The Final Solution to Monty Hall Problem

[Ghost Post]

OK. Here's the solution to the problem we've all been jibbering about for the past week. I'm going to go thru this nice and slow so everyone understands what's going on. I had a math bud of mine with more letters behind his name than the alphabet read thru all the posts and had him give me his response to post here. He says he found all of this quite amusing.

We are all familiar with the first problem with the contestant choosing a door, behind which one of the three holds a big prize. So, let's look at our options.

DOOR 1 2 3

A L L W
B L W L
C W L L

Now, at the start, our contestant has a 1/3 chance of winning. Easy. So, for argument's sake, our contestant chooses A.

So, the argument goes, at this point Monty reveals a losing door (among the remaining 2) and asks the contestant whether he wishes to change his door.

We can see from the table that if the contestant changes, he has a 2/3 of winning as opposed to 1/3 by staying with his original door.

The reason is, and an IMPORTANT POINT to remember for the next problem, is that MONTY KNOWS THE PRIZES BEHIND ALL 3 DOORS.

That is the key. From our table we can see that in Case A, Monty will open door 2. In Case B, Monty opens door 3. In Case C, Monty opens door either door 2 or 3 (doesn't matter because they are both losers).

So remember, if Monty does not know what is behind all 3 doors, the 2/3,1/3 result will not occur.

So let's look at the other problem.

In order to avoid an argument from all the people who post, I will retype the original problem WORD FOR WORD as originally posted.

"There are 3 prisoners A, B, C. One is to be executed the next day and the other 2 set free. A goes to the warden and finds B is to be set free."

That is the problem. The poster offered this solution.

"As in the Monty Hall problem, A has a 1/3 chance to die, and C has a 2/3 chance to die."

No, that is incorrect. Why? Simple. Let's look at a graph of this setup.

Prisoner A B C

Case 1 L L D

Case 2 L D L

Case 3 D L L

Now in order for the 2/3, 1/3 theory to work, we again have to realize that THE WARDEN ALREADY KNOWS WHO LIVES AND WHO DIES. With this in mind, in Case 1, the warden would tell A that B is living. In Case 2, the warden would tell A that C is living. In Case 3, the warden would tell A either B or C was living (it doesn't matter).

You see? The solution is the same as the Monty Hall problem. In the Monty Hall problem, if the contestant switched doors, he increased his probability of winning from 1/3 to 2/3. In the Prisoner problem, the person who asks the question increases his survivability from 1/3 to 2/3.

So, at this stage, I'm sure we all agree. But now everyone says "Well why was Banana making such a big stink about the problem?"

Look at this section of the original problem again.

"A goes to the warden and finds B is to be set free."

That, as they say, is the flaw in the slaw. This statement has changed the complexion of the question. With this statement, The Prisoner problem no longer mirrors the Monty problem. Why? Because you have eliminated possibilities by forcing B IN ALL CASES to be set free. Because of this, we no longer have Case 2 in our Prisoner problem because according to the question, Prisoner B NEVER DIES.

This or course leaves us with just 2 scenarios.

PRISONER A B C

Case 1 L L D

Case 2 D L L

Now you can see why Banana kept saying that "A and C have an equal chance of surviving, it is no longer 1/3 2/3 like the Monty Problem."

You see, as I stated earlier, it was all a question of semantics....the wording of the original puzzle.

In order for the Prisoner puzzle to mirror the Monty puzzle, you would have to word it this way.

There are 3 prisoners A, B and C. One of them is going to die in the morning. If Prisoner A goes up to the warden the night before and asks "Can you tell me, of the remaining two prisoners, which one of the two will you set free?" will saying this statement increase the probability of A going free the next day?

The key again is to remember that THE WARDEN ALREADY KNOWS WHO WILL DIE. Thus, he skews his answers to A and thus A increases his probability as in the Monty problem.

So, overall, we can see that the error in logic was to make the "A is told that B is free" statement in the problem. This eliminates the Case where B is to die on our chart and the warden tells A that C will be set free.

So, in closing, the people who stated that A and C have an equal chance of dying were correct in the terms of the original statement of the problem. The rest did not notice this slip up and hence thought we were disputing the logic of the Monty Hall problem itself, when the wording of the problem eliminated a scenario crucial to obtaining the 2/3 1/3 answer. If B is forced to live, the Monty logic does not apply in the same fashion as if Monty was always forced to open Door number 2 rather than switch between 2 and 3 depending on the circumstance.

[Ghost Post]

Looking back at the original thread, you can see that Drew and I were the only ones who read the question properly.

But, I know it was a simple mistake by everyone else and basically poor wording of the problem by the original poser of the question.

I hope Drew and myself aren't annexed out of the club because of our rants. Let's put all this negativity behind us and get back to some good puzzle solving

[Ghost Post]

The original problem was:

"Three prisoners on death row are told that one of them has been chosen
at random for execution the next day, but the other two are to be
freed. One privately begs the warden to at least tell him the name of
one other prisoner who will be freed. The warden relents: 'Susie will
go free.' Horrified, the first prisoner says that because he is now
one of only two remaining prisoners at risk, his chances of execution
have risen from one-third to one-half! Should the warden have kept his
mouth shut?"

You write:

Look at this section of the original problem again.
"A goes to the warden and finds B is to be set free."
That, as they say, is the flaw in the slaw. This statement has changed the complexion of the question. With this statement, The Prisoner problem no longer mirrors the Monty problem. Why? Because you have eliminated possibilities by forcing B IN ALL CASES to be set free. Because of this, we no longer have Case 2 in our Prisoner problem because according to the question, Prisoner B NEVER DIES.

The statement changes nothing.

In the original problem, this is not something that is going to be repeated over and over again. One prisoner will be executed, the other two will be set free. They'll die someday, but that's beyond the scope of the puzzle. So prisoner A asks to be told one of B or C who will be set free.

Now, here's the problem. This is only happening once. The warden answers the question once. He might answer B, and he might answer C. We don't know which, until he answers. He answers B, and you object. Would you object if he had answered C? I would hope so, because it's the same problem. What answer would you not object to?

What is you complaint about "B NEVER DIES"???

There is no problem with the wording of the problem. The important point is the nature of the question asked, in that it allows only two possible answers: B or C. One of these will be the answer to the question (your upset that it happens to be B). Neither answer changes A's chances of dieing.

Here's "my version" of Monty Hall:
Monty has a prize behind one of doors A, B and C. You pick A. He then tells you that of the other two doors, B and C, he will open one behind which there is no prize, and allow you to switch to the other unopened door. He opens door B, which has no prize behind it. Do you switch to door C? Are you going to argue "B NEVER HAS A PRIZE"?

[araya]

well Banana you seem to crave some congratulations, so here's something for your mantle

code:

---

        __/\__ 


    ----      ----


    \ * Banana * /


     \          /


       \      /


         \  /


         /  \


      __------__

---

Unfortunately, you and your math Ph.D friend are still wrong.

[friskythepig]

This is fun isn't it.
Guess what? I found yet another variation on the Monty Hall problem. It is essentially the same problem but this time there are 3 convicts hidden behind 7 doors, one of which is going to be executed by Monty Hall......

love and kisses, Frisky.

[worm]

Ooh flashback...that trophy's just like what I used to get as a little kid when my Teeball/soccer team didn't do well. Kind of a pity trophy all the parents had to pay for. I know it's hard, Banana, I feel your pain. It's a lot more fun getting a trophy when you actually win.

p.s. Any chance Araya will make a trophy for the Ph.D. buddy?

[HyToFry]

Bannana,The way you explained the problem in your first post, you state that B will never die, but we already know this

Of the three prisoners, at the point when the wardon has tells A that B will not die, the prisoner that will die has already been picked, (lets assume that C will be killed for arguments sake)

At that point these are the TRUE probablity of deaths

A 0%
B 0%
C 100%

Which is to say that 100% of the time when C is picked to die, C will die. See my point? I hope.

All the question is saying is that 66% of the time when A is told that B will live, it will be C that dies, and inadverently 66% of the time when A is told that C will be free, B will die.

This neither increases or decreases A's chances of going free, or dieing, it just gives A a better idea of who out of B and C will die if A lives.

This is because there are only four possible answers that the wardon can give

1. A will die, and Wardon will say C 1/6

2. A will die, and Wardon will say B 1/6

3. B will die, and Wardon will say C 1/3

4. C will die, and Wardon will say B 1/3

now because when the Wardon says B will live, there is a 1/3 chance that C will die, and a 1/6 chance that A will die, C's chances of dieing (WHEN THE WARDON TELLS A THAT B WILL LIVE), are 66%, and A's are STILL 33%, this neither HELPS, nor HURTS A's OR C's chances, because the person who will die has already been selected and has a 100% chance of dieing, the question is only saying that 2/3 times C will be killed if A is told that B will live.

what is very interesting is that if A and C both ask the wardon together, and he tells them both that B will live, the chances of A and C dieing are equall 1/2 times when A and C are told that B will live, A dies, and 1/2 times C dies.

Hope this clears it up for yaz

[Ghost Post]

Ahha!! I have figured out what was confusing me and everyone else about this problem. The post that states the problem gives it in two versions.

First:
"It is essentially the same problem but this time there are three prisoners A,B and C on death row. One of them is to be executed the next day and the other two set free. Prisoner A goes to see the warden and is informed that B is to be set free."


And Second:
"==> decision/prisoners.p <==
Three prisoners on death row are told that one of them has been chosen
at random for execution the next day, but the other two are to be
freed. One privately begs the warden to at least tell him the name of
one other prisoner who will be freed. The warden relents: 'Susie will
go free.' Horrified, the first prisoner says that because he is now
one of only two remaining prisoners at risk, his chances of execution
have risen from one-third to one-half! Should the warden have kept his
mouth shut?"

The second version is the same as the Monty Hall problem. A asks the warden to "tell him the name of one other prisoner."
In the first problem, we are not told what A asked, but just that he is told B will live. For all we know, the warden could have told A that he would live. But Monty would never open the door you originally chose.

All the time we were really discussing two very similar, but different problems.

[Ghost Post]

Yeah, I thought about that, but I though it was pretty clear that the first statement was just talking about the problem and how it is similar to Monty Hall, prior to actually presenting the problem (which is presented verbatim as it is found at the given URL).

Besides, read Banana's first post in this thread, and tell me if that is what he's trying to say. I don't think so. He writes:
"In order for the Prisoner puzzle to mirror the Monty puzzle, you would have to word it this way. There are 3 prisoners A, B and C. One of them is going to die in the morning. If Prisoner A goes up to the warden the night before and asks 'Can you tell me, of the remaining two prisoners, which one of the two will you set free?' will saying this statement increase the probability of A going free the next day?"

Notice the "will saying this statement increase the probability of A going free the next day?" part. He seems to understand the question that A asks the warden. He then raises the following objection:
"the error in logic was to make the 'A is told that B is free' statement in the problem"

Somehow, he accepts that prior to hearing the wardens answer to A's question the problem is like Monty Hall, but that once the answer is given, it's altogether different. The whole point is that, with the question that Banana seems to accept (above), if the answer is B, C will then have a 2/3 chance of dieing, and vice versa.

He simply objects to being informed about the wardens answer to the question, which is irrelevant to A's chances of survival.

[Ghost Post]

He could have done what I think I did. Start to write about the problem and then think a bit. Go back and read the problem again, but this time reading a different version. Then you come to different conclusions and write contrary things in your post.

"the error in logic was to make the 'A is told that B is free' statement in the problem" This statement is in version 1, but not version 2.

Even though the first version was just a prelud eto the actuall question, if you don't realize there is a difference you won't really care which you are reading.


[This message has been edited by Drew (edited 03-29-2000).]

[Terry]

My 2 cents...

Since alot of this discussion seems to be on wording, I think it should be noted that 2 different things are being discussed. The prisoner problem, once started, never increases or decreases A's chance of dying, it simply increases his chance of knowing who will die.
Similar to the Monty Hall situation, A has 1/3 chance of dying and 2/3 chance of survival. The fact that the warden tells him the name of a "to be freed" prisoner does not really help his plight, it simply gives him a better chance at guessing who the dead prisoner will be.

Oh well, don't have time for more right now...

[Wonko the Sane]

I said this on my other post and I'll say it hear. The reason the probability of A surviving is better if in a large number of trials he's told the same person over and over again isn't only because B never dies. It's because B never dies, and 50% of the time that A dies, he's told that C will survive. 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. When it's re-randomized, 50% of the time either B will die or A is told C will live and the case is re-randomzied. So basically, B isn't entirely removed from the puzzle. Rather, using limits, 16.66% of the time, the re-randomization causes C to be chosen. Thus, though as a general rule, it would be 50/50, 16.66% of the time, the re-randomizations cause C to be chosen instead of A. So A's chances of dying are reduced from 50% to 33.33% and C's chances rise from 50% to 66.66%. Make sense yet? It's not a fair system.

[Ghost Post]

Huh? Must be a full moon.

[Wonko the Sane]

Extro...read the post I put on the other thread. It explain it a lot better.

[Ghost Post]

And read my reply.

[Wonko the Sane]

Little message to everyone. Most of the info on this post is on the other one. So we might as well just move all our replies to the other thread.