is difficult to understand but the answer is you switch. There have been many explanations in this thread why that it is so. It is simple logic. I don't mean to be patronizing. It was really difficult for me to understand but it really is very simple logic.

Get 3 cups and a friend and play the game for awhile. The odds will prove it to you. You can play the game here but you will have to trust the programmer.

You need to get Monty Hall out of your minds eye. Monty could only matter if he changed the location of the prize after you chose. That is the only way Monty can change the problem.

So, forget Monty, get a friend, a coin and 3 coffee cups. Have your friend position the coin under one of the cups while your back is turned. Pick a cup and have him show you the one cup that is empty. Make your choice.

Fact: Your choice will always be wrong 2/3 of the time. What do you think Will happen? What is said doesn't matter. Whether you change your mind really doesn't matter.

Your first choice will still always be wrong 2/3 of the time.

If your first choice will always be wrong 2/3 of the time then that means that the coin is in one of the other two cups.

Right?

So if I eliminate one of the cups then there is a 2/3 chance that the coin is under the cup that wasn't chosen.

It is that simple.

Keep working at it, you'll get it.

You only have the choice to switch doors if you're given the choice. And if you're given the choice then the means by which the host decided to give you the choice matters.

You wrote:
"If your first choice will always be wrong 2/3 of the time then that means that the coin is in one of the other two cups. Right?"

Yes, right. Now, what if your friend only gives you the chance to change if your first choice is right? Then 100% of the time in which you're given the choice, you should not switch. Right? Should you switch in that case? No.

If you know that your friend will give you the chance to switch every time then you should switch every time. But in that case you _know_ you're going to get the chance to switch every time, i.e., you know the friends' strategy. OTOH if you don't know your friends' strategy then you don't know what to do.

 

[RandFan]

You only have the choice to switch doors if you're given the choice. And if you're given the choice then the means by which the host decided to give you the choice matters.

No, that is wrong.

You wrote:
"If your first choice will always be wrong 2/3 of the time then that means that the coin is in one of the other two cups. Right?"

Yes, right.

Sorry, that is all that is important. That really is the only salient fact.

Now, what if your friend only gives you the chance to change if your first choice is right? Then 100% of the time in which you're given the choice, you should not switch. Right? Should you switch in that case? No.

Only if you know that he will switch in which case that is not even the Monty Hall Problem. You are talking about something else completely. It's not even a point. It's kind of silly. If I know that he will switch if I'm right then it isn't even a game. He just told me the answer.

If you know that your friend will give you the chance to switch every time then you should switch every time. But in that case you _know_ you're going to get the chance to switch every time, i.e., you know the friends' strategy. OTOH if you don't know your friends' strategy then you don't know what to do.

You are making unwarranted assumptions and introducing something that really has nothing to do with the question. There is no basis for such an assumption. Maybe aliens switched it back after your friend switched it. Perhaps someone drugged both of you and made the switch. Perhaps this is all a dream and for that brief second reality doesn't behave the way it has in the past.

It's fun but silly and beside the point.

Outside of unwarranted assumptions, switching is the only logical thing to do.

But if it helps let me ask you this hypothetical.

The person who knows the answer and exposes the wrong door won't change, now what do you do?

 

[Ashles]

No. As CurtC noted, if, for example, he only offered you the chance to switch because your initial guess was right, then by switching you have a 0% of winning.

Well obviously in that case it would be different odds.

I start that sentence by 'Well obviously' because it is obvious. So I am confused as to why I didn't get that when I posted the question.

Now I get it.

I love this problem because it forces such a wedge between what seems to make sense and what probability dictates.

And I find that incredibly interesting as well - how probability conforms to, well, probability.
It's so theoretical, but it just sort of works.

For example think of the number of individual footballers running around in the last 8 teams of the World Cup. Yet I was convinced that it would be a France Italy final that Italy would win.
And I didn't put any money on it.

 

[RandFan]

If you know that your friend will give you the chance to switch every time then you should switch every time. But in that case you _know_ you're going to get the chance to switch every time, i.e., you know the friends' strategy. OTOH if you don't know your friends' strategy then you don't know what to do.

BTW, I've played this game with friends and family to prove to them that it works. You know what. It workeds. Now this is anecdotal I'd admit. So here is what I suggest. Play the game with friends and family. Just one time with eacth. See how many of them switch on you.

My guess is that it is very low.

Even with your assumption which is possible I would say, switch. The odds are in your favor.

 

[Number Six]

No, that is wrong.

Sorry, that is all that is important. That really is the only salient fact.


Only if you know that he will switch in which case that is not even the Monty Hall Problem. You are talking about something else completely. It's not even a point. It's kind of silly. If I know that he will switch if I'm right then it isn't even a game. He just told me the answer.

You are making unwarranted assumptions and introducing something that really has nothing to do with the question. There is no basis for such an assumption. Maybe aliens switched it back after your friend switched it. Perhaps someone drugged both of you and made the switch. Perhaps this is all a dream and for that brief second reality doesn't behave the way it has in the past.

It's fun but silly and beside the point.

Outside of unwarranted assumptions, switching is the only logical thing to do.

But if it helps let me ask you this hypothetical.

The person who knows the answer and exposes the wrong door won't change, now what do you do?

I realize that if you know you'll always be given the choice to switch you should always switch. Nobody is debating that at this point. But as I recall the original question was, basically "There are three doors, one holds a prize, two hold goats, you pick one, the host opens a different door that holds a goat and then asks if you want to switch. Should you switch?" The answer to that is, it depends on how the host decided to make you the offer. If he is always going to offer you the switch then yes, it makes sense. But in order for you to know that you have to be told beforehand that that is what will happen. Either that or you have to be able to safely _assume_ it.

There are a myriad of strategies the host could use to determine whether to offer you a switch and whether you should switch depends on which strategy he uses. In the example I used, where he only asks you if you want to switch if your initial guess was right, then as you say, it's obvious that you shouldn't switch, but I only picked that example _because_ it was obvious that you shouldn't switch and I was trying to show that whether you switch depends on other factors. I could've used many other examples where you also shouldn't switch.

The strategy where the host only asks you to switch if you guess right the first time means that if you guess right you're asked to switch with 100% probability and if you guess wrong you're asked to switch with 0% probability. But those two probabilities could be anything. If the host always asks you to switch then both probabilities are 100% and you should always switch.

The bottom line is, whether you should switch _depends_ on those probabilities. "Those probabilities" is another way of saying "the hosts' strategy." The same goes for your friend witht he cups and coin. If the rules of the game are that he always asks you, then both probabilities are 100%.

 

[CurtC]

BTW, I've played this game with friends and family to prove to them that it works. You know what. It workeds.

RF, you're not getting what I, Number Six, Ashles, and pgwenthold are saying. I agree that if you play the game as you've described, switching is the best choice. No one disputes this.

The problem is that the way you play the game is just one interpretation of the wording of the puzzle. You're assuming that the host must make you the offer to switch each time. How about this - we'll play the game, for money, with me as the host, and you as the contestant, and play according to the wording of the puzzle. You may be surprised when I don't even offer you the chance to switch on most trials, but just reveal your wrong choice. And when I do offer you the chance to switch, you'll start to notice that if you switch, you lose every time.

You are probably thinking that this isn't in the spirit of the puzzle, but hey, we're pedants here, and this way of play is perfectly consistent with how the puzzle is worded. And it's consistent with the way that the show Let's Make a Deal actually worked. Without the explicit statement that the host is constrained to always offer the choice, the 2/3 answer is not justified.

 

[Ginarley]

Interesting thread - dodging the issues of whether the choice to switch must be offered or not and sticking to the classical trreatment, here is how I got my head around it:

This problem is a classic lateral thinking problem where it is easy to get confused with details that don’t matter. The need to make two decisions and the need to remove a dead door are actually completely irrelevant. Let me rephrase the problem in an entirely different way:

You enter a game show. There are three doors, behind one of them is a prize, and behind the other 2 is nothing. The host offers you 2 options:

1) Choose a door. If the prize is hiding behind it you win it.
2) Choose any 2 doors. If the prize is hiding behind either then you win it.

Now the problem seems much simpler and intuitive that option 2 is clearly the best option. The problem at hand is exactly the same problem, the only difference being that the smoke and mirrors of multiple decisions and removing a dead door are taken away. As with all deception, distracting irrelevant details obscure a simple reality.

Hope this helps someone lol.

 

[RandFan]

But as I recall the original question was, basically "There are three doors, one holds a prize, two hold goats, you pick one, the host opens a different door that holds a goat and then asks if you want to switch. Should you switch?"

And the answer is, of course. Based on what we know at this point, yes. You want to interject all sorts of unwarrented assumptions/

The answer to that is, it depends on how the host decided to make you the offer. If he is always going to offer you the switch then yes, it makes sense. But in order for you to know that you have to be told beforehand that that is what will happen.

You are making unwarranted assumptions that there will be some sort of trickery.

There are a myriad of strategies the host could use to determine whether to offer you a switch and whether you should switch depends on which strategy he uses. In the example I used, where he only asks you if you want to switch if your initial guess was right, then as you say, it's obvious that you shouldn't switch, but I only picked that example _because_ it was obvious that you shouldn't switch and I was trying to show that whether you switch depends on other factors. I could've used many other examples where you also shouldn't switch.

You are creating a different game.

The strategy where the host only asks you to switch if you guess right the first time means that if you guess right you're asked to switch with 100% probability and if you guess wrong you're asked to switch with 0% probability. But those two probabilities could be anything. If the host always asks you to switch then both probabilities are 100% and you should always switch.

The bottom line is, whether you should switch _depends_ on those probabilities. "Those probabilities" is another way of saying "the hosts' strategy." The same goes for your friend witht he cups and coin. If the rules of the game are that he always asks you, then both probabilities are 100%.

But again, there is no reason to assume that there will be any such strategy. You are inserting this without reason.

 

[RandFan]

You're assuming that the host must make you the offer to switch each time. How about this - we'll play the game, for money, with me as the host, and you as the contestant, and play according to the wording of the puzzle. You may be surprised when I don't even offer you the chance to switch on most trials, but just reveal your wrong choice. And when I do offer you the chance to switch, you'll start to notice that if you switch, you lose every time.

But you are adding to the game variables that are not likely. If I ask 100 people to play host how many of them will do as you would? Based only on odds of human nature and the facts of the puzzle I would have to switch.

You are probably thinking that this isn't in the spirit of the puzzle, but hey, we're pedants here, and this way of play is perfectly consistent with how the puzzle is worded. And it's consistent with the way that the show Let's Make a Deal actually worked. Without the explicit statement that the host is constrained to always offer the choice, the 2/3 answer is not justified.

I think the wording of the puzzle is pefectly fine to come to the conclusion that you should switch. And BTW, the puzzle says that the host will offer to switch so to now say that he won't is changing the puzzle. I agree that if you consider that the host might switch prizes then the puzzle changes and that was certainly part of the show but that is simply for imagery. I don't even use the name of the puzzle when I present it. If we forget "Let's Make a Deal" would you change your mind?

 

[Roboramma]

But you are adding to the game variables that are not likely. If I ask 100 people to play host how many of them will do as you would? Based only on odds of human nature and the facts of the puzzle I would have to switch.

Actually, human nature being what it is, I think the opposite is true. If you play for money, particularly with a stranger, you can expect them to take the best strategy possible. And if the rules don't prohibit it, making you the offer only when it's in their favour is the best strategy.