Is Switching the Best Option in the Door Game Show Dilemma?

[kentm]

The movie "21" went over a statistics problem in an attempt to make the main character look smart. I'm not sure though that the Solution was correct, even though it does seem so.

The problem is this: A game show host gives you a selection of 3 doors, two of which enclose goats, and the other encloses a car. Given that you selected the first door, the game show host reveals that door number 3 is a goat. He then offers you the option of changing your choice to door number 2.

The proposed solution: You would want to switch because you're initial chance of being wrong was 66%, but when one of the doors was opened it placed a 66% probability on the 2nd door. It makes sense in theory, but I see a problem with it.

In this situation, you can't place the 66% probability on door number 2, because a conflict arrises if you allow such an algorithm to determine probability. Please, tell me if I am wrong.

Assume the movie is correct and when there is a 2/3 probability of one of two choices being accurate, the probability of either choice becomes 2/3 in the event that the other choice was eliminated as a possibility. That's essentially what the solutions suggests, but look what happens when you apply that concept to doors 1 and 3. Initially the probability of either door 1, or 3 being the door in front of the car is 66%. When Door three is eliminated, there is now a 66% chance that door 1 holds the car. but not both door 1 and 2 can have a 66%; that would be impossible, so the original assumption must not be true.

Am I wrong?

 

[Mute]

This is a well known problem called the "Monty Hall Problem". You can read more about it on the wikipedia article: http://en.wikipedia.org/wiki/Monty_hall_problem

A typical way to get an intuitive feel for the problem is to consider the case of say, 1000 doors. You have a 1/1000 chance of picking the correct door the first time. If the host then opens 998 other incorrect doors, would you switch to the other remaining door that you didn't pick?

In any event, I'm not entirely sure what your objection is. It looks to me like you're saying, "Well, suppose I picked door 1 initially - I conclude a 2/3 chance that it contains the car. But, suppose I'd picked door 1 first, then I'd conclude it had the 2/3 chance of containing the car." That's correct, but you need to realize that in switching your intial choice you've created two different scenarios which lead to the two different conclusions. Of course, the labels really aren't all that important, which appears to be confusing you a little. What the analysis of the problem should say is that no matter what the labels, you have a better chance of winning the car by switching your choice.

 

[nicksauce]

This is known as the Monty Hall Problem.
It is outlined well on wikipedia. http://en.wikipedia.org/wiki/Monty_Hall_problemEdit: Just barely beat me Mute

 

[HallsofIvy]

The movie "21" went over a statistics problem in an attempt to make the main character look smart. I'm not sure though that the Solution was correct, even though it does seem so.

The problem is this: A game show host gives you a selection of 3 doors, two of which enclose goats, and the other encloses a car. Given that you selected the first door, the game show host reveals that door number 3 is a goat. He then offers you the option of changing your choice to door number 2.

The proposed solution: You would want to switch because you're initial chance of being wrong was 66%, but when one of the doors was opened it placed a 66% probability on the 2nd door. It makes sense in theory, but I see a problem with it.

In this situation, you can't place the 66% probability on door number 2, because a conflict arrises if you allow such an algorithm to determine probability. Please, tell me if I am wrong.

Assume the movie is correct and when there is a 2/3 probability of one of two choices being accurate, the probability of either choice becomes 2/3 in the event that the other choice was eliminated as a possibility. That's essentially what the solutions suggests, but look what happens when you apply that concept to doors 1 and 3. Initially the probability of either door 1, or 3 being the door in front of the car is 66%. When Door three is eliminated, there is now a 66% chance that door 1 holds the car.

but not both door 1 and 2 can have a 66%; that would be impossible, so the original assumption must not be true.

That's not what they are saying. What they are saying is that, a priori, each door has probability 1/3 so together there is probability 2/3 that one of those two doors has the prize. Once one has been opened, the entire 2/3 shifts to the door that is left.

Am I wrong?

 

[kentm]

That's not what they are saying. What they are saying is that, a priori, each door has probability 1/3 so together there is probability 2/3 that one of those two doors has the prize. Once one has been opened, the entire 2/3 shifts to the door that is left.

I guess, I am just saying that you could use the same statistics to show that it would be best to stay with your first choice. It's the exact same reasoning as far as I can tell. Each door has a probability of 1/3 so together (door 1 and 3) there is a probability 2/3 that one of those doors has the prize. Once door number three is opened, there is a 2/3 chance that you have selected the correct door.

Obviously, when the reasoning works that way, you end up with a 4/3 chance that the car is in door 1 or 2. It doesn't seem to work.

 

[nicksauce]

Suppose you choose A, and C is shown to have a goat)

Mathematically (| means given)

Before: P(A) = 1/3, P(B or C) = 2/3
After: P(B or C) = P(B or C | C shown to have goat) = P(B | C shown to have goat) = P(B and C shown to have goat) / P(C shown to have goat).

what is P(C shown to have goat)? There are 3 equally likely possibilities for the goats, AB, BC, and AC. In the first case, there is 0 probability, in the second case there is 1/2 probability, and in the third case there is 1 probability. So P(C shown to have goat) is (1/3)*0 + (1/3)*(1/2) + (1/3)*1 = 1/2 (as expected). What is P(B and C shown to have goat)? There are 3 equally likely possibilites for the goats, AB, BC and AC. In the first case, there is 0 probability, in the second case there is 0 probability, and in the third case there is 1 probability, so P(B and C shown to have goat) is (1/3)*0 + (1/3)*0 + (1/3)*1 = 1/3. So P(B | C is shown to have goat) is (1/3)/(1/2)=2/3.

Now what is P(A | C is shown to have goat)? As before, P(C is shown to have goat) is 1/2. What is P(A and C is shown to have goat)? There are 3 possibilities for the goats, AB, BC and AC. In the first case, the probability is 0, in the second case, the probability is the second case is 1/2 and the probability in the third case is 0. So P(A and C is shown to have goat) is 0*1/3 + 1/2*1/3 + 0*1/3 = 1/6. So P(A | C is shown to have goat) = (1/6)/(1/2) = 1/3.

Thus we have mathematically shown that
P(A | C is shown to have goat) = 1/3 and P(B | C is shown to have goat) = 2/3.

Please let me know what, if any, steps in my reasoning you don't understand.

 

[kentm]

Edit: Apparantly I screwed up the quote tag. It actually looks fine to me, but w/e.

Suppose you choose A, and C is shown to have a goat)

Mathematically (| means given)

Before: P(A) = 1/3, P(B or C) = 2/3
After: P(B or C) = P(B or C | C shown to have goat) = P(B | C shown to have goat) = P(B and C shown to have goat) / P(C shown to have goat).

what is P(C shown to have goat)? There are 3 equally likely possibilities for the goats, AB, BC, and AC. In the first case, there is 0 probability, in the second case there is 1/2 probability, and in the third case there is 1 probability. So P(C shown to have goat) is (1/3)*0 + (1/3)*(1/2) + (1/3)*1 = 1/2 (as expected). What is P(B and C shown to have goat)? There are 3 equally likely possibilites for the goats, AB, BC and AC. In the first case, there is 0 probability, in the second case there is 0 probability, and in the third case there is 1 probability, so P(B and C shown to have goat) is (1/3)*0 + (1/3)*0 + (1/3)*1 = 1/3. So P(B | C is shown to have goat) is (1/3)/(1/2)=2/3.

It makes sense that AB would have 0 probability (Not both A and B can be goats). And the probability of BC being goats is (yes) 1/2. We know C is a Goat, but there are two possibilities for B so you are correct, but I don't see why AC is 1. The goat could just as easily still be in B. Anyways, I actually do understand the math. I was never asking for an explanation of why the math is right. I need to know why my method for showing the probability of staying with door 1 is inaccurate. I've tried to word it as closely to the movie, and other people here so that you can see why I think there is no difference in the algorithm. And if my math is accurate, logic would telll us that 2/3 is wrong in this situation.

 

[kentm]

I understand the issue with my thinking now. The movie never really explains the algorithm. Ben simply says, "before my chances wer 33.3%, and now they are 66.7%. When I looked up the solution, I found explanations that act as though the 2/3 combined probability from doors 2 and 3 transfer completely to door 2 when door 3 is closed. That is not the reason it is best to switch. If it were, you could also prove that it would be best not to change your choice (I think I showed why already). The solution is much more simple.
There are three possible Goat/car combinations:

CGG
GCG
GGC

Switching in the first case looses. Switching in the 2nd and 3rd case wins, so switching has a 2/3 chance of wining.

 

[HallsofIvy]

I guess, I am just saying that you could use the same statistics to show that it would be best to stay with your first choice. It's the exact same reasoning as far as I can tell. Each door has a probability of 1/3 so together (door 1 and 3) there is a probability 2/3 that one of those doors has the prize. Once door number three is opened, there is a 2/3 chance that you have selected the correct door.

Obviously, when the reasoning works that way, you end up with a 4/3 chance that the car is in door 1 or 2. It doesn't seem to work.

No, because the MC NEVER opens the door with the prize behind it. You are using the extra knowledge the MC has.

 

[Otropablo]

I agree with Before: P(A) = 1/3, P(B or C) = 2/3

I think, for example, if car is behind A:

P(win) = P(Ben chooses A, host shows C, Ben stays) + P(Ben chooses A, host shows B, Ben stays) + P(Ben chooses B, host shows C, Ben switches) + P(Ben chooses C, host shows B, Ben switches)
P(win) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1*1/2) + P(1/3*1*1/2) = 1/2

and:
P(loose)= P(Ben chooses A, host shows C, Ben switches) + P(Ben chooses A, host shows B, Ben switches) + P(Ben chooses B, host shows C, Ben stays) + P(Ben chooses C, host shows B, Ben stays)
P(loose) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1*1/2) + P(1/3*1*1/2) = 1/2

Repet it for car in B and C and you'll get to P(win)=P(loose)= 1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2

Suppose you choose A, and C is shown to have a goat)

Mathematically (| means given)

Before: P(A) = 1/3, P(B or C) = 2/3
After: P(B or C) = P(B or C | C shown to have goat) = P(B | C shown to have goat) = P(B and C shown to have goat) / P(C shown to have goat).

what is P(C shown to have goat)? There are 3 equally likely possibilities for the goats, AB, BC, and AC. In the first case, there is 0 probability, in the second case there is 1/2 probability, and in the third case there is 1 probability. So P(C shown to have goat) is (1/3)*0 + (1/3)*(1/2) + (1/3)*1 = 1/2 (as expected). What is P(B and C shown to have goat)? There are 3 equally likely possibilites for the goats, AB, BC and AC. In the first case, there is 0 probability, in the second case there is 0 probability, and in the third case there is 1 probability, so P(B and C shown to have goat) is (1/3)*0 + (1/3)*0 + (1/3)*1 = 1/3. So P(B | C is shown to have goat) is (1/3)/(1/2)=2/3.

Now what is P(A | C is shown to have goat)? As before, P(C is shown to have goat) is 1/2. What is P(A and C is shown to have goat)? There are 3 possibilities for the goats, AB, BC and AC. In the first case, the probability is 0, in the second case, the probability is the second case is 1/2 and the probability in the third case is 0. So P(A and C is shown to have goat) is 0*1/3 + 1/2*1/3 + 0*1/3 = 1/6. So P(A | C is shown to have goat) = (1/6)/(1/2) = 1/3.

Thus we have mathematically shown that
P(A | C is shown to have goat) = 1/3 and P(B | C is shown to have goat) = 2/3.

Please let me know what, if any, steps in my reasoning you don't understand.

 

[Otropablo]

HallsofIvy: Why the 2/3 shifts? Didn't you already know that the first door to be opened will be one with a goat? So who cares what door you're already choose?. There is a 50% chance between the door who you just pointed and the one the hosts just pointed. Am I wrong?
I think the host showes you 3 doors and tells you "there's 2 goats", you say "open that one" and he opens another one, with a goat. So you must think: Thats good, now my chance is simply 2 agains 1, without taking chances. Host is basically telling you, before your final (and only one real) determination which door not to choose

Table of truth attached

The rules, if you take the labels from the doors are:

1) You are going to choose between 3 doors
2) Host is going to open a door (not the one you've choose, useless information) and that door will have a goat inside. Then he'll make you choose between the others 2 doors, one of them will have a goat, the other a car.

That's not what they are saying. What they are saying is that, a priori, each door has probability 1/3 so together there is probability 2/3 that one of those two doors has the prize. Once one has been opened, the entire 2/3 shifts to the door that is left.

 

[Otropablo]

MC NEVER opens the door you choose either. So why there's a 1/3 chances of winning? You are afraid of your first choice, when you have 1/3 for car. But MC don't kick you if you choose the goat at fist stage either.

Sorry for being this annoing, but I've read many threads about this, non of them has a clear logical explanation, and many doubt (specially those who simulate it).

No, because the MC NEVER opens the door with the prize behind it. You are using the extra knowledge the MC has.

 

[CRGreathouse]

How about this. You and I simulate this game with dice.* We play 100 times. If the original choice has the car more than 44% of the time, I give you $100; otherwise you give me $100. Sounds like a good deal, right? That is, since you think the original has a 50% chance of having the car.

* Dice: To ensure against cheating, we each roll a 6-sided die. We add the two values and reduce mod 3. A 0 means that the car is behind the original choice, while 1 and 2 are the other choices. If the car is not the original choice the revealed choice is the one out of 1 and 2 that isn't a car; otherwise we roll dice and reduce mod 2, then add one, to get the number of the choice that is revealed. The motivation behind this die-rolling procedure is that if either of us uses a fair die, the result is uniformly random (even if the other is cheating with loaded dice). That way we don't need to trust each other.

 

[Otropablo]

Thanks for your explanation GreatHouse. I've finally get it (thanks to the explanations from physforums members). I was a hard night (i couldn't sleep until I've understood it), but thanks to this community now I'm just a little less ignorant ;).

Thanks for your time, to all of you!

 

[CRGreathouse]

Glad to have helped. It's a tricky problem, for sure -- don't feel bad for taking some time to think it through!

 

[whitesteve22]

I apologize if someone has said this already, I didn't read every post but think of it like this:

-by choosing one door out of three, you have a 1/3 chance of being correct and a 2/3 chance of being wrong

-by opening one incorrect door out of the two you did not choose, you still only have a 1/3 chance of being correct and a 2/3 chance of being wrong

-by switching to the last remaining door the odds then flip flop in your favor

-yes, in the end, only 2 doors remain but the only reason one of them is still there is b/c it was originally your choice and doesn't quite make it a 50/50 chance.

-the example used with the 1000 doors is a great way to think of it as well, the fact that there are only 3 doors makes it what it the hot topic that this has become really.

 

[pallidin]

The problem is this: A game show host gives you a selection of 3 doors, two of which enclose goats, and the other encloses a car. Given that you selected the first door, the game show host reveals that door number 3 is a goat. He then offers you the option of changing your choice to door number 2.

At that point, under those end circumstances, the odds are 50-50.

 

[pallidin]

Because of this, the value then becomes one of extended entertainment using statistical probability.

 

[glen922]

One of three things can happen:
1. player picks car, host shows goat A or goat B, switching loses
2. player picks goat A, host shows goat B, switching wins
3. player picks goat B, host shows goat A, switching wins

2/3 times switching wins. Seems simple.

it's harder to believe this, but it must be true:
1/3 chance initial pick correct
2/3 chance one of unchosen doors is correct
0 chance revealed unchosen door is correct
2/3 - 0 = 2/3 chance of remaining unchosen door being correct.

 

[cpeterson]

You have all done an outstanding job of describing the solution to the problem but unless I missed reading a post (I apologize if this is the case) you have not properly addressed the most important part of this problem: the premises.

Premise 1: The game show host knows what’s behind each door. (Otherwise he has a 1/3 chance to open the door with the car and the statistics change.)

Premise 2: The game show host always opens a door with a goat behind it. (Otherwise he could use the offer to trade only when he felt he had an opportunity to trick you.)

Premise 3: You are always given the opportunity to switch doors. (Otherwise he could only offer you a chance to switch to the remaining goat.)

These 3 premises are what assure you that you get the benefit of the 2 door statistical switch instead of a 50/50 or worse.

 

[nafunkiii]

This is my Doubt.

Suppose door 1 is picked. and the player always switches.

There is a probability of 1/3 that the car is behind either doors.

so if car is behind 1, host can open door 2 or 3 with probability of 1/2.

so switching results in loss of 1/3 (1/6+1/6).

so if car is behind 2, host has to open door 3 with probability of 1.

switching results in a gain of 1/3

same case with door 3
so winning probability is 2/3.

NOW, if the host does not know where the goat is and he opens a random door and finds a goat, the question that arises is, Should the player switch?

considering door 1 is picked. and the player always switches.

if the car is behind door 1. probability of which is 1/3. then host opens either door with probability of 1/2. so in either case he losses. so results in a loss with probability of 1/3.

if car is behind door 2 probability of which is 1/3. probability of choosing the second door is 1/2. so his chance if winning is 1/6

same with door 3.

so. the probability that the player wins after swicthing in this case is 1/3.
So he shouldn't switch.

I know i made a blatant error somewhere which i am not able to pick up. So please help.

 

[nafunkiii]

Another explation to the problem that leads to the right answer

assume car in 1 ,goat in other 2.. the man picks door 1 (host opens 2,3) TWO CASES. mans picks door 2 and host opens three and the other way round..
so in 2 cases out of 4 the man will win the car.. so equal probablity

 

[jerrylam123]

The key to the problem (professor Rosa in 21 clearly states it) is that the game show host knows where the car is. Switching to the other door (which by the way is the total of all remaining doors after the other door reveals a goat) doesn't guarantee you the car but it does increase your chances two fold. If there were a thousand doors you would also increase your chances by switching - i.e. host shows you one door with goat. Do you believe the car is behind the door you picked or one of the remaining 998 doors?!

 

[jp79]

The problem is confusing. On top of being hard to figuere out it's even harder to explain. I thought it was wrong first. When it clicked for me it was this scenario that helped me.

Lamens terms.

You have a 2 out of 3 chance of picking out a goat. Which means after you choose your door
the host has a 2 out of 3 chance of only having 1 door to choose from because 2 out of 3 times the door he can't choose is a car. Which gives you an extra 33.3 percent chance if you switch.

Since the host has to show you a door and only has 2 doors to choose from (because you already chose 1 door with a 66.6% chance you chose a goat) the host has a 66.6% to be forced to choose 1 door (the one with the goat) with the remaining door being the car. His 66% chance of being forced to choose a door gives you an extra 33.3% chance advantage to switch doors to have a 66.6 % chance of being correct.

 

[dennynuke]

I look at this problem more simply. I initially have a 33% chance of finding the car. Regardless of which of the three doors I choose, the host always has a 100% chance of being able to reveal a door with one of the two goats. Because he knows where they are, and regardless of whether I initially choose the car or a goat, there is still one goat available for him to show me.
Once he shows me a door with a goat, then I'm faced with a new choice. It doesn't matter if he knows where the car is, I still don't know, and my choice is now between two doors which as far as I know have an equal chance of having the car. 50/50.

Comments?

 

[pallidin]

Denny, I tend to agree that the "end-result" is 50/50 even though the math, and repeated experiments, shows differently.
I consider myself reasonably intelligent and this STILL messes with me.
Maybe some day I will awake from being brain-dead, but that's my problem.

 

[jp79]

If the host reveals a door randomly (him not knowing where the car is) and he opens the door and reveals a goat it is a 50 % chance either way. Since the judge has to open a door and know's where the car is the 66.6 % chance that he is holding the car falls on the door you are offered to switch to.

 

[CRGreathouse]

If the host reveals a door randomly (him not knowing where the car is) and he opens the door and reveals a goat it is a 50 % chance either way. Since the judge has to open a door and know's where the car is the 66.6 % chance that he is holding the car falls on the door you are offered to switch to.

I don't think so.

You choose door A and the MC shows you door B (without knowing what it is).

1/3 of the time you have the car and you see a goat. 1/3 of the time you have a goat and see a goat. 1/3 of the time you have a goat and see the car.

If you switch, you win 2/3 of the time. (Obviously, you switch to the door that you don't know has a goat behind it.)

 

[jp79]

In reply to CR.

You choose a door thus two doors left. The host has two doors to choose from to reveal. If the host has two doors to choose from not knowing what is behind either of the doors and opens a door with a goat in it. It only leaves two doors left the one you chose and the one with the car in it therefore 50/50 chance.

The variable change is the host does know where the car is and 66.6 % of the time he has the car in one of his two doors. He can't show you the car right.

 

[Mensanator]

Why do you guys keep going on about this ad nauseum?

A simple simulation would clear it up in 5 minutes.

 

[jp79]

Some people can see the simulation and still not understand. Seeing why
and knowing why are two different things.

 

[Gib Z]

My preferred explanation: You pick a door, say Door 1. Note everything is still closed. Door 1 has 1/3 chance. Hence; Doors 2 and 3 have 2/3 chance together. So at this point you agree that if you could somehow pick the option of "Both door 2 and door 3", that would give you 2/3 chance. All opening the door does is allow you to pick the "Both door 2 and door 3" option because Monty opened one of them, now if you open the other, together you opened both. And we agreed before, that this has a 2/3 chance.

 

[Mensanator]

Some people can see the simulation and still not understand. Seeing why
and knowing why are two different things.

Sure, I can see that you still may not understand why, but certainly you must understand that if mathematical reality does not match your premise then your premise is false. No argument, no matter how clever, can change mathematics.

 

[elibj123]

Since you know definitely know what's behind door 3 ( a goat) the following reasoning occurs:

The chance of winning by switching=the chance you've picked a goat in the first place=66%

The chance of losing by switching=the chance you've actually picked the car=33%

 

[Mensanator]

Since you know definitely know what's behind door 3 ( a goat) the following reasoning occurs:

The chance of winning by switching=the chance you've picked a goat in the first place=66%

The chance of losing by switching=the chance you've actually picked the car=33%

Yet people STILL think the probability is 50/50.

 

[vector03]

As I understand probability, if I can "count" the TOTAL possible ways the game can play out and I can "count" the TOTAL number of ways to WIN (or lose), If I divide the number of ways to win by the total number of ways the game can be played then I would be able to calculate exactly the theoretical probability of winning.

Making a table of the 24 possible ways the game can be played out and then counting the 12 ways in which the game can be won, I would suggest the probability of a random selection of winning is 0.50.

 

[Indefinite]

As I understand probability, if I can "count" the TOTAL possible ways the game can play out and I can "count" the TOTAL number of ways to WIN (or lose), If I divide the number of ways to win by the total number of ways the game can be played then I would be able to calculate exactly the theoretical probability of winning.

Making a table of the 24 possible ways the game can be played out and then counting the 12 ways in which the game can be won, I would suggest the probability of a random selection of winning is 0.50.

You are suggesting that if one selects a single door at random from three doors, one of which contains a car, that the theoretical probability of selecting the car is 0.5.

The fact that the host eventually opens one of the doors containing a goat does not alter the fact that there is a 1/3 probability that you will select the door that hides the car.

Surely, you can agree that if the host does not open one of the doors and provide the opportunity for switching, there is a 1/3 probability that you have selected the winning door.

If you accept this, but still assert that there is a .5 probability that you have selected the correct door, you are asserting that if the host opens a door that he knows to contain a goat after you have made your selection, the probability that you initially made the winning selection increases.

If we accept your assertion that after the goat has been revealed, the probability that your door is the winning door increases to .5, then we would also expect that in a very large number of trials, you would likely win more often if the host opens a door after you have made your selection than if he does not, even if you stay with your selection after he reveals one of the goats.

This result is clearly absurd. If you stay with your initial selection, the fact that the host shows one of the goats will not change the frequency with which you make the correct selection.

While the "paradox" may appear initially counter-intuitive, the notion that if the host opens a door that he knows to contain a goat, the probability that you made the correct selection will increase to .5 should seem even more absurd.

We know that if the host never opens the door after you make your selection, and you have no opportunity to switch, your chances of winning are 1/3. How could the event of the host opening a door he knows not to be the winning door after you make your selection possibly increase your chances of winning? Would you really expect to win more often if you make your selection, and choose to stay with that selection after the host opens the door hiding a goat, simply because the host opened the door?

Suppose you roll a die, but do not look, and I look to see what it is. Suppose you guess what number was rolled, and I tell you which number it is not, and ask you if you wish to switch your guess. Suppose you choose not to switch. Will the fact that I told you which number it is not in any way increase the probability that you initially guessed right? Will the probability that you initially guessed right increase from 1/6 to 1/5 simply because I told you something I know after you guessed? Will the event of me telling you something I know after you guess cause you to guess correctly more often if you do not ever change your guess?

 

[pallidin]

I like that explanation. Very clear.

 

[vector03]

You are suggesting that if one selects a single door at random from three doors, one of which contains a car, that the theoretical probability of selecting the car is 0.5.

Not actually suggesting what you have stated above.
Suppose the player makes his door choice and the "host" opens a losing door. The player decides to switch doors as recommended.

Now suppose that a new player walks in and knows nothing about what has transpired between the host and original player. The new player simply sees two doors. The host asks the new player to make a choice of doors as well.

According to previous assertions, the original player has 2/3 chance of winning while the new player has a 1/2 chance of winning. What I'm suggesting is that one player cannot have a 2/3 chance of winning at the same time another player has a 1/2 chance of winning given exactly the same set of circumstances.

This implies that a "new" game ensues after the host opens a door and the chance of winning is simply changed from 1/3 to 1/2.

 

[trambolin]

That's not correct, you have caused the door open by the host, since you made a choice. That's why we need the premise of having the host aware of the things behind the doors.

The newcomer will have a new game, but not your game since he did not cause the open door to open.

Think it with the 1000 doors or lottery tickets nationwide. If I show you all the numbers except the last one. it might happen that all of them are the same but you don't know about the last one. Because you did not know you will miss it by one number initially. Your initial chance to win does not change. You still hold the ticket that you bought with a very small chance of winning.

Now I say, I am going to remove all the the dummy tickets and leave you with one. You are standing there with your ticket and slim chance. But the one I have left has a big chance because all the possibilities of the dummy tickets are stacked on that ticket. Why? Because you might as well picked up a dummy one which is very likely to happen...

 

[vector03]

If I understand, Your suggesting it is possible for 2 different players at a single point in time given an equal set cirumstances to have 2 different probabilities of winning?

 

[trambolin]

No, I am saying that one of them is cheating since there were 3 doors for one and afterwards 2 doors for the other.

 

[vector03]

Considering that point in time, what are the probabilities for both the original player and the new player?

 

[trambolin]

That's crux of this question!

The newcomer has %50.

The original player has 1/3 as he started but given a chance to modify the odds or switch sides. Moreover, they reduced the options down to one. So if I had 1/3, and I have one more option to choose, then that option has to have the probability 2/3.

That's why it is neither a multiplication of probabilities nor resetting to %50 after he made one decision. Hence, a "dice has no memory" argument does not apply here. Because the event is not repeated it is altered.

 

[vector03]

We're almost there... :-)
The new player has 1/2 chance of winning, it seems we agree.
What is the probability of winning for the original player at the point in time under consideration?

 

[trambolin]

Do you understand what I wrote there?

 

[jgens]

If we accept your assertion that after the goat has been revealed, the probability that your door is the winning door increases to .5, then we would also expect that in a very large number of trials, you would likely win more often if the host opens a door after you have made your selection than if he does not, even if you stay with your selection after he reveals one of the goats.

vector03, I suggest that you read this part very carefully.

This implies that a "new" game ensues after the host opens a door and the chance of winning is simply changed from 1/3 to 1/2.

This isn't true. Just because the host opened a door to reveal a goat doesn't mean that we can ignore the circumstances before it.

 

[vector03]

Do you understand what I wrote there?

Yes. I understand all the explanations presented and I understand the mathematics behind the calculation. Each one who has attempted an explanation has presented what I would consider a reasonably good explanation (for whatever that's worth). My big "hang-up", and I'm not sure I agree, at least yet, that it's possible for a player to have a 1/2 chance of winning while another player at the same point in time with exactly the same set of conditions can have a 2/3 chance of winning.

Either something wrong with the theory or there is something wrong with the application that doesn't consider (or account for) the "hypothetical" new player (3rd observer).

I think to achieve 2/3 chance of wiinning requires and depends on an event that has a 100%chance of occurring --> "mechanically" requiring the original player to "switch" everytime which takes away from, in my opinion, some of the "randomness". An event which must occur 100% of the time is, in my opinion, not random.

So bottom line, yes... I've understood your explanations and appreciate them yet I'm just having a hard time "wrapping" my thoughts 100% around them.

 

[DrGreg]

My big "hang-up", and I'm not sure I agree, at least yet, that it's possible for a player to have a 1/2 chance of winning while another player at the same point in time with exactly the same set of conditions can have a 2/3 chance of winning.

When you calculate a probability, your answer depends on how much information you have. The second player is lacking some information, so the probability they calculate is 1/2. The original player has more information and calculates the probability as 2/3. The game show host has even more information and will calculate the probability to be either 1 or 0.

The probability represents how often you would win if you repeated the experiment many times using the information you have and assuming all the information you don't have is random.

 

[vector03]

Personally, I would vote for the host's chances as 0 since the host is generally not allowed to play.

I note the qualifications of "long run" or "repeated many times" and respectfully submit that the theory is based on the "long run" assumption. In this particular case, that assumption is not met. This experiment is setup as a one time chance of winning. If the player had hundreds of chances, in the long run, his chances would approach a limit of 2/3. However, the player only get's 1 chance and that invalidates any use of the "repeated many times" assumption. The player only has one chance.

Applying any theory that is based on certian assumptions being met to the solution of a problem where those assumptions are not being met does not seem cosistent