MONTY HALL LOGIC - Not adequate for real life situations.

[Page 2]

I can't believe I am honoring this thread with my presence...

[Ghost Post]

Wonko, I may have criticized what you wrote, and I understand what you wrote. I'm trying to understand what you mean, because it's obviously not what you wrote. You said I misinterpreted a lot of things, and that you'll go through them all, and then you go off in a totally new direction. You address none of the questions I asked you.
Specifically, I wrote the puzzle that I was talking about above. Steps 1, 2, 3, 4 and 5, clearly spelled out. And I asked you to answer two questions after step 4 (before step 5). Then step 5 completes the puzzle, and asks another question. Why not play nice and answer them, with the understanding that you are answering questions about "my version" of the puzzle?

Then, I asked you to spell out the problem you are trying to solve. But you don't. Just state "your version" of the puzzle.

And AGAIN, I point out that you wrote "the question assumes that A will receive the same answer EVERY TIME"

Yes, I dont understand why you think that, and I criticized the conclusion. You said you would go through it all, but you've ignored this. You repeat things like:
"the guard has to say B...every single time"
and
"he can't say C and have the puzzle still work"

You might know what you are talking about, but do you expect anyone else understands when you say things like:
"A's probability is .50, but the reset prob is 1/3, not 1/2"
and
"But he can't be told that C will survive by the logic of the problem, thus that case must be removed, or, in a more realistic case, re-randomized."

Regarding me having NO RIGHT to flame it to hell: It's a freedom that I have (this is the Internet), not a right, but I didn't do that. I've seen flaming to hell, and that wasn't it. It was mild criticism, and quite civil. Maybe I should use more smiley faces. Would that help? I was and still am critical of your inability to read what you write as others might, and see how confusing it is ("gibberish" ), and of your refusal to clarify why you hold some particular position, such as "the question assumes that A will receive the same answer EVERY TIME". There is ONLY ONE TIME. If there were other's, perhaps things would be different.

(see what you started, frisky.)

[Evil Empire]

Good to see some things never change

Banana,

I'm going to randomly kill someone and you are one of the potential targets. I'll give you 2 options.

1) I can randomly choose between you, Wonka and extro...(Sorry). After I have randomly choosen I will set one of the lucky ones free(not you).

2) I can randomly choose between you and Wonka.

The clock is ticking, option 1 or 2?

[worm]

I think you're wrong, extro...!! You do have the right (as well as the freedom) to flame this character. You've earned it w/ the way you've defended your position so well & so many times. It seems like as soon as you convince one another comes in telling you what the answer really is. And you just keep on going.

Your patience is truly inspiring, but I think it's time to kill this thread. Please!!!

BTW, leave the smilies to HyToFry

[Prisoner B]

Yippee!!!!!!!!!!

I just found out that I am to be set free!!!!!

[Ghost Post]

Sorry, but I can't resist. I just noticed it was Wonko who wrote:

"Feel free to flame, I've used some intentionally flawed logic in places."

then later wrote:

"Just because you don't understand something gives you NO RIGHT to go on and flame it to hell."

[Wonko the Sane]

Extro, entirely different post. And flame and discredit without proof. A bit different. And I say again, I haven't disagreed with anything you've said. And no, it doesn't work in a large number of trials if the guard can say C. Hence why I said before. In a large number of trials

When the guard says B will go free
A dies: 3333
C dies: 6667
When the guard says C will go free
A dies: 3333
B dies: 6667

Combined, it's still 1/3,1/3,1/3. However, when you cut out those cases from C, you do have 1/3-2/3. The problem is that isn't realistic because you can't just remove the problem if your data is wrong. Thus you'd have to get new data until it all conforms. Is that simple enough for you? Or should I keep it to words under 5 letters next time?

[Ghost Post]

If you say you agree with me, I'll have to take your word for it. Anyway, you have the numbers right for a "large number of trials", before discarding C. You have:

When the guard says B will go free
A dies: 3333
C dies: 6667
When the guard says C will go free
A dies: 3333
B dies: 6667

So, that means that if there is a SINGLE trial (as in the puzzle), and the guard says B will go free, that A has a 1/3 chance of execution, and C has a 2/3 chance of execution.

But I still don't know what you mean when you say:
"it doesn't work in a large number of trials if the guard can say C"

What doesn't work? Why not? (Am I the ONLY one having trouble understanding this?)

And I'd still like to know why you said:
"the question assumes that A will receive the same answer EVERY TIME"
and
"the guard has to say B...every single time"

You write "The problem is that isn't realistic because you can't just remove the problem if your data is wrong. Thus you'd have to get new data until it all conforms. Is that simple enough for you?"

No, not simple at all, and it has nothing to do with the length of the words. Who is removing a problem because some data is wrong? What data? Conforms to what?
You came up with the numbers above with a large number of trials, so you tell us what A's chances and C's chances are in a SINGLE trial when the guard happens to say "B will go free".

It's just like:
"A's probability is .50, but the reset prob is 1/3, not 1/2"
and
"But he can't be told that C will survive by the logic of the problem, thus that case must be removed, or, in a more realistic case, re-randomized."

This is a simple problem, and you purport to make it simpler by talking about "reset probabilities" and more realistic when "re-randomized".

"The problem is that isn't realistic because you can't just remove the problem if your data is wrong."

Who is removing the problem??? What data is wrong??? There is a SINGLE "trial", and the warden will say B or C. He CAN'T say BOTH in a SINGLE trial. It just so happens he says B. We didn't "remove" the trials where he says C.

Flame and discredit without proof? Here's proof:
I said: "A asks a question, and no matter what answer he gets, his chance of execution increases? Don't you see a problem with that?"

You said: "There's no problem with that. It's perfectly logical."

There's your proof, and you'll probably provide more proof by still failing to see the problem. In the post that upset you so much, I tried to show you just what you are saying is "perfectly logical", and what the implications are. You did not address it AT ALL. In closing, I'll cut and paste the following (which you've ignored) from that post, and which demonstrates what you consider "perfectly logical":

.
.
.
First, how can getting an answer to a question change our estimate of the probability, if we knew before even asking that the answer would not matter? Before even asking, we know that after getting the answer, no matter what the answer is, we will conclude the probability is P. Do we need to ask then? Apparently we do, because P is not the probability we have in mind before getting the answer.

If you don't realize you just said "there's no problem with that", and don't realize that there IS a problem with that, then you're just plain Wonko.

It's exactly like me telling you "I will flip a coin. If it comes up heads, tomorrow I will shoot you. If it comes up tails, tomorrow I will shoot you", and then you waiting in suspense for the coin to be flipped to see whether or not you will be shot tomorrow. If both heads and tails lead to the same outcome, then you should know the outcome before the toss.

Similarly, if every possible answer to a question leads to the same conclusion, then we don't need to ask the question. But you just claimed there is no problem with having one conclusion before the question, and a new conclusion after getting the answer, when in fact every answer leads to the same "new" conclusion. You said "It's perfectly logical."

[HyToFry]

Extro when wonko says

quote:

---

But I still don't know what you mean when you say:
"it doesn't work in a large number of trials if the guard can say C"

What doesn't work? Why not? (Am I the ONLY one having trouble understanding this?)

---

He is only saying that if the gaurd says "C" then that scenerio WOULDN'T FIT THE QUESTION, not that its not possible, just that It wouldn't fit the question.

think about it this way..

If the guard says "C Will go free", then what percent of the time will A die, and What percent will C die assumeing that the guard tells A that B will go free.

this is pretty easy math because A will die 100% of the time..

All Wonko was trying to prove is that A will die 33% of the time that the guard says "B will go free"

for this to be true, you can only look at the results where the guard actually says B will go free.... i know this sounds confuseing, but if you think about it, it makes sense.... You and Wonko are on the same side (the correct side), Wonko is just cutting out the results that won't plug DIRECTLY into the question.. get it? i hope?



Oops sorry, got a little carried away thinking about worm

[This message has been edited by HyToFry (edited 03-30-2000).]

[Ghost Post]

Look, I've been saying this crap all along!

By stating that "B never dies" the whole Monty comparison goes out the window. You can't skew the probability like that.

It would be the same as if you forced Monty to reveal what is behind door #2 all time, even if it was the winning door.

Therefore you will NEVER get the 2/3 1/3 result!

I left space below so Extro can show the failings of the American educational system.

[friskythepig]

I've re-read my original posting, and I can't see any part that states that "B never dies", or even that this situation is going to happen more than once. If you want to assume this you're more than welcome, but it seems to me that it's a different problem, so it's not surprising you're coming up with a different answer.

[Ghost Post]

Where does it say "B never dies"? It does not! Go back and see.

Where does it say it will happen more than once? It does not!

If you are going to assume it happens more than once, and also assume that the guard gives the same answer every time, then you have changed the problem.

With Monty Hall, the contestant picks one of three doors, and Monty Hall always opens one of the other two doors behind which there is no prize.
If you were a contestant (and you get to be a contestant only once), and you picked A, then Monty must open either B or C. If he opens B this time, revealing no prize, what would be the probability that the prize is behind A, and what would be the probability that the prize is behind C?

It's still A = 1/3, C = 2/3. I think you agree with this, no?

Now, the prisoner problem is identical. The only assumption is the the guard always answers questions truthfully. It does not matter if this happens only once. When prisoner A asks the guard: "Of B and C, tell me one who will go free", it's the same situation. He must say either B or C. If he says B, then C has a 2/3 chance of execution, and if he says C, then B has a 2/3 chance of execution, and in either case A's chance remains 1/3.

He must say either B or C. He says B, and you start freaking out. Would it matter if he had said C instead? Or if B had asked the question, and the guard had said A?

This happens only once!

It's like me flipping a coin once, it comes up heads, and you conclude it's not a fair coin because it always comes up heads.

You say "Look, I've been saying this crap all along!"

I say "Look, you've been saying this crap all along!"

I'm trying to figure out why. Where does it say "B never dies"? You even put "B never dies" in quotes, as if you imagine having read it somewhere. You didn't. Go back and see.

[HyToFry]

According to THIS RIDDLE AND THIS RIDDLE ONLY, B will never DIE, never never never, no matter how many times you run this test, to see if B ever DIES ever ever ever ever... B WILL NEVER DIE WHEN THE GUARD SAYS THAT B WILL GO FREE, I don't care how you try to justify it, B never dies when the guard says B will Live, never never never Die, never, B will always go free, WHEN THE GUARD SAYS THAT B WILL LIVE.. GET IT?

So to run a fair test, to see how many times C will die and how Many times A will die WHEN THE GUARD SAYS THAT B WILL LIVE, you have to exclude the times that B dies, and the times that A dies and the guard says that C will live, that way your ALLWAYS using the same criteria...

The riddle is only asking (In its rawest form) "What are the Chances of A dieing, and the chances of C dieing when, AND ONLY WHEN, the guard tells prisoner A that B will survive.

Anybody that knows how the Monty Hall problem works (and doesn't get confused by the fact that there are two winners now instead of just one), will know from the question that A will die 1/3 of the time and C 2/3 of the time WHEN THE GUARD SAYS THAT B WILL LIVE, however if you need to do 10000 Test runs to prove it, then you need to exclude the times that the Guard says "C Will die", Because in the riddle, (if you follow the guidelines of the riddle) the guard HAS to say that B will live, otherwise the testrun is a failure.

I hope this helps extro see where Wonko is comeing from.

[HyToFry]

Sorry extro, after re-reading your's and Bannanas post, I think its Bannana that you are argueing with, not Wonko... hehe my bad

Bannana, extro is right, it never says B never dies, just that the guard says B doesn't die, it would be like this in the Monty Hall logic...

You pick door A, Monty opens Door B, and as extro says, "at this point your complaining because there was never a prize behind door B"

If you look at it this way, then yes, EVERY TIME YOU PICK A, AND MONTY SHOWS YOU THAT THERE IS NO PRIZE BEHIND DOOR B, THERE WILL NEVER NEVER EVER BE A prize behind door B.. Never, no matter how many times you pick door A, and Monty shows you there is no prize behind door B will there ever be a prize behind door B.

run some tests if you like.. hahahhahaha LMAO

oh p.s. extro... this was meant to go against what Bannana is saying, not that you didn't already know that

Oh and <---that ones for worm

[mithrandir]

i really can't believe this is still being discussed... this has been proven using probability theory AND several programs testing large numbers of cases, and it is quite clear to most everyone that the the 1/3-2/3 case is right. Give it up Banana.

[HappyMutant]

2 thought are arguing about a puzzle. If one tells the other it is wrong and vice versa, what's the probability I win a mango?
I just couldn't help it %)

[HyToFry]

No Offence Happy Mutant.. but... well... in basic terms


HUH??????

[Wonko the Sane]

I'm giving this one more shot and then I'm throwing in the cards but a few things to say first.

Extro, sorry for the grating post, I made the mistake of trying to debate something on a day that something really, really bad happened to me and I leaked in to my mood. So I apologize for that.

On to the topic at hand.
Okay, everyone is making some right points and some wrong ones. But nobody is making all of the right points at the same time. Here's a list of conflicting arguements

B never dies: sort of correct
A is never told C: sort of correct
B never dies in applicable cases: correct
If A can be told C, the chances are 50/50 in a large number of trials: correct
A and C's chances are always 50/50: wrong

Now, let me explain the first two points
B will die sometimes, and A will sometimes be told C. The thing is, in this problem, the cases where A is told C or when B dies must be ignored. The problem states that A will be told B. That's why in a large number of trials it's 2/3-1/3 survival chances. Now let me explain the other cases.

First, I'm going to work it backwards.

in a case where the chances are 33%/33%/33%
33% A dies
16.6% A is told B
16.6% A is told C
33% B dies
33% A is told C
33% C dies
33% A is told B
So, 50% of the time A is told B, 50% of the time A is told C
This doesn't work because there are two cases that don't work since A must be told B 100% of the time

in a case where the chances are 50%/0%/50%
50% A dies
25% Guard says B
25% Guard says C
50% C dies
50% Guard says B

in a case where the chances are 33%/0%/66%
33% A dies
16.6% A is told B
16.6% A is told C
66% C dies
66% A is told B

Seeing a pattern? Working backwards, anything works if you remove A is told C cases. Of course laws of logic say we can't do this inversly and look for a case where A is told B 100% of the time unless C dies 100% of the time. Obviously, this isn't true. This does tell us that B lives every time though.

Now to the important stuff. So how often does the cop actually choose B or C? Is it 50/50? Well, here's what we get

if B dies, the guard says C
if C dies, the guard says B
if A dies, the guard randomly chooses either C or B

Now, assuming B and C carry equal weight, which the puzzle says they do (the prisoner is randomly selected), you get 50% guard says C, 50% guard says B

So
50% guard says B
50% A dies
50% C dies
50% guard says C
50% A dies
50% B dies

The big IF is here. Are the chances that A dies or C dies really 50/50 if the guard says B? Well, not according to the logic of the problem. We should have to use

50% guard says B
16.6% A dies
33.3% C dies
50% guard says C
16.6% A dies
33.3% B dies

Why do we have to use this? I'll explain
33% B dies 33% guard says C
33% C dies 33% guard says B
33% A dies 16.6% guard says C
16.6% guard says B

Except two cases have to get removed. Since A was told B goes free the first chance (33% B dies) is removed. Also the case (A dies 16.6% guard says B) must also be removed leaving.

33% C dies guard says B
33% A dies 16.6% guard says B

So, in 33% of possible cases, C dies
in 16.6% of possible cases A dies

So C is twice as likely to die as A if A is told that B will survive. The same applies for B. That's why there is division. That's also why this only works when A can only be told be. If you can have

50% guard says B
16.6% A dies
33.3% C dies
50% guard says C
16.6% A dies
33.3% B dies

Then you end up with
33.3% A dies
33.3% B dies
33.3% C dies

Even chances. You'll never have 50/50 if the same prisoner asks about the other two prisoners and he is told the same prisoner will survive each time. That's the key to this puzzle. That good enough? Because that's the last time I say it.

[Ghost Post]

I'll go along with that.

[Wonko the Sane]

Glad to hear we're finally no longer at odds then. =>

[Ricky]

I just couldn't resist.... http://www.dartmouth.edu/~chance/course/topics/Monty_Hall.html

[Ghost Post]

I read the article at link, but find it somewhat hard to beleive that people can argue over this problem, only to find that their real difference is in the interpretation of the wording of the problem.

[Ghost Post]

I need to know why the answer is not 50%. All this prisoner jibberish is throwing me off, because it is not even the same problem.
There are three doors, one of them has a good prize behind it. Odds of getting that prize are 1/3. Right?
You pick a door. Odds of being right are 1/3 still.
You are shown a door that doesn't have the prize behind it.
Now there are two doors, one has the prize behind it, one doesn't. Whichever of these doors you pick, you have a fifty percent chance of getting the good prize. So, logically (to me at least), THE ODDS THAT CHANGING YOUR MIND WILL GET YOU THE PRIZE IS 1/2.
Wasn't that the original question? Isn't that the logical answer? Tell me what is wrong with my logic without bringing in math programs and talking about prisoners and stuff I don't understand. Just help me out please. Also, don't be mean in your response, I'm new to this site.

[friskythepig]

Hiya,
Yep, I agree it's not obvious why the probability of being right if you switch doors is 2/3 and if you stick with your first choice it's only 1/3. The easiest way I have found to picture what is happening is to imagine that there are not 3 but say, 100 doors with a prize behind only one of them. In this case you can see that if you pick one door, the chance that you are right is very small; 1/100, and it is much more likely that the prize is behind one of the other 99 doors. If Monty Hall then opens 98 of the other doors which do not contain a prize (he knows where the prize is) you are then left with 2 doors; your original choice and one other. You should be able to see now that the chance that the other door contains the prize is much greater than your original choice.
In this case, if you stick with your original choice, the chance of getting the prize is 1/100, whereas, if you switch the chance of winning is 99/100. The same is true with the 3 door problem but the odds are not so markedly different.
I hope this explanation make sense. If not, there quite a good simulation of the problem at:-
http://www.greylabyrinth.com/Puzzles/answer020.htm

which may help you to picture what is going on.

[worm]

sorry if the page i referred has confused you, kristen. i thought the prisoner problem was a very good analogy, but, but...

well,i usually don't quote myself but i thought my own explanation of Monty Logic was pretty simple (go figure):

quote:

---

if you pick the wrong door at the start, a 2 in 3 chance, he has to reveal the one remaining bad door. so switching gives you the right door.

if you pick the right door, a 1 in 3 chance, he can reveal either of the remaining doors. switching in this case equals losing.

switching will double your chances of winning.

---

does that make sense?

[worm]

kristen, i know the discussion about the prisoner problem didn't help you. just in case you are at all curious about the connection between it and the Monty Hall problem, i'll explain my view.

prisoner to be executed = winning door.

prisoner who asks the warden the question = door you pick at the beginning.

prisoner the warden says will be freed = losing door that Monty reveals.

they are essentially the same problem...except you can't really switch doors in the prisoner problem.

if you had no interest in this, please ignore this entire message

[Ghost Post]

Hello all,
I'm new here and have enjoyed reading your posts. I hate to open up what appears to be a healed wound but think that some of you are arguing different points. Firstly, the Monty Hall and Prisoner puzzles are much different puzzles. The Prisoner puzzle is viewed from a single context and the Monty Hall puzzle is viewed from two different contexts.

In the Prisoner puzzle everything is constant, you cannot change which prisoner you are and you cannot change the choice of the warden. The probably is calculated when the warden makes his random choice and it does not change. Regardless of whether he tells you who is going free or not. Your probability for being executed is 1/3 and going free is 2/3. The warden can tell you who one or both of the prisoners are to be freed and who will be executed, it doesn't change the probably of it happening, it *does* change the probability of you knowing the outcome of what actually did happen. I hope that's clear.

Wonko - I think you may have been trying to get this point across in an earlier post.

As FriskyThePig suggested, it might be easier to think of the Monty Hall puzzle if you use 100 boxes with one containing the prize. In the Monty Hall puzzle, the probability of a box containing the prize is calculated twice but the first is irrelevant. The first probability is calculated when Monty is setting up the stage and chooses the correct box with the prize (in this case 1/100). Initially you have a 1/100 chance of the box you choose containing the prize. When Monty then shows you one of the boxes that does not have the prize and gives you a chance to re-choose, this is an altogether new scenario or 2nd context which is not dependant on the first (he may as well have not asked you to choose a box the first time at all). When Monty has eliminated one box from the equation, you now have a choice of 99 boxes, one with a prize (1/99 chance that a box has a prize). If this is difficult, think of it this way - Monty may have just asked the first time: "there are 99 boxes and one has a prize - choose one, oh yeah ... there is another box over here but it is empty." The probability of the prize being in a box starting out was really always 1/99 as long as it was a given that Monty was going to take away one of the boxes. So in the 3 box version, the answer is, when you make your second choice (which is the only choice that applies) there is a 1/2 probability that it is correct (the fact that there were originally 3 choices is irrelevant). The first probability of 1/3 has no bearing on your second choice. The Java script (nice app BTW) never really takes into consideration that you were never really making a decision based on 3 boxes but are actually making it on based on 2.

[Sofis]

Oh, look guys. It's a fifty-fiftyer. Let's beat him up.

[Ghost Post]

bink: "The warden can tell you who one or both of the prisoners are to be freed and who will be executed, it doesn't change the probably of it happening"

A probability is not something tatooed on the ass of reality. It's an estimate based on information, and new information warrants a new estimate. So if the warden tells you the other two prisoners are to go free, it sure as hell changes the probability that you will the one of the three that gets executed.

[Ghost Post]

And another thing, binky: You missed the whole point of friskythepig's example.

Suppose there are 100 doors, and behind one is a prize. You get to pick a door. Then, at least 98 of the remaining 99 doors have no prize behind them. Monty will then open 98 of those 99 doors (but never opens the door with the prize). Now there are two doors. You will lose by switching if and only if your initial pick of one door out of the 100 was correct. How often (with what probability) will your initial pick of one door out of 100 be correct? You will win by switching if the prize was behind any one of the 99 doors you didn't pick. Isn't it obvious that it is far more likely that the prize was behind one of the 99 doors than behind the one you picked?

I'm willing to hop on a plane and fly to wherever you are and gamble big bucks on this (local restrictions may apply). We'll do 100 rounds, $10,000 each round, as follows:

Each round, I put up $9000, you put up $1000.
First round, I place a prize in one of 100 boxes. You pick a box. I open 98 of the remaining boxes. You don't switch. You win the $10,000 if the prize is in the box you picked.
Next round, you place a prize in one of 100 boxes. I pick a box. You open 98 of the remaining boxes. I switch to the one you didn't open. I win the $10,000 if the prize is in the box I switched to.

According to you, we'll both win about an equal number of times, so about 50 times I'll win $1000, and about 50 times I'll lose $9000. And about 50 times you'll lose $1000, and about 50 times you'll win $9000. That should sound good to you. You'll come out about $400,000 ahead (which I'll lose).

Of course, I believe I'll lose $9000 about once, and win $1000 about 99 times. I'll be about $90,000 ahead (which you'll lose).

Also, you wrote "The Java script (nice app BTW) never really takes into consideration that you were never really making a decision based on 3 boxes but are actually making it on based on 2." Explain. There are three doors, you pick one, then one without a prize is opened, and you are allowed to switch. Just like in the Monty Hall problem. If you switch, you win 2/3 of the time, and if you don't, you win 1/3. How is that different than the Monty Hall problem?

[Wonko the Sane]

I'm going to try and end the arguing on the prisoner problem because I went through the painful process of realizing WHY you have a 2/3 probability of surviving after you ask.
You don't really have a 2/3 probability of surviving after you ask. You have a 2/3 probability of surviving after you ask and are told that prisoner B will survive. That's different. If you just ask the question point blank and you can be told either prisoner B or C, you still have a 1/3 change of survival. However, that's not how it goes. Prisoner B MUST survive. Except prisoner C was only chosen to be executed 1/3 times, so part of the time that you're told B will survive, you'll die, and part of the time, C will die. The trick is figuring out what it is. Well, as it turns out, all the times that prisoner B dies are irrelevant. So you lose some cases. let's look at a breakdown

Possible cases:
1/3 You die
1/6 You are told B
1/6 You are told C
1/3 B dies
1/3 You are told C
1/3 C dies
1/3 You are told B

Now, we've got to remove all cases where you aren't told B, right? Here goes
1/6 You die and you are told B
1/3 C dies and you are told B

Now we have to reduce this ratio
1/3 You die and you are told B
2/3 C dies and you are told B

Which in practical language means
1/3 You die
2/3 C dies

See? It's the being told a specific person will live that matters. If you are told C will survive every time, then it'll be
1/3 You die
2/3 B dies

But if you can be told both then you have to put these two cases together. Let's change the ratios back to 1/6 and 1/3 and you get
1/6 You die
1/3 C dies
1/6 You die
1/3 B dies

Which is
1/3 You die
1/3 B dies
1/3 C dies

What's important is that you can't be told that prisoner C will survive. Ever. B will survive every time and you will never be told that C will survive, ever. The real question isn't "does your chance of survival increase" but "do you have a better chance of surviving if you will be told B when you ask who of the other two will survive". Make sense? if not, I'll try again.

[Ghost Post]

I don't understand why B always dies? And does it matter? You still have a 2/3 chance of surviving whether your told B or C dies, which is the same as before you asked. I think. Is that right?

[HyToFry]

Thats right, and B doesn't always die, (actually the way the riddle is told B always lives), its just that B always has a chance of dieing.

[Ghost Post]

Hey, I was nice enough to give you back the worm password, wasn't I?

[HyToFry]

Uh... no.. not me

I don't wants worms password, and neither does he

unless you are worm.. oh wait.. your registered, you can't fool me

[Ghost Post]

First - would whoever screwed this thread up kindly fix it.

Second - "how can the probability of something being behind one door with two choices be anything but 50%??"

Well, suppose one door was an itsy-bitsy teeny-tiny door, and the other was a really big door, and I opened them both, and while blindfolded, threw the prize in the general direction of the two doorways, and repeated this until the prize went in one of them?

It's really no different with the Monty Hall puzzle. You pick door A. Doors B and C are then really like one bigger door - call it door BC. There is a 1/3 chance the prize is behing A, 1/3 behind B, and 1/3 behind C. Or, another way to look at it: 1/3 chance it's behind A, and 2/3 it's behind BC. Because if it's behind either B or C (2/3 chance), you will win if you switch to door BC (whichever door of B and C does not get opened by Monty).

Suppose you could choose to open one door (A, B or C), or two (B&C, A&C or A&B)? Wouldn't you choose to open two doors? It's the same thing. You pick A. If you switch, it's like being able to choose doors B AND C. Monty opens one for you, you open the other, and IF EITHER HAS A PRIZE, YOU WIN. See how simple?

[Ghost Post]

Sofis,
I'm not really a fifty-fiftyer in the sense that you are thinking of.

Extro,
With all due respect, probability *is* something that is "tattooed on the ass of reality." It is a finite thing that can be calculated over and over to an exact degree of accuracy.

Probability is a quantitative description of the likely occurrence of a particular event. The actual outcome of an event is not dependent on the probability that the outcome will occur and the probability of an outcome is not dependent on the actual outcome. Both exist completely independent of each other. Probability is a tool used to make an educated guess as to what the outcome of an event will be.

Additional knowledge on what the outcome actually is, is also a *different* tool used to make an educated guess as to the outcome of an event. The two, however, are not measuring the same thing and cannot be added or subtracted. A probability is calculated only once when the conditions of the experiment are defined and the sample space is determined. The only time that it can be recalculated is if the conditions of the experiment are redefined or the sample space has *changed*. In the Monty Hall puzzle the sample space does change. In the Prisoner puzzle the sample space does not change nor do the conditions of the experiment change.

You have a friend that flips a coin. I say (and assume you would agree) that the coin has a 1/2 probability of landing heads and a 1/2 probably of it landing tails. Your friend tells you that the actual *outcome* of the toss was not heads. Would you now tell me that the probability needs to be recalculated, and that the coin had a 2/2 probability of landing tails? I would say no, the coin had a 1/2 probability of landing tails, the actual *outcome* was tails, but you have a 2/2 chance of guessing that the actual outcome was tails. The sample space (or possible outcomes) never changed (heads, tails) and the condition of the experiment never changed.

In the case of the Prisoner puzzle: Just because the warden tells you that prisoner c is going free does not mean that he did not have a 1/3 chance of being executed like you and prisoner b. It does not mean that you can adjust his probability to 0/3 because you know he has been let free. Adding "information about the outcome" to "the probability that a given outcome is likely to happen" is like adding apples and oranges. I understand your math and I understand what it is showing. The problem is it is comparing unlike things. It gives you the result you are seeking but not because of the reason you think it is.

Try doing an experiment and calculate the outcome of how often either of the remaining 2 prisoners (that weren't eliminated by the warden) were chosen to live. But ... have them loose their identity a little.

If Prisoner A and B are left over consider
prisoner A = "left prisoner" and prisoner B = "right prisoner"
If Prisoner A and C are left over consider
prisoner A = "left prisoner" and prisoner C = "right prisoner"
If Prisoner B and C are left over consider
prisoner B = "left prisoner" and prisoner C = "right prisoner"

I bet your outcome will be close to left prisoner = 1/2; right prisoner = 1/2.
Which hopefully will show you that it always was:
1/3 probability of death; 2/3 probability of freedom.
The chance of freedom was not affected by information from the warden.

I understand that this is a complicated idea - this is why it is considered a paradox and so many intelligent people take different sides. This is also why so many people want so badly to believe that the answer is 1/2, their logic tells them so and their logic is mostly right.

Like I said earlier, the Monty Hall Puzzle is a much different problem:

Say you and your significant other were planning a wedding. You make a guest list with 100 people on it. Your SO sends out the invitations and tells you that your friends don't like you much and there was only one person who returned an RSVP (assume this is random). She asks you to guess who it is. You guess, but then she tells you that she forgot to tell you that you can only afford to feed 75 people so she only sent out 75 invitations but that your guess was one of the 75 people she sent invitations to. She asks if you want to re-guess. Should you re-guess? This is the same problem as the Monty Hall Puzzle.

The 25 people that you did not send invitations to did not have a chance to RSVP because they were never sent an invitation - they were never really in the sample space in the first place. Every event must have an outcome. The original guess over the 100 people never had an outcome, so it was never really an event. The only real event was the guess that was made over the 75 people. The only real sample space was the folks that received an invitation and were allowed to RSVP. I hope this is clear.

So carry this idea over to the 3 box puzzle. When Monty identifies one of the empty boxes, remove it from the table. The sample space for your event is now 2. Then add up how many times the remaining right box and left box have a prize. I will bet you all $90,000 that it will end up 1/2. There will always be even distribution of the sample space in a simple event.

I won't argue that you get the result you want using your math -- even though it is not proper. Math is funny that way, sometimes bad equations give us good solutions. The point is: even distribution of 1/2 in a simple event with a sample space of 2 is not a wrong answer.

[friskythepig]

Binky, I'm not sure you understand how the Monty Hall game is played. In your game, there seem to be three boxes one of which contains a prize. If Monty removes one of the empty boxes then yes, there is a fifty-fifty chance that the remaining boxes contain a prize.
However the way the Monty Hall game is played is that you pick one of the boxes first. Monty then removes one of the other two boxes that does not contain a prize. 2/3 of the time you will pick an empty box initially. Monty removes the other empty box, which means that you will win if you switch, 2/3 of the time.
If you want to play the game this way I'll happily take your $90,000

[Ghost Post]

binky wrote: With all due respect, probability *is* something that is "tattooed on the ass of reality." It is a finite thing that can be calculated over and over to an exact degree of accuracy.

I disagree, and let me explain what I mean. Suppose we are standing in a room with a black and white checkered floor. The floor has a checkerboard pattern of black and white squares, about 6 inches square. Half the area of the floor is black, half is white. The room has a very high ceiling, and I take a small steel ball, toss it high into the air, and, while it's in the air, ask you the probability that it will land on a white square. I think you'll agree that it's 1/2.

But at this point, while the ball is in the air, it has a very definite trajectory. If we knew it's exact position, direction and velocity at any given moment, we could calculate exactly where it would land, and then we would either say that the probability of it landing on a white square is 1, or we would say it is 0.

So it isn't that the ball has, in some real and objective sense, a 1/2 probability of landing on a white square. Probabilities are just estimates made using available information. The same may be true of a coin toss. I toss a coin, and it comes up heads. Can you tell me that it might have been otherwise? Sure, other tosses came up tails, and will come up tails. But can you tell me that that particular toss was not "predestined", in a sense, to come up heads? No, you can't. And it doesn't matter. Maybe the event was predestined to happen a particular way (heads), but we have no way of knowing, so we estimate the probability. But we do so out of ignorance of the trillions of factors that will determine the outcome. We do so out of ignorance of reality. It may be that the only probabilities "tattooed on the ass of reality" are 0 and 1, but we don't get to see them.

This has nothing to do with math, but is more a philosophical concept of what a probability is. I think I agree with the way you do your math, but I think you've misread the Monty Hall puzzle. With the Monty Hall puzzle, you pick a box first. There is a 1/3 chance you picked the prize. Now, Monty always does the following: of the two boxes you didn't pick, he always opens one that does not have a prize. There is still a 1/3 probability that the box you picked has the prize, right? If you are not sure, consider: do 100 trials, where you pick one of three boxes. How many times will you pick the correct one? Now do 100 trials where you pick one of three boxes, and then Monty opens an empty box (and never the one you picked). How many times was the box you picked the correct one? It's the same - about 33 out of 100. So if 33 times out of 100 you pick the correct box, then 67 times out of 100 the prize is under one of the other two boxes, and specifically, in the box that you didn't pick and Monty didn't open (he never opens a box with the prize).


[This message has been edited by extro... (edited 05-09-2000).]

[Tom]

Nicely put, extro..., I suppose probability is (in a deterministic universe) more to do with the observer than the event. (Hence the puzzles which say "given such-and-such what is the probability that some past event happened" - in this case, the outcome has already been chosen.)

One question - how does quantum fit into this? As I remember, (for example) some atom decays in t time with probability p .. in this case there is no way to get the information, as it is not a matter of us just not knowing the information. That said, quantum is horrribly observer dependent .. does anyone know about this? (Is anyone reading this thread an more?)