Three doors

The first guess is irrelevant since the host always will eliminate a non-winning door. The second decision is what counts, and you make that choice (switch or not to switch) based on two doors.

It follows that regardless what you do for your second choice, switch or not, you have a 0.5 probability of winning. Therefore, the thought that switching improves your odds of winning is false.

Originally posted by warewiz
The first guess is irrelevant since the host always will eliminate a non-winning door. The second decision is what counts, and you make that choice (switch or not to switch) based on two doors.

It follows that regardless what you do for your second choice, switch or not, you have a 0.5 probability of winning. Therefore, the thought that switching improves your odds of winning is false.

Wrong. The host eliminates a non-winning door conditional on your first choice.

In your first choice:
You have 1/3 probability of having chosen the correct door.
You have 2/3 of choosing an prizeless door.

If you have chosen the right door and the host opens another door, then you have 0% chance of winning if you change and 100% chance of winning if you stay.

If you have chosen an incorrect door, then when the host opens a door he opens the other incorrect door. This gives you a 0% chance of winning if you keep your door and a 100% chance of winning if you change.

Sum the two cases and you have:
Keep: 1/3*1+2/3*0=1/3
Switch: 1/3*0+2/3*1=2/3

Originally posted by doodinthemood
edit: I've just thought, as the two given logical answers both rely on assumptions, and both give opposite conclusions, can it not be said that maybe the likelihood is 50/50 after all? Think about it.

True, if the probabilities were both arranged appropriately.
But since one of them is zero... 🙂

Well my proposal makes that 100% of the time you should stick.
Yours makes that 66% of the time you should switch.

While it's difficult to judge probability of assumptions, if mine is 2/3 as likely as yours, then we get a 50/50.

Originally posted by doodinthemood
Well my proposal makes that 100% of the time you should stick.
Yours makes that 66% of the time you should switch.

While it's difficult to judge probability of assumptions, if mine is 2/3 as likely as yours, then we get a 50/50.

You still haven't realized the problem with your assumptions, have you?

Your assumption leads you to game theory. The contestant knows that the host is trying to manipulate him, the host knows that the contestant knows and the contestant knows that the host knows.

The best the host can do is choose a course of action where the contestant cannot deduce from his actions which door is more likely. But the funny thing is that this implies that the best the host can do is split the probabilities 50/50 between switching or not switching or his strategy wouldn't be in sub-game equilibrium.

This means that your proposal makes it that the contestant is indifferent between switching and not switching. If p is the probability your assumption is correct and 1-p is the probability your assumption is incorrect, then for all p different than 1, switching is still the best strategy.

😛

Originally posted by AThousandYoung
This is called the "Monty Haul Problem". You can find proofs easily online by searching for that phrase.

It's the Monty Hall Problem, inspired by the TV game show "Let's Make a Deal." And yes, switching doors does increase your chances of winning.

Originally posted by Palynka
This means that your proposal makes it that the contestant is indifferent between switching and not switching. If p is the probability your assumption is correct and 1-p is the probability your assumption is incorrect, then for all p different than 1, switching is still the best strategy.

😛

Not so, you must take into account that in my situation, sticking is 100% the best option, and in yours, it is 66%. Thus for any p 0.67-1, sticking is best, at p=2/3, it is 50/50 and p0.4 and, assuming it's variable, on average we get pretty smack on 50/50.

Game theory wouldn't work when attributed to this, as gut followers cannot be swayed by anything the host does, and will stick, and those who work it out to be better as sticking are in a minority (this board is an example, with numerous deducing logically that switching would be better, and only me arguing otherwise)

Originally posted by doodinthemood
Not so, you must take into account that in my situation, sticking is 100% the best option, and in yours, it is 66%. Thus for any p 0.67-1, sticking is best, at p=2/3, it is 50/50 and p

You have to read my whole post to see why that isn't true.

If, under your assumptions, the host doesn't want you to win then why would he choose a strategy that can give you 100% chance of winning?

Edit - "Gut followers"? Ok, now I know you know you're wrong and are just messing with me. 🙂

hang on, my posting's not working

edit: ok it's worked now.

Originally posted by Palynka
Wrong. The host eliminates a non-winning door [b]conditional on your first choice.

In your first choice:
You have 1/3 probability of having chosen the correct door.
You have 2/3 of choosing an prizeless door.

If you have chosen the right door and the host opens another door, then you have 0% chance of winning if you change and 100% chance of winn ...[text shortened]... f you change.

Sum the two cases and you have:
Keep: 1/3*1+2/3*0=1/3
Switch: 1/3*0+2/3*1=2/3[/b]

You two need to relax a bit--I think you're both sort of right, since the original question was not clearly worded, hence my first post.

Palynka's proof is correct, IF it is actually part of the rules that the host will always offer you a chance to switch (this was not clearly stated in the first post of the thread, but it is the standard assumption for the Monty Hall problem). If the host does not always offer the option to switch, then the proof is no longer valid and there is no solution.

Originally posted by castlerook
You two need to relax a bit--I think you're both sort of right, since the original question was not clearly worded, hence my first post.

Palynka's proof is correct, IF it is actually part of the rules that the host will always offer you a chance to switch (this was not clearly stated in the first post of the thread, but it is the standard assumption fo ...[text shortened]... always offer the option to switch, then the proof is no longer valid and there is no solution.

I'm quite relaxed.

If the host does not always offer the option to switch, then the proof is no longer valid and there is no solution.

This is blatantly false, though. At the moment of the contestant's choice, we are conditional on the fact that the host DID open the door.

Originally posted by Palynka
I'm quite relaxed.

[b] If the host does not always offer the option to switch, then the proof is no longer valid and there is no solution.

This is blatantly false, though. At the moment of the contestant's choice, we are conditional on the fact that the host DID open the door.[/b]

I'm relaxed too, more or less. 🙂

I think there are two issues here worth discussing, one of which has been brought up before--if the host does not always switch, then I don't see how you can assume that psychology will not play a role. I also don't think you can make the opposite assumption either, that the host will only offer to switch if you chose the correct door initially. But the point is, there is certainly the possibility that your choice could impact whether the host offers you the chance to switch, in which case our events are not independent and calculating conditional probabilities won't work.

However, let's assume for now that the host's actions are random and not set up to trick you. The other issue (and I wasn't clear about this earlier) is whether when the host opens a door, he always shows a non-winning door, or whether the choice is random and sometimes the open door will reveal the prize. This is actually crucial, I think. If a door is opened at random and it does not contain the prize, but it could have, then the two remaining doors will each have probabilty 1/2 of winning. The key is that your proof includes the assumption "If you have chosen an incorrect door, then when the host opens a door he opens the other incorrect door." However, if the host always decides to open a door which he knows does not win, then the door you picked has probability 1/3 of winning and the other door is 2/3.

Originally posted by castlerook
I'm relaxed too, more or less. 🙂

I think there are two issues here worth discussing, one of which has been brought up before--if the host does not always switch, then I don't see how you can assume that psychology will not play a role. I also don't think you can make the opposite assumption either, that the host will only offer to switch if you chose win, then the door you picked has probability 1/3 of winning and the other door is 2/3.

My assumption is based on that it makes no sense for the host to open a door with the prize. But I agree with your calculations in that case. It's just that, like I said, unless we keep to the minimum sensible assumptions, we cannot solve any problem of this sort ("What if we assume that there's a probability God can intervene and place prizes in all doors? What if there's a probability of the show cheating and there are no prizes behind the doors? Etc." )

Note that the opposite (that he can open the prize door) is also an assumption. The question is, does that make sense in the context of a game show? I don't think so.

My feeling is that people who start with such "what if" considerations that they only thought about after realizing they were wrong are just compensating. But maybe that assumption isn't sensible! 😉

Edit: And I'm not ashamed to admit that I followed my gut the first time I heard about it.

Originally posted by Palynka
My assumption is based on that it makes no sense for the host to open a door with the prize. But I agree with your calculations in that case. It's just that, like I said, unless we keep to the minimum sensible assumptions, we cannot solve any problem of this sort ("What if we assume that there's a probability God can intervene and place prizes in all doors? : And I'm not ashamed to admit that I followed my gut the first time I heard about it.

Fair enough. Again though, I think it's fairly easy to remove any ambiguity by saying something like "after the contestant picks a door, the host always reveals a door that does not contain the prize" or something like that.

Edit: Just clicked on your profile--loved it 🙂

First time I heard this, I said "50/50" out of gut instinct.
Second time I heard this, I said "Switch is 66%" out of calculation
Third time I heard this, I had worked out myself "Stick is 100%" out of nitpicky logic.
Now, after this debate, I'm willing to argue that the extrapolation into whether or not the host has to open a door or not is based on such huge incalculable variables that it is as good as being 50/50 after all.