The Monte (Door # X) Dilema

Great Debates
21

Whether or not Monte always eliminates the looser is certainly germane to the puzzle. If he does your odds are 66 percent for a winner if you always take the door he does not eliminate, and 33 percent for the door you originally chose. If he does not, your chances are 33 percent each for the one he leaves, the one you chose originally, or that he threw away the winner.

Try this. Deal out cards, face down, to another person. They pick a card at random. Then you peek, toss the lower, and check the results. Two thirds of the time, the card you have not eliminated will be the higher of the remaining cards (not considering ties.) That reflects the odds if you always take the door Monte does not eliminate. Run it for multiple choices, and you will find that the rule for the first pick, if you always change is “every time you would have lost, you win, every time you would have won, you loose.” Since the straight up odds of one and three are the short side, you swap every time, and take the two out of three.

If Monte chooses at random, or the winner is not “revealed” in the sense of being identifiable, or if the “revealed” choice is not eliminated, the odds remain one in three, despite any dancing girls, or dog and pony show.

Tris

Imagine my signature begins five spaces to the right of center.

22

Tris, I suspect you and I agree on the substance of this problem, but I totally lost you the last time round. I have always assumed that Monty never eliminates any door, only that he opens one of them. Before my last post, I always assumed that he would only open losing doors, but I changed that assumption for the sake of argument.

Am I reading Hippocrates right?
He’s not just a big box for storing large semi-aquatic mammals, is he?

23

I have a related problem, which I will state to the consternation of all and the probable embarrassment of myself.

Suppose person A is at a reverse raffle with 999 other people, all of whom has one ticket. It’s obvious that, at any point, person A would be acting rationally in trading their single ticket for all the tickets left unpicked. This is true because A divides the tickets into two groups:[list=1][li]A’s ticket (with 1/1000 chance of winning)[]all other tickets (with 999/1000 chance of winning)[/list=1]and realizes that the chances of one of the tickets in group 2 winning is not affected by the shrinking size of group 2. As a corollary, I will note that the other participants in the raffle can use the same reasoning that A did to come to the same result.[/li]
Now, the disturbing conclusion. The raffle gets down to two tickets, one held by A and the other held by B. A reasons as above and concludes that A would like to swap tickets. B reasons in the same manner, and concludes the same thing. That is, both A and B conlude that the other has a much better chance of winning. We have three different odds estimations:[list=a][li]A has a 1/1000 chance (A’s belief)[]B has a 1/1000 chance (B’s belief)both A and B have a 1/2 chance (belief of every other rational being in existence)[/list=a]Everybody has the same information, everybody followed the rules, and they got different answers.[/li]
If this isn’t a paradox, it’s real close. Anybody want to take a stab at what went wrong?

24

Tominator2, there is a flaw in your scenario. Your “leap of faith” or your “magical thinking” or whatever was when you seemingly innocently say “the raffle gets down to two tickets”. What is that supposed to mean? I read on and found out that, not only did the raffle somehow “get down” to two tickets, but “every other rational being in existence” not only knows it is down to two tickets but also know that either A or B hold the winner. Yet in your scenario, poor A and B are the only rational beings in existence still stumbling along believing their ticket has a 1 in 1000 chance of winning. You suggest that A and B still reason that “all other tickets” have a 999 out of 1000 chance of winning.

It’s not logical that A and B would not also be aware that it was “down to” two tickets and therefore their tickets have a 1/2 chance of winning.

Is this sufficient to precipitate your “probable embarrassment”, Tominator2?

25

Perhaps this added wrinkle will help people understand the concept.

Monte Hall told you there was a car behind one of three doors. You picked the first one. He then opened door #3, and shows you there’s nothing behind it. He then asks you if you want to change your mind. You say “no,” reasoning (wrongly) that you have a fifty-fifty chance.

Well, suppose Monte then picks another person out of the audience and says “Tell you what, guy, YOU can have what’s behind door #2.”

NOW, you get outraged, because you can see that this newcomer has a better chance of winning than you had initially. HE has a 50% chance of winning the car, but you didn’t!

26

The flaw in the raffle problem is that presumably the other people were eliminated through a random process. If that’s the case, each person has exactly a 50/50 chance of having the correct ticket, and nothing changes.

Or, if Monty Hall makes the choices, the fact that he picked YOU out of the thousand changes your odds to 50/50 on the spot. You know he’s going to pick the winner and one random loser. If you are picked, there’s a 50/50 chance that you’re the winner, and you gain nothing by switching.

27

Remember, in the original scenario, there is one good prize, and 2 crap prizes,(not like the real show)

We also “seem” to be assuming that Monty knows what the hell he’s doing, he knows which door has the prize, and he would never show a winner door, because that would bring things to an end too quickly. (and I have a feeling others may have been assuming even more ) So…

There are actually 2 different ways to try for an eventual odds advantage, a purely mathmatical, and a psycological read of a “tell”. If the pyscological fails, the mathmatical advantage will always be there. I’ll try to keep em seperate…

When you first choose, your chances are 1 in 3

The only way to improve them is for one or more of 3 things to happen.

#1 Monty must show you that a door you didn’t pick, was a loser. Which he is always able to do,“if he chooses to” because there are 2 loser doors. (but removing a loser door from the equation is important, if all psycological reasoning or patterns fail)

#2 He must allow you to change doors(if you wish)

#3 (most importantly if trying for pyscological advantage) He must have offered a “tell”.

The only way he can offer a tell, is if you know the following,

#1 Does he want you to win or lose?

#2 Does he “always” open a loser door?
if he does, then opening the door, tells you nothing.
If he only shows the loser door when the contestant is correct, then your odds of winning if you stay are 100%.
If he only shows a loser door when the contestant is wrong, then your odds are 100% if you switch.
If he varies his pattern 50/50, it tells you nothing.

#3 does he “always” offer the chance to switch?
if he always does, then this tells you nothing. (the rest same as #2)

If you appear on the first show, so there is no track record, then you must know if he wants you to win or lose.
If you know he wants you to lose,
When he offers you the chance to switch, don’t do it. why else would he have offered it. At best he is planning on always offering this choice, and your chance remains the same . At worst he is trying to get you to change your door so you lose you dufus!

If you know he wants you to win, and he offers you a chance to change, then do it.
At worst your odds remain the same, and at best he’s trying to give you an “extra” chance to win which would only be nessesary if you picked wrong the first time.

On the other hand,if there is no pattern to showing doors, allowing contestants to change, and monty cant make up his mind wether or not he wants you to win or lose. In other words no “tell” Then we go purely mathmatical…
…the chance you picked the right door the first time is only 1 in 3, in other words there’s a “good” chance (2 in 3) you made a bad choice, if another bad choice is then removed from the scenerio,(because its the rule , and not because of game show host psycology) and your given a chance to switch,(again, because its the rule…) then Go For It!!! Yes your better off.

So there! Who did I piss off? :wink:

I could be wrong…it happend once before…

28

OK, I’m usually good with these things, but I’m having a lapse today. I read Cecil’s column when it first came out and was in full agreement with him. But I have a couple of questions for those of you who know the solution to this thing backwards and forwards:

Is it presupposed that Monty has a hidden agenda, or is his showing of the loser door completely random?

If it is random, is it not true that when you are faced with the second choice (either keep your original door or take the remaining one) you can (maybe should?) look at the choice as a completely new choice?

That is, what if after you made your initial choice a boulder fell on your head and you were killed. You are then replaced by someone else who steps into your place and is faced with the choice… Keep this door or take that door? Isn’t it 50/50 for him?

It seems to me that this is only untrue if Monty does in fact have a hidden agenda. If the process were controlled by a random machine, would the answer be different?

I thought I had this all figured out, but could someone tell me what I’m forgetting in the above?

Thanks,
PeeQueue

29

It’s a reverse raffle. In a regular raffle, the first ticket picked wins. In a reverse raffle, tickets are picked (and thrown away) until only one ticket remains. That ticket is the winner.

This is true because of the way a reverse raffle works. Somebody has to win, so it must be either A or B, and a bystander has no reason to prefer one over the other.

A and B are aware of this, but they’re also aware of their own reasoning that the odds aren’t 50-50. That’s the paradox.

30

Why does it being a random process change the odds? One could imagine Monty Hall picking the door to show at random (given that it doesn’t contain the car, which is exactly what happens in a reverse raffle) and it wouldn’t change the odds.

I agree, but this is a different problem. In this case any person picked is a bystander in terms of the original problem.

31

This raffle problem is hardly a paradox. You have made the assumption that…

… which is just incorrect. The chance for one ticket in group 2 to win is initially 1/1000, then 1/999 after the next draw, 1/998 after the next, and so on and so on. By the last draw, when you only have B left, the chance is 1/2.

As for the OP, it’s a dead horse as far as I’m concerned…

32

It’s a simple riddle so don’t make it more complicated than it is. I had the chance to study this in a British magazine while flying cross the Atlantic.

In a one time only event, Monty Hall offers you a choice of three doors - two empty, one with a prize. You indicate your selection and then Monty opens one of the unselected (and empty) doors and offers you the opportunity to change your selection. The riddle is: “Do you improve your odds by switching your selection?”

The trick that makes this riddle so fascinating is the seemingly scientific analysis that says (and you can draw charts and use all kinds of logic to support this) “if your selection was empty door A and you switch you win; if your selection was empty door B and you switch you win; if your selection was the prize door and you switch you lose; therefore switching improves your odds of winning to 2:1 in your favour.”

As Vincent Bugliosi is so fond of saying “there is no relationship between intelligence and common sense.”

The trick of the riddle is implying that there is a difference between empty door A and empty door B. This is true, they are different empty doors. BUT for all practical purposes, to our grunt in the trenches faced with this decision, they are the same - empty doors, exactly equal to each other and to zero.

Theorists can make a pretty good argument that you improve your odds by switching (I repeat: “if your selection was empty door A and you switch you win; if your selection was empty door B and you switch you win; if your selection was the prize door and you switch you lose; therefore switching improves your odds of winning to 2:1 in your favour.”) BUT the problem with this is too much unnecessary information -empty door A or B is irrelevant. To the contestant an empty door is an empty door is an empty door. The reality is that the contestant is now looking at two doors, one winner and one loser, and has to decide whether to switch or not - a 50/50 proposition.

It’s a great riddle and a bit of a brain twister and the trick is the introduction of the concept of empty door A and empty door B in the analysis of the correct answer. For practical purposes (making the decision) the difference between the two doors is irrelevant.

In the magazine article there was another brain twister:

You are a scientist testing a population for a disease. Your test is 99.9% accurate and the disease occurs in one person in a thousand. One of your subjects tests positive. What are the odds that person has the disease? Apparently something like 75% of all doctors posed this question answered 99.9% because that is the accuracy of the test. The true answer is 50/50 because, out of 1000 subjects, one will test positive for the disease and one will test positive because the test is only 99.9% accurate.

Once again the trick is in too much information - you are told that the disease occurs in 0.1% of the population. If you were testing the population to establish the percentage of occurrence of the disease rather than to identify individuals then a positive subject has a 99.9% chance of having it.

33

I thinking JJ was the smart one. He posted once and then left.

Some posters are apparently confused about the Monty Hall problem works. (Some posters are apparently confused about how the Monty Hall problem is spelled, but that’s another issue.) So to make it clear, the contestant is given a choice of three doors, one of which has a prize behind it. Monty Hall knows where the prize is and the contestant doesn’t. After the contestant picks a door but before he opens it, Monty picks a different door that he knows doesn’t have the prize and opens it. Monty then offers the contestant the choice of switching from his first pick to the other unopened door.

So Monty doesn’t pick doors at random. He always opens a door and offers a switch regardless of whether the contestant picked the right door or not. Monty knows where the prize is, and never opens that door while the game is going on.

Now seeing as there are some people who don’t believe in the logic of the correct answer, you might want to try some empirical evidence. Take a sheet of paper and a die. Mark down door 1, 2, and 3. Using the die to generate random numbers, put the prize behind a door. Rolling the die again to simulate the contestant, pick a door. Pretending to be Monty Hall, eliminate a door as described above. Stay with your original choice. Repeat this one hundred times and record the results. Now flip the page over, do all the steps above another hundred times, but this time always switch doors.

Now look at your results. If you’ve followed instructions, you’ll see that you have drastically improved your results by switching. If you don’t see this, you may conclude one of the following:

1 - You screwed up and didn’t do it right.
2 - You just ran through a phenomenal run of unlikely die rolls. Too bad you weren’t playing craps in Vegas, huh? Try again.
3 - You were unconsciously influencing the results through your telekinetic powers. Post the results and annoy David B.

34

Apparently so. :smiley:

35

Ahh, you’re right I didn’t know how the riddle worked at all. (Actually it seems I forgot how the riddle worked.) I also forgot what the riddle was asking. Thanks for clarifying.

PeeQueue

36

I apologize if I wasn’t as clear as needed. Perhaps I should have said “The chances of the winning ticket being one of the tickets held by someone in group 2 is not affected by the shrinking size of group 2”. This insight is the basis of the original solution.

37

Little Nemo you were doing great but then, in my humble opinion, you lost it. This is not a flame but in the spirit of a great debate. You said:

“Take a sheet of paper and a die. Mark down door 1, 2, and 3. Using the die to generate random numbers, put the prize behind a door. Rolling the die again to simulate the contestant, pick a door. Pretending to be Monty Hall, eliminate a door as described above. Stay with your original choice. Repeat this one hundred times and record the results. Now flip the page over, do all the steps above another hundred times, but this time always switch doors.”

The question being “is it better to switch doors” - I don’t need the die. I will take a sheet of paper and mark three rows of three doors putting the prize behind a different door in each row. Then I will draw 3 sets of the rows illustrating the contestant selecting one of three doors in each of the three sets (in the first set I pick the first door in all three possible prize door combinations, second set the second door…). I now have the only nine (9) possible combinations frozen in time at the precise moment that Monty steps forward and opens a door in all nine different possible scenarios. If you look at the nine possible combinations on your paper you will see that in six of the nine possible scenarios there is only one door Monty can open because the other one is either the one you picked or the prize door. In three of the possible combinations there are two doors Monty can open because the contestant has picked the right door, so that means six more possible doors for Monty to open.

So, mathematically predetermined before the game even begins is the fact that Monty is going to step forward and, out of the nine different possible combinations of prize doors and contestant door selections, open one of twelve possible doors (being a non-winning door and also not a selected door).

If the contestant switches in all twelve possible door openings by Monty the contestant will win 6 and lose 6. If the contestant stays the results are the same, 50/50. It’s right there on paper in front of you.

There is no advantage to switching.

Sorry to hammer on this but I like trying to nail down a convincing and understandable analysis. Is this not a fight against human ignorance?

38

Tominator2:

That’s still wrong–the chances of someone in group 2 winning do go down as group 2 shrinks. It starts at 999/1000, then 998/999, 997/998, all the way till just B (and A, from group 1) is left with a 1/2 chance of winning (along with a 1/2 chance for A, as well).

39

Al Zheimers:

OK, you established there are 9 different ways the car can be hidden, along with the contestant having selected one of the doors–no problem there.

If you’ll look on your paper, in three of those cases, switching will always force you to lose, regardless of which door Monte opens. In 6 of those cases, switching will always force you to win.

Chances of winning if you switch: 6/9

Chances of winning if you don’t switch: 3/9

Your mistake was when, for example, you pick door #1 and the car is behind door #1. There’s a 1/9 chance of this happening. The fact that Monte can open either door does not, in any way, double the chances that you have already picked the correct door, it’s still that 1/9 chance, which was already set before Monte opened either of the doors, and cannot change. Same if you correctly pick door #2, or door #3. There’s a 1/9 + 1/9 + 1/9 = 1/3 chance of picking the correct door, and if you do, you switch you lose. Leaving a 2/3 chance of you switch, you win (2/3 chance of picking the wrong door).

40

Rereading it just now, I wanted to make sure my last parenthetical statement was clear. It should have read (2/3 chance of having picked the wrong door in the first place).