MONTY HALL LOGIC - Not adequate for real life situations.

[Ryoushi]

quote:

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Ok, one more time!
I decide in advance that I'm going to choose door 'A' first and always switch.

By doing this, the only way I can lose is if the prize is behind door 'A'.

There is a 1/3 chance that the prize is behind door 'A'.

Therefore, I have a 1/3 chance of losing and a 2/3 chance of winning.

Right?

---

But what if Monty adds another page to the thread, so there are now 4?

[mathgrant]

I'm sure someone will love to read this thread full of stuff. BTW, the answer is it doesn't matter if you switch. Monty will put chocolate bunnies behind all three doors anyway.

Wow, I thought surely this thread was less than a year old.

------------------
GL DPWestley: I had my backspace key removed and a beverage dispenser pur in it's place. See? I can't go back and fix "put."
GLmathgrant: I once was banned from a M:TG tourney for bringing over 50 banned cards! They were Ace of Clubs, Ace of Diamonds, Ace of Hearts, Ace of Spades, 2 of Clubs. . .

[This message has been edited by mathgrant (edited 09-24-2002 08:24 PM).]

[extropalopakettle]

Boy, some people had way too much free time on their hands back then.

[Brian]

No, I'm not trying to re-open old wounds. But this seems like the right thread to ask this.

Way way back when I was in my 1st year at university, my statistics lecturer taught me all about the Monty Hall puzzle. She also taught me about this set of 4 six-sided dice, all four of which had different configurations on the faces (one had 1,1,1,1,6,6 but I don't recall the other three).

The interesting thing was that which ever die you picked of the four, there was another one that statistically should roll higher, so she would challenge us to pick a die and roll off 10 times against her, and she would pick the one that would beat the one we picked.. 1 beats 2 which beats 3 which beats 4... which beats 1!

It was pretty cool, and when I did the math it all worked out right (somehow).

Sadly, that was years ago, and I have been unable to remember what they were called or what the face configurations were. Can anyone help?

[Chuck]

They're called nontransitive dice.

http://www.sciencenews.org/20020413/mathtrek.asp

[mith]

Yeah, one of the nifty things you can do is take a magic square:

code:

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834


159


672

---

Then take each row (or column), and label one dice per row, like so:

A - 883344
B - 115599
C - 667722

Dice B beats A 5/9 of the time, Dice C beats B 5/9 of the time, and Dice A beats dice C 5/9 of the time.

[kthejoker]

Why is this whole thread so incongruous? Here are the probabilities in simple state:

Before talking to the warden:

P(A dies) = 1/3
P(B or C dies) = 2/3

After talking to the warden:

P(A dies) = 1/3
P(C dies) = 2/3

These are the actual odds. The reason for this is the WARDEN himself has knowledge of who will die. Thus, he is Monty Hall, and he knows what is behind all of the doors.

Now, to equate this to the Monty Hall dilemma:

There are 3 prisoners in front of you. One has been chosen to die. Your job is to try to guess which one that is.

You select Prisoner A (or, to be imaginative, you *become* prisoner A.)

Now Monty (the warden) tells you, "Prisoner B has not been chosen to die." (ie nothing behind door #2)

You would be wise to switch to Prisoner C because 2/3 of the time he is the one who had been selected to die. This because he is among the pool of P(not A dies), which is much higher (increasingly so with larger numbers of prisoners) than P(A dies.)

And the fallacy in the earlier argument that to play Monty Hall with a deck of cards would be to simply show you one non-ace and ask you to switch is:

The real Monty Hall odds are based on paring down the choices to two. With 3, obviously, the logistics are simpler: remove one bogus choice. With a 52 card deck, you would have to remove 50 bogus choices. The second caveat to that is that Monty Hall *KNOWS* where the ace of spades is. That's the only way he could legitimately *NOT* pick the ace of spades from the deck.

Ok, to restate the simple logic of this whole conundrum:

Monty is offering you a chance to take every other door except the one you chose. If there were a million doors, the odds that you picked the right door on the first shot are very slim. Even when there's only 3, you only had a 33% chance. Switch doors!

[mathgrant]

I believe the card version had three cards to start with. . .

[Strebo]

As far as the Monty Hall thing goes. You do win over time more often if you switch. And the odds are better (1/3 vs 2/3) if you switch. But realisticly you get one shot at it, and you are not out anything (except dignity being on national telivision in that ridiculous costume) if you guess wrong, so why switch. If you are going to doubt yourself why are you playing in the first place. (also, I've always wanted a goat. I have a couple of nice recipes.) I guess what I'm saying is whatever the probabilities are when you only have one shot, go with your gut.

As for the prisoners, "A" should just bribe a trustee to replace the name on the desk with either b or c (whichever he likes least) to be sure that "A" get's to go free.

but that's reality, not theory. (at least my reality.)

[ewan]

if there are 52 prisioners and prisoner 1 asks the jailer which one is going to live..

Jailer: who do you think it will be?
prisioner 1: *gulp* me?
Jailer: well... if I tell you that prisioners 2 though 51 _WILL_ be executed, would you like to change your guess to prisioner 52 or stick with your origional choice?
prisioner 1: ermmm... wait no I guess prisioner 52 first, do it again..
Jailer: OK prisoners 1 through 50 _WILL_ be executed, would you like to change to your guess to prisioner 51?
prisioner 1: Arse! it doesnt work!!!! why?! why me?!?!
Jailer: so your sticking with 52 huh?

[cha]

Consider this...
The only odds that matter are those that exist, based on information available to you the contestant, when it is time to decide to stay or switch. When you originally select, you don't know if there are one or two possible incorrect doors remaining. Monty shows you an incorrect door, now you know there is at most one incorrect door remaining. The problem has changed. In the end, there are eight potential relevant scenarios.

If "A" is the prize door, the possibilities come down to:
Guess A, Monty shows you B, you stay, you win
Guess A, Monty shows you B, you switch, you lose
Guess A, Monty shows you C, you stay, you win
Guess A, Monty shows you C, you switch, you lose
Guess B, Monty shows you C, you stay, you lose
Guess B, Monty shows you C, you switch, you win.
Guess C, Monty shows you B, you stay, you lose
Guess C, Monty shows you B, you switch, you win.

Stay or switch doesn't matter - it's 50/50.
The flaw with the 2/3 solution is this: Adam chooses A, Bart chooses B. Monty shows C. Whether they both switch or both stay, only one will win. 50/50.

[Chuck]

What if they both choose the same door to start with? If that's not allowed then one of them would have only two doors to choose from. That makes the situation different for him. Also, Monty might not have an empty door to open after the choices are made.

[mathgrant]

They're not the same probability. If you choose A (1/3), Monty has a 1/2 chance of opening either door. Choose B or C and Monty must open the other. Thus the chances of the pick-B-open-C scenario occuring are 1/3, and same for pick-C-open-B. But pick-A-open-B and pick-A-open-C have 1/6 chances each.

[QuikSand]

Trouble is, cha, that your table of 8 different events are not equally likely to happen, as mathgrant has explained. (Assuming that your 8 possible cases are equally likely is tantamount to assuming that we have a 50% chance of initially guesing the right door out of three - it's a tautology, basically assuming the very fact you're trying to prove) So, when we correctly add the various probabilities in, we get this:

- - -

If "A" is the prize door, the possibilities come down to:

Guess A, Monty shows you B, you stay, you win
Guess A, Monty shows you B, you switch, you lose
Guess A, Monty shows you C, you stay, you win
Guess A, Monty shows you C, you switch, you lose
Probability of these must sum to 1/3

Guess B, Monty shows you C, you stay, you lose
Guess B, Monty shows you C, you switch, you win.
Probability of these must sum to 1/3

Guess C, Monty shows you B, you stay, you lose
Guess C, Monty shows you B, you switch, you win.
Probability of these must sum to 1/3

- - -

Since there is a 1/3 chance that you have originally picked the right door in the first place (the cases represented by the first set of four possible outcomes) then we know, from your analysis and a simple application of probability, that:

It's 1/3 likely that we're in the situation you described as "you stay, you win"


Since there is a 2/3 chance that you have originally picked a wrong door in the first place (the cases represented by the second or third set of two possible outcomes each) then we know that:

It's 2/3 likely that we're in a situation you described as "you stay, you lose."


And so, we can easily abbreviate to the correct result:

Staying gives you a 1/3 chance of winning
Switching gives you a 2/3 chance of winning

q.e.d.

[jeep]

That isn't the same problem because there isn't necessarily a losing door to open. The key is that you get to pick Orig or (Both the others). If Monty does NOT open a losing door that you didn't pick, then the problem changes. For example, if Monty randomly opened a door, then you gain nothing by switching (sometimes he'll open up the winning door and you'll know that you've lost before you get the chance to switch AND sometimes he'll open up the door you picked, so you'll know if you should switch to another door or not).

Even if he opened up a random door that you didn't pick, you gain nothing. The fact that he always opens a losing door, that is why you gain by switching.

Anyone who doesn't believe it, I'm perfectly willing to do this:
I'll put $10 behind a "door". You pick a door. I'll expose a losing door. Then give you the opportunity to switch. You always decline. If you get the right door you win the $10. Otherwise, you pay me $6. We'll repeat 1000 times. If you win over 50%, we'll work out a payment schedule. If you are correct and your expected payout is 50% then you should expect $5000-$3000 = $2000. If I'm right and you have a 33% chance, then you can expect to lose $666.

I'm also willing to do it the other way around. You put $6 behind a door. I'll pick a door. You expose a losing door. I'll always switch. If I'm right I get the $6 and if I'm wrong, I pay you $10.

If either of us loses a lot, then we can work out a payment schedule.

-JEEP

[cha]

IT FINALLY MAKES SENSE TO ME!
No matter what door I select initially, 2/3 of the time I'll be wrong. So after Monty opens a door, 2/3 of the time the remaining door will be the correct one.

[jeep]

<== is sad

No one has ever taken me up on that and I have offered that deal to several people.

-JEEP

[mikegoo]

Jeep...I'm willing to play your game. You never specified number of doors in the description of your game (assuming the same rules as Monty's probelm won't work as ytou never stated this). One door sounds good to me...and I'd even play with 2 doors

[The Fatman]

Hello

Now, I don't know if this prisoner thing has been put to rest or not, due to I can't be bothered to read all thses posts (they're doing my head in, I'm not very mathmaticly inclined) but from what I can figure out is this:

You have three prisoners, A B and C.
One of them is going to die, the other two will live.
The warden has already made the choice.
A is going to ask the warden which of B and C will live.

Before A asks, the odds seem as so:

A = 1/3 chance of dying, 2/3 chance of living.
B = 1/3 chance of dying, 2/3 chance of living.
C = 1/3 chance of dying, 2/3 chance of living.

Now A is told B will live, and the puzzle asks does this effect A's chance of survival?

I would say no.

None of the prisoners have control over who the warden chooses, nor can they change his mind. They are aware only one will be killed, so thier odds of survial? 2/3.

I can't see how A knowing that B will live has any effect upon the odds.

To relate it to Monty Hall:

There are three players A B and C.
There are three cards, face down.
Each choose a card. One is an Ace, the other two are jokers.
Whoever has the ace wins. Monty knows which one is the ace.
A asks Monty, "which of the other two has a joker?" He replies "B".

Well B is screwed, but by knowing this cannot alter the fact that either A or C has the ace, and the odds of this are 1/3 due to those being the odds when they chose.

[jeep]

Uh, goo:

[tjg]

This post is for the prisoner dilemma.

That question has nothing to do with the Monty Hall problem, and that's why you people are confused.

In Monty Hall, when the producers chose a door to put the car behind, each had a 33% chance of being correct, assuming it is choosen randomly. Pretend the producers put it behind #2. Later, the game show contestant picks #1, and #3 is opened to reveal no car. This does not mean that when the producers were originially choosing a door, they had a 0% chance of putting it behind #3, a 66% chance of putting it behind #2, and a 33% chance of putting it behind #1! This process was done before the contestant ever had a chance to guess anything! The contestant's guess doesn't now redetermine the percentage that the producers had in choosing a door to place the car behind hours earlier! It would be absurd to think the producers were somehow magically kept from putting the car behind #3. This is obvious because if they placed the car behind #3, then Monty would have simply opened #2 instead. He would not have opened #3. Therefore, the probability that any one door was picked to have a car behind is still 33%. The fact that Monty revealed a door with no car merely makes it easier for the contestant to now guess which door has the car, but doesn't change the percentage that any of the doors was choosen to be the winner in the first place.

Therefore, the prisoner still has the same chance that the warden chose him to die. Just because A knows that B will live doesn't mean that the warden somehow COULDN'T HAVE PICKED B TO DIE before the choice was actually made.

Look at the Monty question like this: if you switch your choice, it is the equivalent of chosing simultaneously both the door you originally didn't choose (but are now switching to), AND the door that Monty revealed to have nothing. So in essence, when you switch, you are choosing two doors, while if you stay the same, you are choosing one door. Each door still had a 33% chance of being correct, but by choosing two doors, you upped your chances of being correct to 66%.

Maybe this sounds confusing, but it's quite logical. If you could choose doors #2 and #3 together, you would have a 66% chance of being correct. If you did that, and Monty opened one of your doors to reveal no car (even if one of your doors has the car, that means one doesn't, and Monty knows which one doesn't), then you would STILL HAVE A 66% CHANCE OF BEING CORRECT, since one of your doors obviously doesn't have a car, even if another does. Well, that's what you're doing when you switch doors. You're saying, "I'm switching doors to #2 and #3, and Monty just opened #3 to reveal nothing, but my probability of winning has not changed...it's still a 66% chance, as if I had choosen #2 and #3 to begin with."

Well, the same goes with the prisoners. Each prisoner had a 33% chance of dying from the beginning. But, A has a better chance of guessing who will die by guessing both B and C as one guess (obviously). Once B is revealed to live, then A still has a 66% chance of being correct that C will die. This is NOT because C has a 66% chance of dying, but because B and C combined have a 66% chance that one will die. Even if one will die, that means one will live. The warden has simply told you which will live, of the two you chose.

Simple...

[tjg]

Ok, after reading what I wrote, I think I need to also explain something else.

I can picture some of you asking, "If I choose #1, and #3 was opened to reveal nothing, why can't I say my decision was the equivalent of choosing #1 and #3 together, giving me a 66% chance of winning? Why is it only the same as choosing two doors when I switch, and not when I stay?"

Well, the reason is because a door has not yet been revealed to have nothing. You can't say, "choosing door #1 becomes the same as choosing #1 and #3 after #3 is revealed to have nothing" because #3 might not have been the one that's revealed. If #3 has the car, then #2 would be opened instead. It would be like saying, "if #3 has nothing, then my choice is #1 and #3, but if #2 has nothing, then my choice is #1 and #2." That would be the equivalent of saying, "my choice is always the first door I choose, and whichever door is the first door opened that reveals nothing." If that's the case, then your odds are always 33% since you only actually have a chance at winning with the door you first chose; choosing two doors in that situation won't increase your odds at all.

I know what you're thinking: well, with the above logic, if I picked #1, and then #3 is revealed to have nothing, then #2 should only have a 33% chance of winning also, because one of your doors, even if you picked both #2 and #3 together, would still have a loser. Now that #3 is revealed to be a loser, then #2 should only have a 33% chance of winning. This is not true because when you switch, it's the same as picking #2 and #3 BEFORE you know which is the loser. Therefore, it's not the same as picking #2 and a loser, or #3 and a loser (as in my example above: picking #1 and a sure loser)...instead, it's the same as picking #2 and #3 together, knowing that one will lose, but that one could win. This gives you a 66% chance of winning, even after one is revealed to lose...you already knew one would lose.

[CrystyB]

umm

[extropalopakettle]

[What if...]

First of all, quite simply, I have yet to find a meaningful difference between the prisoner and Monty Hall problems. The prisoner has the same information, and the same probabilities. Think of it this way. Since at least one of B and C are going to be executed, A's question provides him with no information regarding his own execution or freedom. Since A still has a probability of 1/3 of being executed, the remaining prisoner now has 2/3 probability from A's standpoint. this is because having no information about his situation makes A's situation different from the others', from A's viewpoint.
Another way to think about it is that A knows there is a 2/3 chance one of the other prisoners will be executed, and the guard tells him that one is definitely not him, so the remaining one will have 2/3 chance. If C asks the same question, the answer might be A or B. So A cannot deduce that he has a 1/2 chance just because C could say the same thing.
Two different "contestants" don't negate the 1/3 or 2/3 chances, unless they can communicate, or the Monty gives them the same information, because they could be told different wrong doors, and not know what the other was told.