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Javier Gil
30-08-2004, 07:04 PM
Please don't submit a reply if you know the correct answer to this problem

A little math/logic problem for you: There are three closed doors from
which you are asked to select only one. Behind two of the doors are
worthless junk but the behind the other is a tremendously valuable prize
which you would like to win. You get to keep whatever is behind the door
you select. This process is monitored by a master of ceremonies who knows
what is behind each door. He asks you to pick a door and you choose door
one. Rather than open the door you've chosen, the MC opens door two
revealing worthless junk. He then asks you if you would like to change
your selection. Would you?

Goughfather
30-08-2004, 07:13 PM
Just out of interest, where do you play Tetris?

Rincewind
30-08-2004, 07:41 PM
This is the controversial Monty Hall problem (which I mentioned briefly in the early days of the rec math thread). I'm not sure if the way you worded the puzzle has produced the "desired" problem. There was lengthy discussion on this in the letters to the editor of The Skeptic (Aust) a few years back including some luminary remarks from a Professor of Statistics from ANU or UoC (from memory). I'll try to dig it up.

It basically boils down to recreational mathematicians are quite happy to make some small assumptions to produce a nice puzzle. "Real" mathematicians however, try to never assume anything and this is basically where the problems begin. :)

However, the well formed puzzle is quite nice and anyway, the way you worded it would be acceptable to most puzzlers.

Javier Gil
30-08-2004, 07:47 PM
Personally, I think the answer to this problem can be worked out without knowing a single thing about mathematics, using deductive logic.

Goughfather: I used to play tetris at home, on my pc.

Goughfather
30-08-2004, 07:56 PM
It basically boils down to recreational mathematicians are quite happy to make some small assumptions to produce a nice puzzle. "Real" mathematicians however, try to never assume anything and this is basically where the problems begin.

Yeah, maybe I'm somewhat pedantic, but I found the question a bit ambiguous. Is "Door One" actually the first door in the line, or is it simply suggestive of the door you pick, indicating that it could be any of the three doors. Is "Door Two", the door that has one of the worthless junk in it, the door in the middle, or simply one of the doors that doesn't contain the grand prize?

Kevin Bonham
30-08-2004, 08:09 PM
Not sure if I have seen this before or not, so what I am going to do is post a solution and delete it so that it can be revealed later. In any case I am fairly experienced at puzzles like this and others should be given a chance to get it right/wrong without my input.

Kevin Bonham
30-08-2004, 08:16 PM
I would split.

The original guess has a 1/3 chance of being right. This is not changed by the turning of an empty box because whether the guess is right or wrong, an empty box can always be turned.

However if the original guess is wrong (2/3 chance) then the remaining door must be correct.

The only ambiguity in the question that matters is that it is not stated whether the location of the prize is determined randomly. It could, for instance, be determined based on psychology - that people were more likely to choose particular numbers out of 1,2,3 than others. For a better wording the door which the prize is behind should be chosen randomly by the show, and there is no need to specify which number door is chosen.

Javier Gil
30-08-2004, 09:25 PM
Yeah, maybe I'm somewhat pedantic, but I found the question a bit ambiguous. Is "Door One" actually the first door in the line, or is it simply suggestive of the door you pick, indicating that it could be any of the three doors. Is "Door Two", the door that has one of the worthless junk in it, the door in the middle, or simply one of the doors that doesn't contain the grand prize?

Ok, let's say each door has the number drawn on it.

Rincewind
30-08-2004, 09:28 PM
Personally, I think the answer to this problem can be worked out without knowing a single thing about mathematics, using deductive logic.

I have never met anyone who did not know a single thing about mathematics. Not one who was old enough to speak, anyway. Further to that, Russell and Whitehead (in Principia Mathematica) almost proved that mathematics is just one area of deductive logic anyway. :)

I think the main problem with the way this puzzle has been expressed is it is not clear whether the host's actions are entirely predictable. It should be clear that he will ALWAYS show a door without a prize after the first choice is made and offer the contestant the option of changing their pick. If he only does it sometimes or if this a once-off experiment, then you can say less about the probabilities.

To demonstrate this idea, consider what happens if the host behaves this way:

When the contestant has picked correctly he always shows them another door, when they have not picked correctly he never shows them another door.

Obviously if this is the case then you should never change as you only get the option to do so when you have it right. (IE You can only go wrong).

However, if you assume the host always shows a losing door and always gives the cnotestant a chance to change their mind, then you start gettnig the counterintuitive solution you are after.

Javier Gil
30-08-2004, 09:44 PM
The problem has been expressed with as much information as is needed. Whether the master of ceremonies will always show what is behind the door (repetitive behaviour) or if this is a "one time only" event makes no difference whatsoever.
You can ignore the MOC altogether and imagine similar real life situations. You could even program a computer to behave unpredictably when this situation arises, and that wouldn't change a thing.
No extra information is needed.

Rincewind
30-08-2004, 10:05 PM
The problem has been expressed with as much information as is needed. Whether the master of ceremonies will always show what is behind the door (repetitive behaviour) or if this is a "one time only" event makes no difference whatsoever.
You can ignore the MOC altogether and imagine similar real life situations. You could even program a computer to behave unpredictably when this situation arises, and that wouldn't change a thing.
No extra information is needed.

Assuming the host behaves unpredictably is not a counterargument to my point. I was talking about the host behaving predictably but dependant on whether the contestant correct or not with their first pick. Remember the host knows what is behind each door and has the power to act based on that knowledge.

Write a program to simulate the case where the host will always show you a door when you have it correct and only show you a door 50% of the time when you have it wrong. The fact that you have been shown a door means there is a 50/50 chance your first guess was right. Therefore changing will not change your odds of being correct at all.

Javier Gil
30-08-2004, 10:44 PM
Well, what I meant is that the master of ceremonies doesn't care whether you win or lose. The past and future matters not in this situation. There are no hidden interests. Your answer must be based on this situation alone, no additional information is required.
It's pretty much like a chess position seen on a diagram, it doesn't matter who plays it, previous results, etc. there is a best move, and you must find it.
Any hypothetical data will only confuse matters.

Rincewind
30-08-2004, 11:08 PM
Well, what I meant is that the master of ceremonies doesn't care whether you win or lose. The past and future matters not in this situation. There are no hidden interests. Your answer must be based on this situation alone, no additional information is required.
It's pretty much like a chess position seen on a diagram, it doesn't matter who plays it, previous results, etc. there is a best move, and you must find it.
Any hypothetical data will only confuse matters.

Yes, but you have to either state that or assume it. Those familiar with the work of Eddie McGuire from Australian Who Wants to be a Millionaire, might not leap to that assumption. ;)

Basil
11-03-2008, 10:59 PM
And all of this before the arrival of Manga! Here you are mate.

Aaron Guthrie
11-03-2008, 11:04 PM
And all of this before the arrival of Manga! Here you are mate.I have encountered this one before. Can't be bothered reading the thread right now. I will note that usually there is an unstated assumption imported that I do not like. That is that the host won't open a door with the car behind it (or something like that).

Aaron Guthrie
11-03-2008, 11:10 PM
I think the main problem with the way this puzzle has been expressed is it is not clear whether the host's actions are entirely predictable. It should be clear that he will ALWAYS show a door without a prize after the first choice is made and offer the contestant the option of changing their pick. If he only does it sometimes or if this a once-off experiment, then you can say less about the probabilities.Right on.

(You can tell I am tired by the way I didn't make an reply to the Rusell and logic comment.)

Desmond
11-03-2008, 11:13 PM
I voted "It makes no difference", making me the second person to vote in almost 4 years.

Rincewind
11-03-2008, 11:13 PM
Right on.

(You can tell I am tired by the way I didn't make an reply to the Rusell and logic comment.)

K. I did say "almost".

Aaron Guthrie
11-03-2008, 11:16 PM
K. I did say "almost".Alright then, I can't resist. Probably not very controversial comments though.

That it cannot be deduce from a finite set of axioms doesn't show it isn't a subset of logic.

Russell actually tried a tad more than proving maths is logic, he tried proving all is logic! (Via his logical atomism.)

Rincewind
11-03-2008, 11:30 PM
That it cannot be deduce from a finite set of axioms doesn't show it isn't a subset of logic.

True, and only three negatives in that sentence. :)

Miguel
11-03-2008, 11:40 PM
FYI, Jason Rosenhouse over at ScienceBlogs is in the process of writing a book about the MHP (http://scienceblogs.com/evolutionblog/2007/11/ancestral_monty.php).

TheJoker
12-03-2008, 12:32 AM
I think the main problem with the way this puzzle has been expressed is it is not clear whether the host's actions are entirely predictable. It should be clear that he will ALWAYS show a door without a prize after the first choice is made and offer the contestant the option of changing their pick. If he only does it sometimes or if this a once-off experiment, then you can say less about the probabilities.

Actually the event only need be a once-off as the outcome probabilities for that particular event are fixed and independant of any subsequent games played.

The additional information might change your strategy (i.e. if the host only offers the option when the player first choice is right) but in the absence of such information your strategy should always be to switch if you want to increase your chance of winning.

Aaron Guthrie
12-03-2008, 03:43 AM
The additional information might change your strategy (i.e. if the host only offers the option when the player first choice is right) but in the absence of such information your strategy should always be to switch if you want to increase your chance of winning.If the host acts randomly, i.e. just opens a door at random, switching doesn't help. If the host acts according to some rule set (knows therefore deceives, knows therefore makes sure not to show the car etc.), you have a bunch of rule sets to choose from. So two questions A) how do you assign probabilities to the chances of each rule set occurring? B) if you can answer A, where is your proof that this leads to switching increasing the chances?

Rincewind
12-03-2008, 07:57 AM
Actually the event only need be a once-off as the outcome probabilities for that particular event are fixed and independant of any subsequent games played.

If the host doesn't always do it then you start to think, maybe he is doing this to trick me.

The host also has to know that he is showing you an empty door. If the host chooses at random and only happens to choose an empty door in this case then there is no advantage in switching.

Garrett
12-03-2008, 07:57 AM
I have no idea what the correct answer to the problem is because I know nothing of the MC's motives (assuming he has a complete poker face - or no face at all).

As far as I am concerend, no matter which door I originally chose, the MC could open a garbage door.

To my mind the chance has simply reduced from 1 in 3 to 1 in 2.

I wouldn't be bothered changing my mind, I would rather miss out through an incorrect guess then by procrastination and indecision.

Cheers
Garrett.

Rincewind
12-03-2008, 08:01 AM
I wouldn't be bothered changing my mind, I would rather miss out through an incorrect guess then by procrastination and indecision.

I feel the same way about exams. That's why I never check my work or change answers once I've written them down. ;)

Garrett
12-03-2008, 08:09 AM
I feel the same way about exams. That's why I never check my work or change answers once I've written them down. ;)

Yes I was like that at Uni too.

I changed when I entered the workforce though.

85% at Uni gets you a HD (7).

85% at work get you a kick in the arse.

Cheers
Garrett.

Rincewind
12-03-2008, 08:28 AM
85% at Uni gets you a HD (7).

85% at work get you a kick in the arse.

I guess that is a measure of how intellectually challenging work ismeant to be.

At school or uni you are trying to grade people so you design the exams so only a few people are capable of getting >85%. If everyone got a HD the exam would be useless.

At work people are given tasks (it is believed) they are capable of achieving. To give them anything more is counterproductive.

Basil
12-03-2008, 09:41 AM
If everyone got a HD the exam would be useless.
Indeed. HDs are special and those that have access are most fortunate ;)

TheJoker
12-03-2008, 10:44 AM
The host also has to know that he is showing you an empty door. If the host chooses at random and only happens to choose an empty door in this case then there is no advantage in switching.

I believe this statement is incorrect. Because in the cases that the host reveals the door containing the prize both strategies lose if you are not allowed to switch to the revealed prize, or if you are allowed switch then switching always wins when the host reveals the prize.

The only requirement is that the host randomly chooses a door other than your initial selection.

There a three possible positions for the prize A, B or C
There are two possible outcomes of a random selection by the host.
1. The host reveals an empty door
2. The host reveals the prize

Assume you choose Door A (and you are not allowed to switch to the door revealed by the host)


Here are the possible scenarios:


Prize behind A
Host reveals an empty door: Switch = L, Stay = W
Host reveals prize: N/A (host cannot choose your door)

Prize behind B
Host reveals an empty door: Switch = W, Stay = L
Host reveals prize: Switch = L, Stay = L

Prize behind C
Host reveals an empty door: Switch = W, Stay = L
Host reveals prize: Switch = L, Stay = L


Switching results in a win in 2 of the 5 scenarios.
Staying results in a win in 1 of the 5 scenarios.

Therefore it is not important that the host knows where the prize is.

Kevin Bonham
12-03-2008, 11:34 AM
My answer was:

I would split.

The original guess has a 1/3 chance of being right. This is not changed by the turning of an empty box because whether the guess is right or wrong, an empty box can always be turned.

However if the original guess is wrong (2/3 chance) then the remaining door must be correct.

The only ambiguity in the question that matters is that it is not stated whether the location of the prize is determined randomly. It could, for instance, be determined based on psychology - that people were more likely to choose particular numbers out of 1,2,3 than others. For a better wording the door which the prize is behind should be chosen randomly by the show, and there is no need to specify which number door is chosen.

However Rincewind's reservations about the wording are quite correct and the ambiguity I mentioned is not the only ambiguity that matters at all!

TheJoker
12-03-2008, 12:30 PM
The only ambiguity in the question that matters is that it is not stated whether the location of the prize is determined randomly. It could, for instance, be determined based on psychology - that people were more likely to choose particular numbers out of 1,2,3 than others. For a better wording the door which the prize is behind should be chosen randomly by the show, and there is no need to specify which number door is chosen.[/i]

However Rincewind's reservations about the wording are quite correct and the ambiguity I mentioned is not the only ambiguity that matters at all!

Considering this information is unknown (e.g. is the prize biased in its positioning) it can not be considered, therefore based on the available information a person can deduce that the best strategy is to switch.

Note that empirical studies of the Monty Hall's paradox show that switiching is indeed a more profitable strategy.

Capablanca-Fan
12-03-2008, 12:58 PM
Considering this information is unknown (e.g. is the prize biased in its positioning) it can not be considered, therefore based on the available information a person can deduce that the best strategy is to switch.

Note that empirical studies of the Monty Hall's paradox show that switiching is indeed a more profitable strategy.
Yes. I explained this a few years ago to some incredulous American former colleagues, and they tested it many times, and found that switching really did approximately double their winning chances over a large number of tries. I didn't know it was called Monty Hall though; just heard it as the three door problem.

Garrett
12-03-2008, 01:45 PM
Wow

Nice puzzle !!

Got me - and I even thought about it for a little while.

Thanks !

Garrett
12-03-2008, 03:05 PM
I guess that is a measure of how intellectually challenging work ismeant to be.

At school or uni you are trying to grade people so you design the exams so only a few people are capable of getting >85%. If everyone got a HD the exam would be useless.

At work people are given tasks (it is believed) they are capable of achieving. To give them anything more is counterproductive.

Hi Rincewind

What I meant was that at Uni I stopped when I thought I'd done 'good enough'.

The standard of work which got me HD's at Uni would get me a kick in the pants now (computer programmer).

Desmond
12-03-2008, 07:19 PM
Yes. I explained this a few years ago to some incredulous American former colleagues, and they tested it many times, and found that switching really did approximately double their winning chances over a large number of tries. I didn't know it was called Monty Hall though; just heard it as the three door problem.
:hmm: I don't get it. :doh:

After the host has revealed the booby prize, we have 2 doors, one of which has the prize, one of which does not. Surely the odds are even either way.

Capablanca-Fan
12-03-2008, 07:47 PM
Here are the three doors, with the probabilities of 1/3 each:

A 1/3
B 1/3
C 1/3

S'pose ya pick A. You have 1/3 chance of being right. Then imagine you were asked whether to change to look in both B&C. Then you would probably pick them, since the chance of it behind one of the two is 2/3:

A 1/3
B or C 2/3

Now in the problem, the host opens door B to reveal nothing. It's important to note that this does nothing to promote p(a) from 1/3 to 1/2 — it still stays as 1/3. And neither has it done anything to demote p(B or C) from 2/3 to 1/2. Rather, all the host has done is eliminate one of the two, although both of them put together still have p = 2/3. So the probabilities become:

A 1/3
B 0
C 2/3

It might be easier to see if it were done backwards, which makes no difference to the calculations. That is, being allowed to pick both B&C *before* the host opens the door to B. So picking C has exactly the same chance of being right as picking both B and C in the case above.

The situation can be made clearer if, say, you had a bag of 100 marbles, only one of which was black and the others were white. You are asked to pick the black one blindfolded to win a prize. After you make your choice, you are given the chance to pick all the other 99. Presumably you would. And nothing would be changed whether you were shown that the other 98 were all white or not, or whether before or after.

Here is a global experiment (http://delicategeniusblog.com/?page_id=134) which you can take part in, and which the score of >13,000 swaps is about 2–1, while the is about 2/3–1/3 in favour of swappers. I swapped this one (http://www.userpages.de/monty_hall_problem/) 50 times and won 33–17.

Kevin Bonham
12-03-2008, 07:59 PM
Here is a global experiment (http://delicategeniusblog.com/?page_id=134) which you can take part in, and which the score of >13,000 swaps is about 2–1, while the is about 2/3–1/3 in favour of swappers. I swapped this one (http://www.userpages.de/monty_hall_problem/) 50 times and won 33–17.

On the latter I won the first five in a row by staying but by 100 goes (it doesn't take long to do 100 goes by repeatedly clicking on the same square) it was down to 39%.

Rincewind
12-03-2008, 08:12 PM
Prize behind A
Host reveals an empty door: Switch = L, Stay = W
Host reveals prize: N/A (host cannot choose your door)

Prize behind B
Host reveals an empty door: Switch = W, Stay = L
Host reveals prize: Switch = L, Stay = L

Prize behind C
Host reveals an empty door: Switch = W, Stay = L
Host reveals prize: Switch = L, Stay = L


Switching results in a win in 2 of the 5 scenarios.
Staying results in a win in 1 of the 5 scenarios.

Therefore it is not important that the host knows where the prize is.

This analysis is incorrect. There are actually 6 options and the way to view it is this.

Lets say you always pick door A and the host picks randomly door B or door C.

Prize behind A : Host picks B : Switch = L, Stay = W
Prize behind A : Host picks C : Switch = L, Stay = W
Prize behind B : Host picks B : Automatic loss
Prize behind B : Host picks C : Switch = W, Stay = L
Prize behind C : Host picks B : Switch = W, Stay = L
Prize behind C : Host picks C : Automatic loss

So Switch wins 1/3 of the time, Stay wins 1/3 of the time and the host wins by default 1/3 of the time. Therefore if the host is choosing randomly there is no advantage to swapping.

In your solution you collapse case #1 and #2 into one case which is not valid as it is not of equivalent likelihood to the other cases you consider (it is actually twice as likely).

Rincewind
12-03-2008, 08:17 PM
:hmm: I don't get it. :doh:

After the host has revealed the booby prize, we have 2 doors, one of which has the prize, one of which does not. Surely the odds are even either way.

Another way of thinking about it is by open specifically a booby door he has given you information that is the information which improves your chance to 2/3 if you change in light of the change. When the host doesn't know what he is doing and randomly shows you a door which happens to be empty, your chances don't improve because there was a 1/3 chance he would have given the game away by accidentally revealing the prize.

Don't worry if this doesn't make sense, Jono's explanation is probably better.

Aaron Guthrie
12-03-2008, 08:51 PM
This analysis is incorrect. There are actually 6 options and the way to view it is this.

Lets say you always pick door A and the host picks randomly door B or door C.

Prize behind A : Host picks B : Switch = L, Stay = W
Prize behind A : Host picks C : Switch = L, Stay = W
Prize behind B : Host picks B : Automatic loss
Prize behind B : Host picks C : Switch = W, Stay = L
Prize behind C : Host picks B : Switch = W, Stay = L
Prize behind C : Host picks C : Automatic loss

So Switch wins 1/3 of the time, Stay wins 1/3 of the time and the host wins by default 1/3 of the time. Therefore if the host is choosing randomly there is no advantage to swapping.

In your solution you collapse case #1 and #2 into one case which is not valid as it is not of equivalent likelihood to the other cases you consider (it is actually twice as likely).Just to note that in this specific case, where he didn't show the pirze, you don't need to worry about what happens if the host shows the prize. That is to say, it doesn't matter if the host showing the prize wins or loses or resets the game, the odds of winning if switching when he doesn't are the same. (That is, the odds for switching or staying are the same.)

Rincewind
12-03-2008, 09:00 PM
Just to note that in this specific case, where he didn't show the pirze, you don't need to worry about what happens if the host shows the prize. That is to say, it doesn't matter if showing the prize wins or loses or resets the game, the odds of winning if switching when he doesn't are the same. (That is, the odds of for switching or staying are the same.)

Yep. That was my original meaning when I said

The host also has to know that he is showing you an empty door. If the host chooses at random and only happens to choose an empty door in this case then there is no advantage in switching.

Since TheJoker's way of accounting made no difference I stuck to his layout.

TheJoker
12-03-2008, 09:07 PM
This analysis is incorrect. There are actually 6 options.

Your right:doh:

The situation I originally had in mind was that the host is choosing randomly but has already revealed a junk door. As in the stated problem the junk door has been revealed.


See if you agree wwith me here.

Again assume we choose A

Here are the possible scenarios that result in the host revealing a "junk door"

Prize behind A : Host picks B : Switch = L, Stay = W
Prize behind A : Host picks C : Switch = L, Stay = W
Prize behind B : Host picks C : Switch = W, Stay = L
Prize behind C : Host picks B : Switch = W, Stay = L

Since the prize will appear behind each door with a prob of 1/3 the two options with prize behind A each have a 1/6 prob.

So even if the host was chosing randomly the fact that he revealed a junk door means you should switch. Am I right?

TheJoker
12-03-2008, 09:14 PM
Just to note that in this specific case, where he didn't show the pirze, you don't need to worry about what happens if the host shows the prize. That is to say, it doesn't matter if the host showing the prize wins or loses or resets the game, the odds of winning if switching when he doesn't are the same. (That is, the odds for switching or staying are the same.)

Thank Manga that was my initial thoughts that once the host had already revealed a blank door it did not matter if it was the result of a random choice or knowledge based choice, switching was still more likely to produce a winning result.

Somehow between that idea an my solution I got totally lost :lol:

Rincewind
12-03-2008, 09:22 PM
Your right:doh:

The situation I originally had in mind was that the host is choosing randomly but has already revealed a junk door. As in the stated problem the junk door has been revealed.


See if you agree wwith me here.

Again assume we choose A

Here are the possible scenarios that result in the host revealing a "junk door"

Prize behind A : Host picks B : Switch = L, Stay = W
Prize behind A : Host picks C : Switch = L, Stay = W
Prize behind B : Host picks C : Switch = W, Stay = L
Prize behind C : Host picks B : Switch = W, Stay = L

Since the prize will appear behind each door with a prob of 1/3 the two options with prize behind A each have a 1/6 prob.

So even if the host was chosing randomly the fact that he revealed a junk door means you should switch. Am I right?

Nope. If the host chooses randomly and just happens to not pick the prize door then there are four possibilities. You have a 50/50 regardless if you switch of if you stay.

Another way of looking at it is every door had a 1 in 3 chance before the host took a pot shot. After the pot shot the chances of the other doors having the prize dropped (from 1 in 3 to 1 in 6). The original door preserved it original odds of 1 in 3. So your original door is 1 in 3 and the combined chance of the other unopened door is now 1 in 3. Hence it makes no difference which strategy you choose.

Basil
12-03-2008, 09:25 PM
Can we work clocks into this thread?

Capablanca-Fan
12-03-2008, 09:31 PM
On the latter I won the first five in a row by staying but by 100 goes (it doesn't take long to do 100 goes by repeatedly clicking on the same square) it was down to 39%.
I just got 67-33 by switching, but the percentage was in the mid 70s before.

Desmond
12-03-2008, 09:34 PM
It's important to note that this does nothing to promote p(a) from 1/3 to 1/2 — it still stays as 1/3. And neither has it done anything to demote p(B or C) from 2/3 to 1/2. Rather, all the host has done is eliminate one of the two, although both of them put together still have p = 2/3. I don't follow the rationale behind this part. Why shouldn't the probabilities change? The are now only 2 doors behind which the prize can be.

Aaron Guthrie
12-03-2008, 10:02 PM
Here is my explanation of what is happening in the switching is always best case.

Assume the host must choose not your door, and not a prize door.

2/3 you didn't choose the prize. In this case the host has only one choice, the non-prize door. This means the other door that you didn't choose must have the prize, so you win if you switch.

So 2/3 chance you win if you switch (I won't bother with the 1/3 chance you did pick the prize to begin with).

Aaron Guthrie
12-03-2008, 10:06 PM
And here is my explanation of the tricky bit of the host chooses randomly (but again assume he doesn't choose your door).

2/3 you didn't chose the prize. If so, host has 1/2 chance of showing non-prize
1/3 you did choose the prize. If so, host has 2/2 chance of showing non-prize

(Italics really don't work well for numbers.)

Rincewind
12-03-2008, 10:13 PM
I don't follow the rationale behind this part. Why shouldn't the probabilities change? The are now only 2 doors behind which the prize can be.

Ok. If you think it makes no difference imagine there are 100 doors not just 3.

You pick a door and then the host reveals that 98 of the doors you didn't pick didn't have the prize. Now there are just 2 doors, the one you originally picked (1/100 chance, right?) and the other door the host has not revealed. What do you think the odds are in this case?

TheJoker
12-03-2008, 11:58 PM
Nope. If the host chooses randomly and just happens to not pick the prize door then there are four possibilities. You have a 50/50 regardless if you switch of if you stay.

Another way of looking at it is every door had a 1 in 3 chance before the host took a pot shot. After the pot shot the chances of the other doors having the prize dropped (from 1 in 3 to 1 in 6). The original door preserved it original odds of 1 in 3. So your original door is 1 in 3 and the combined chance of the other unopened door is now 1 in 3. Hence it makes no difference which strategy you choose.

Original door has a probablility of 1/3 of conatining the prize and the remaining doors have a combined porability of 2/3, after the host revelas a junk door it has a probability of zero (it cannot possibly contain the prize). Since the prize is now behind one of the remaining doors they must have a total probability equaling 1 (not 2/3 as in your solution). Since the initial door retains its 1/3 probability the other door must have a probability of 2/3.

I'll reiterate that once the host door has been revealed as junk you should always switch.

Assume initial selection door A

Prize Behind A (1/3)

Host reveals junk door B (1/3*1/2) = 1/6
Host reveals junk door B (1/3*1/2) = 1/6
Host reveals junk door C (1/3*1/2) = 1/6
Swaping = loss

Prize Behind B (1/3)

Host reveals junk door C (1/3 * 1)
Revealing junk door A is not an option as the problem state the host reveals one of the other doors (0)
Revealing door B is not an option as the problem states the host reveals a junk door (0)
Swapping = win

Prize Behind C (1/3)

Host reveals junk door B (1/3 * 1)
Revealing junk door A is not an option as the problem state the host reveals one of the other doors (0)
Revealing door B is not an option as the problem states the host reveals a junk door (0)
Swapping = win


In any instance wheere the host has reveals a junk door the best option is to switch regardless of whether his decision was random or not.

Rincewind
13-03-2008, 01:03 AM
In any instance wheere the host has reveals a junk door the best option is to switch regardless of whether his decision was random or not.

No you are still missing the point. By excluding the options where the host MIGHT reveal the prize you are working out the probabilities as if the host is not acting randomly and so coming up with the same probabilities.

If the host acts randomly he has a certain possibility of spoiling the game by revealing the surprise. The fact that he did was chance and did not tell us anything more about the location of the prize. If he acts with knowledge and specifically shows us an non-winning door then that is the information which brings the switch strategy up to 2/3 probability.

In both cases there are six possibilities as outlined in my first reply to you. The reason I expressed the probability as 1 in 3 and 1 in 3 is because there is a final 1 in 3 of the host picking the winning door. The fact that it is excluded by the way the puzzle is worded does not invalidate the possibility at the time the host picked the door.

If you want to test this you could write an experimental program.

1. Designate a winning door at random
2. Pick a door at random for the contestant.
3. Pick one of the remaining two doors at random for the host.
4. If host picked winning door go to 1
5. Increment total attempt count
6. If contestant door is winning increment stay counter
7. If unopened door is winning increment swap counter
8. if total attempt is less than sample size go to 1
9. end

You should get a roughly 50/50 split on the stay/switch counters.

Compare this to

1. Designate a prize door at random
2. Pick a door at random for the contestant.
3. If contestant door has the prize pick one of the remaining two doors at random for the host, else pick the specific no-prize door for the host
4. Increment total attempt count
5. If contestant door is winning increment stay counter
6. If unopened door is winning increment swap counter
7. if total attempt is less than sample size go to 1
8. end

In this case you should get a 33/67 split on the stay/switch counters

Rincewind
13-03-2008, 01:19 AM
To illustrate this point I have written two programs. Firstly MHP-random implements the first algorithm.


#include <stdio.h>
#include <stdlib.h>
#include <time.h>

main()
{
int prize, contestant, host, total=0, ctrstay=0, ctrswitch=0;

srand( time(NULL) );

while (total < 1000)
{
prize = rand() % 3;
contestant = rand() %3;
host = rand() % 2;
if (host >= contestant)
host++;
if (host != prize)
{
total++;
if (contestant == prize)
ctrstay++;
else
ctrswitch++;
}
}
printf("Stay: %d; Switch: %d; Total: %d",ctrstay,ctrswitch,total);
}

Five runs of this code produces the following output...


Stay: 488; Switch: 512; Total: 1000
Stay: 488; Switch: 512; Total: 1000
Stay: 533; Switch: 467; Total: 1000
Stay: 512; Switch: 488; Total: 1000
Stay: 512; Switch: 488; Total: 1000

Pretty close to 50/50 considering rand() is probably a fairly unsophisticated RNG.

Then I implemented the classic MHP using the following code


#include <stdio.h>
#include <stdlib.h>
#include <time.h>

main()
{
int prize, contestant, host, total=0, ctrstay=0, ctrswitch=0;

srand( time(NULL) );

while (total < 1000)
{
prize = rand() % 3;
contestant = rand() %3;
if (contestant == prize)
{
host = rand() % 2;
if (host >= contestant)
host++;
}
else
for (int i = 0; i < 3; i++)
if (prize != i && contestant != i)
host = i;

total++;
if (contestant == prize)
ctrstay++;
else
ctrswitch++;
}
printf("Stay: %d; Switch: %d; Total: %d",ctrstay,ctrswitch,total);
}

The results of 5 consecutive runs were


Stay: 334; Switch: 666; Total: 1000
Stay: 334; Switch: 666; Total: 1000
Stay: 357; Switch: 643; Total: 1000
Stay: 336; Switch: 664; Total: 1000
Stay: 336; Switch: 664; Total: 1000

I think this illustrates the both variants of the problem and also why the fact that the host has to be acting with knowledge about which door contains the prize for the classic interpretation to hold.

Aaron Guthrie
13-03-2008, 04:00 AM
I think this illustrates the both variants of the problem and also why the fact that the host has to be acting with knowledge about which door contains the prize for the classic interpretation to hold.That is the same thing I thought about my 48 and 49. ;)

I'm gonna try again with my explanation of the random host. This time I will say that the game is abandoned if the host shows the prize, this just simplifies the wording, but makes no real difference otherwise.

i) 1/3 you pick the prize. If so, 2/2 chance you get the choice of switching. Switching always loses.
ii) 2/3 you pick junk. If so, 1/2 chance you get the choice of switching. Switching always win

So, you are twice as likely to start off by picking junk. But if you pick junk you are only half as likely to get to make the switch/stay choice. This means that if you end up having a choice, you are just as likely to be in a case i situation as a case ii situation. Thus, switching and staying are just as likely to win.

Desmond
13-03-2008, 07:16 AM
Ok. If you think it makes no difference imagine there are 100 doors not just 3.

You pick a door and then the host reveals that 98 of the doors you didn't pick didn't have the prize. Now there are just 2 doors, the one you originally picked (1/100 chance, right?) and the other door the host has not revealed. What do you think the odds are in this case?
OK I think I get it now.

BTW can this be applied to the Deal or No Deal show?

Garrett
13-03-2008, 07:29 AM
OK I think I get it now.

BTW can this be applied to the Deal or No Deal show?

probably not - as no-one knows where the 200k is.

PS - I don't actually watch this show (lame) so not 100% sure of how the show works.

Rincewind
13-03-2008, 08:13 AM
BTW can this be applied to the Deal or No Deal show?

I believe in the deal or no deal show the host (and possibly the bank though I'm not sure on this one as I only caught snippets of the show occasionally) doesn't know which cases contained which amounts. Secondly the contestant is picking all the cases in any case so that part is always happening randomly.

So I don't believe you can apply anything much from the Monty Hall problem.

TheJoker
14-03-2008, 04:53 PM
No you are still missing the point. By excluding the options where the host MIGHT reveal the prize you are working out the probabilities as if the host is not acting randomly and so coming up with the same probabilities.

If the host acts randomly he has a certain possibility of spoiling the game by revealing the surprise. The fact that he did was chance and did not tell us anything more about the location of the prize.

No you are missing my point. You must ignore the cases where the host where the host reveals the prize as they are not applicable to the strategy for the given situation (i.e. the host reveals a junk door).

I agree that if the host is picking randomly the overall result of the game will 50/50 between split and switch. But this only because in some instances the host will reveal the prize.

In our particular case the host has not revealed the prize. If you look at all cases in either of your experiments where the host revelas a junk door and then apply a switching strategy you will find that the switching strategy is twice as successful as staying with your original selection.

Possibilities that could have happened but cleary did not happen should not influence your strategy. You should only apply the the scenarios could result in you current situtation.

For example assume a standard 52-card deck. I randomly draw 2 cards from the deck. In order to win the game I need two draw 2 ACES. There are a total of 1326 different two-card hands including 6 contaning a pair of ACES

I have probability of drawing two ACES of 1/221. I am offered odds of 100-1

I calculate the probabilities and it clear that the game as whole does not favour me so I conclude that my best strategy is not to play.

My freind however has already drawn his first card which is an ACE he asks me if doubling his bet is a good strategy. Of course I say yes as he has a 1/17 chance of drawing another ACE at odds of 100-1.

I don't base my calculation on all the other possible cards he could have drawn with his random drawing process when considering whether the best strategy is to double the bet or not.

Same applies to the 3 doors problem given that the host has revealed a junk door I need not consider that with his random process he may not have opened a junk door.

TheJoker
14-03-2008, 05:04 PM
i) 1/3 you pick the prize. If so, 2/2 chance you get the choice of switching. Switching always loses.
ii) 2/3 you pick junk. If so, 1/2 chance you get the choice of switching. Switching always win.

Where you solution fails is that we have been told that host already opened a junk door. Therefore the probability of being offered the switch is of course 1 in both cases.

The question ask what is the best strategy in the given scenario, not for the game as a whole.

People inncorrectly state that if the situation (host reveals a junk door) resulted as consequence of a random action by the host then switching does not improve your chances in the given scenario.

This is because when trying to prove their case they erroneously include events that could not possibly have given birth to the scenario described (i.e. the host reveals the prize).

In summary regardless of how it come about if the host reveals a junk door you should switch your choice.

Rincewind
14-03-2008, 05:23 PM
No you are missing my point. You must ignore the cases where the host where the host reveals the prize as they are not applicable to the strategy for the given situation (i.e. the host reveals a junk door).

I agree that if the host is picking randomly the over result of the game will 50/50 between split and switch. But this only because in some instances the host will reveal the prize.

Yes that is exactly the problem. The fact that he did not does not tell us anything. After he reveals a junk door randomly the odds have gone up from 1/3 to 1/2 for both doors.


In our particular case the host has not revealed the prize. If you look at all cases in either of your experiments where the host revelas a junk door and then apply a switching strategy you will find that the switching strategy is twice as successful as staying with your original selection.

But you cannot count those cases as wins for the switching strategy as you never get to switch. When you ignore those cases the odds are 1/2 each.


Possibilities that could have happened but cleary did not happen should not influence your strategy. You should only apply the the scenarios could could result in you current situtation.

True but what has happened before does play a role in the probabilities since it is a compound problem. The distinction here is that in the classic case the host decision is informed on the location of the prize in the second case it is not an information choice. It is this information which the host reveals through his choice of door which provides the contestant with the ability to increase his odds by switching. Without any additional information the contestant cannot improve his odds to any better than a guess. As there are then only 2 doors his odds have gone from 1/3 to 1/2 but no more.


For example assume a standard 52-card deck. I randomly draw 2 cards from the deck. In order to win the game I need two draw 2 ACES. There are a total of 1326 different two-card hands including 6 contaning a pair of ACES

I have probability of drawing two ACES of 1/221. I am offered odds of 100-1

In begging I calculate the probabilities and it clear that the game as whole does not favour me so I conclude that my best is not to play.

My freind however has already drawn his first card which is an ACE he asks me if doubling his bet is a good strategy. Of course I say yes as he has a 1/17 chance of drawing another ACE at odds of 100-1.

I don't base my calculation on all the other possible cards he could have drawn with his random drawing process when considering whether the best strategy is to double the bet or not.

Same applies to the 3 doors problem given that the host has revealed a junk door I need not consider that with his random process he may not have opend a junk door.

I fail to see how that example is relevant at all. The point is the present problem is that the host (by choosing randomly) could spoil the game by revealing the prize. The fact that he didn't was just luck and forms a part of the probability calculation. When factored in correctly this give odds of stay: 1/2, switch: 1/2.

If you still disagree I suggest you offer a modification the the algorithm I posted earlier:

1. Designate a winning door at random
2. Pick a door at random for the contestant.
3. Pick one of the remaining two doors at random for the host.
4. If host picked winning door go to 1
5. Increment total attempt count
6. If contestant door is winning increment stay counter
7. If unopened door is winning increment swap counter
8. if total attempt is less than sample size go to 1
9. end

The key feature of this algorithm is that step #4 accounts for the host spoiling the game by starting again. Once the process flow reaches step 5 we are into the scenario of the Monty Hall Problem with a random host as posed. The odds from there are equal Stay: 1/2, Switch: 1/2 as confirmed by an implementation of this algorithm in C++, also given above.

TheJoker
14-03-2008, 07:19 PM
Yes that is exactly the problem. The fact that he did not does not tell us anything. After he reveals a junk door randomly the odds have gone up from 1/3 to 1/2 for both doors.

True but what has happened before does play a role in the probabilities since it is a compound problem. The distinction here is that in the classic case the host decision is informed on the location of the prize in the second case it is not an information choice. It is this information which the host reveals through his choice of door which provides the contestant with the ability to increase his odds by switching. Without any additional information the contestant cannot improve his odds to any better than a guess. As there are then only 2 doors his odds have gone from 1/3 to 1/2 but no more.

You are right I think my workmates might have slipped some crack in my coffee:uhoh:

However I can propose a change to the game where the host can have a random strategy and the switching can still be the best strategy, but it's kinda cheating:lol:

1. Designate a winning door at random
2. Pick a door at random for the contestant.
3. Pick one of the remaining two doors at random for the host.
4. If host picked winning door go to 2
5. Increment total attempt count
6. If contestant door is winning increment stay counter
7. If unopened door is winning increment swap counter
8. if total attempt is less than sample size go to 1
9. end

Aaron Guthrie
14-03-2008, 09:12 PM
Where you solution fails is that we have been told that host already opened a junk door.And from this we can work out that our chances of being in case i (you started by picking the prize) are 1/2, and our chances of being in ii (you didn't start by picking the prize) are also 1/2.
People inncorrectly state that if the situation (host reveals a junk door) resulted as consequence of a random action by the host then switching does not improve your chances in the given scenario.If the host was forced to choose a non-prize door, the chances of being in i (you started by picking the prize) are 1/3, and the chance we are in ii (you didn't start by picking the prize) are 2/3.

Aaron Guthrie
14-03-2008, 09:42 PM
However I can propose a change to the game where the host can have a random strategy and the switching can still be the best strategy, but it's kinda cheating:lol:

snip...

4. If host picked winning door go to 2I don't think this changes things at all. If the go to was to step 3, then this would change things to the same odds as host knowing. This is as he would be forced to choose a non-prize.

Rincewind
14-03-2008, 11:11 PM
I don't think this changes things at all. If the go to was to step 3, then this would change things to the same odds as host knowing. This is as he would be forced to choose a non-prize.

I agree. Go to step 3 is basically the same as the classic MHP. Go to step 2 is the same as the random MHP. The point is what happens after the contestant has chosen his door. If the contestant gets to choose his door again then we are effectively in a new run of the whole test.

One reason for this is that step 1 of the algorithm is really superfluous. We could set the winning door to the same value each time and it would make no difference since the contestant and host always act at random.

TheJoker
15-03-2008, 10:50 AM
I don't think this changes things at all. If the go to was to step 3, then this would change things to the same odds as host knowing. This is as he would be forced to choose a non-prize.

Yep you're right another stuff up

Capablanca-Fan
03-11-2015, 09:03 AM
Walter T. Herbranson and Julia Schroeder
Are Birds Smarter Than Mathematicians? Pigeons (Columba livia) Perform Optimally on a Version of the Monty Hall Dilemma (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3086893/)
J Comp Psychol. 124(1): 1–13, Feb 2010 | doi:10.1037/a0017703

Abstract

The “Monty Hall Dilemma” (MHD) is a well known probability puzzle in which a player tries to guess which of three doors conceals a desirable prize. After an initial choice is made, one of the remaining doors is opened, revealing no prize. The player is then given the option of staying with their initial guess or switching to the other unopened door. Most people opt to stay with their initial guess, despite the fact that switching doubles the probability of winning. A series of experiments investigated whether pigeons (Columba livia), like most humans, would fail to maximize their expected winnings in a version of the MHD. Birds completed multiple trials of a standard MHD, with the three response keys in an operant chamber serving as the three doors and access to mixed grain as the prize. Across experiments, the probability of gaining reinforcement for switching and staying was manipulated, and birds adjusted their probability of switching and staying to approximate the optimal strategy. Replication of the procedure with human participants showed that humans failed to adopt optimal strategies, even with extensive training.

Rincewind
03-11-2015, 11:34 AM
Part of the issue with human behaviour could be that they are too smart. Humans are reasonably good at adapt strategies from one context to another and if the reason humans stick with the initial guess is because they are suspicious of being tricked by the experiment designer then even after training on the experiment protocol may be insufficient to overcome the more ingrain general suspicion of deals too good to be true.