ISF Logo   IS Forum
Forum Index Register Members List Events Mark Forums Read Help

Go Back   International Skeptics Forum » General Topics » Science, Mathematics, Medicine, and Technology
 


Welcome to the International Skeptics Forum, where we discuss skepticism, critical thinking, the paranormal and science in a friendly but lively way. You are currently viewing the forum as a guest, which means you are missing out on discussing matters that are of interest to you. Please consider registering so you can gain full use of the forum features and interact with other Members. Registration is simple, fast and free! Click here to register today.
Reply
Old 21st February 2019, 07:30 AM   #41
bruto
Penultimate Amazing
 
bruto's Avatar
 
Join Date: Jun 2005
Location: Way way north of Diddy Wah Diddy
Posts: 24,201
Originally Posted by Dave Rogers View Post
Which, if I recall correctly, Monty himself stated was not the case.

Dave
In the version I heard, which came from Marilyn Vos Savant, and which made quite a stir, it is clearly stated, as I recall, in the form "Monty, who knows what is behind the doors...."

But it doesn't matter anyway. If Monty's choice were random, he would have a 50/50 chance of getting the prize door and thus ending the game. If the game continues, it means his choice was the same as if he knew what was behind the doors, and the odds for the player remain the same.
__________________
I love this world, but not for its answers. (Mary Oliver)

Quand il dit "cuic" le moineau croit tout dire. (When he's tweeted the sparrow thinks he's said it all. (Jules Renard)

Last edited by bruto; 21st February 2019 at 07:37 AM.
bruto is online now   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 07:31 AM   #42
The Great Zaganza
Maledictorian
 
The Great Zaganza's Avatar
 
Join Date: Aug 2016
Posts: 8,650
Originally Posted by p0lka View Post
if you do the adding before the multiplication you get 5 + 1 = 6, then 6 x 10 = 60.
so you need to make at least two mistakes to get to 60.

thanks.
__________________
Careful! That tree's bark is worse than its bite.
The Great Zaganza is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 07:44 AM   #43
abaddon
Penultimate Amazing
 
abaddon's Avatar
 
Join Date: Feb 2011
Posts: 18,721
Originally Posted by arthwollipot View Post
The puzzle seems to me (like the second one that MEequalsIxR posted) to be a deliberate trap. I don't know what it's purpose is. Perhaps to make people feel stupid. Why someone would want to do that is beyond me.
I suspect you pretty much nailed it with one caveat. The purpose seems to be no more than to make the originator feel all smug and self satisfied.

The commonality with all of these is that some critical information is either concealed or entirely omitted.
__________________
Who is General Failure? And why is he reading my hard drive?


...love and buttercakes...
abaddon is online now   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 07:57 AM   #44
The Great Zaganza
Maledictorian
 
The Great Zaganza's Avatar
 
Join Date: Aug 2016
Posts: 8,650
Well, such tests can highlight flaws in our perception, like this one:

How many F'S can you find in the following sentence?

FINISHED FILES ARE THE RESULT OF YEARS OF SCIENTIFIC STUDY COMBINED WITH THE EXPERIENCE OF YEARS.

Answer:
6
__________________
Careful! That tree's bark is worse than its bite.

Last edited by The Great Zaganza; 21st February 2019 at 07:58 AM.
The Great Zaganza is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 08:14 AM   #45
alfaniner
Penultimate Amazing
 
alfaniner's Avatar
 
Join Date: Aug 2001
Posts: 20,019
Originally Posted by abaddon View Post
I suspect you pretty much nailed it with one caveat. The purpose seems to be no more than to make the originator feel all smug and self satisfied.

The commonality with all of these is that some critical information is either concealed or entirely omitted.
Right -- I think it's missing a couple of "given"s.
__________________
Science is self-correcting.
Woo is self-contradicting.
alfaniner is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 09:21 AM   #46
RecoveringYuppy
Philosopher
 
Join Date: Nov 2006
Posts: 8,858
Originally Posted by pgwenthold View Post
As the problem is typically stated, you have to assume that Monte is playing fair.
I see others have addressed this so I'll add a link.

https://en.wikipedia.org/wiki/Monty_Hall_problem

The problem is stated in a way that requires no assumptions. Some people get hung up on the idea that Monty didn't allows follow the assumptions in the problem.
__________________
REJ (Robert E Jones) posting anonymously under my real name for 30 years.

Make a fire for a man and you keep him warm for a day. Set him on fire and you keep him warm for the rest of his life.
RecoveringYuppy is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 09:34 AM   #47
JoeMorgue
Self Employed
Remittance Man
 
JoeMorgue's Avatar
 
Join Date: Nov 2009
Location: Florida
Posts: 18,774
I think the Monty Hall Problem variation "The Three Prisoner Problem" actually addressed the counter-intuitiveness better because in it the "Monty" part of the problem is being honest rather than dishonest (or ulterior I guess) so we can just look at the information given rather then second guessing a third party's second guessing.

Ted, Bill, and Harry are arrested. They are told that tomorrow morning, two of them are going to be executed and one is going to be set free, but none of them are told who it is going to be. They are then put in separate cells to await the morning.

As the night goes on Ted becomes more and more nervous. He asks, begs, pleads with the guard to please, please, tell him if he's one of the ones who is going to die.

The guard thinks for a moment and says, and we will assume a truthful guard for our purposes, that he can't tell Ted if he is going to die or not. But he will tell him this. Bill is definitely one of the two who is going to be executed.

Ted is pleased. From his POV his chances of survival have raised from 1 in 3, to 1 in 2.

But that's not correct. He's not factoring in the variable that the answer the guard was going to give him was already at 50/50 since the guard couldn't tell him his fate the information given to him only showed that Bill and Harry each have a 1 in 3 chance of dying and he already new that. It doesn't matter (for our mathematic purposes here for Ted's POV) whether it's Bill or Harry that die, he knows at least one of them is going to and no new information has been added to his perspective that really changes anything.
__________________
- "Ernest Hemingway once wrote that the world is a fine place and worth fighting for. I agree with the second part." - Detective Sommerset
- "Stupidity does not cancel out stupidity to yield genius. It breeds like a bucket-full of coked out hamsters." - The Oatmeal
- "To the best of my knowledge the only thing philosophy has ever proven is that Descartes could think." - SMBC

Last edited by JoeMorgue; 21st February 2019 at 10:30 AM.
JoeMorgue is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 09:46 AM   #48
Thermal
Philosopher
 
Thermal's Avatar
 
Join Date: Aug 2016
Location: Currently Dismembered
Posts: 7,622
Originally Posted by JoeMorgue View Post
I think the Monty Hall Problem variation "The Three Prisoner Problem" actually addressed the counter-intuitiveness better because in it the "Monty" part of the problem is being honest rather than dishonest (or ulterior I guess) so we can just look at the information given rather then second guessing a third party's second guessing.

Ted, Bill, and Harry are arrested. They are told that tomorrow morning, two of them are going to be executed and one is going to be set free, but none of them are told who it is going to be. They are then put in separate cells to await the morning.

As the night goes on Ted becomes more and more nervous. He asks, begs, pleads with the guard to please, please, tell him if he's one of the ones who is going to die.

The guard thinks for a moment and says, and we will assume a truthful guard for our purposes, that he can't tell Ted if he is going to die or not. But he will tell him this. Bill is definitely one of the two who is going to be executed.

Ted is pleased. From his POV his chances of survival have raised from 1 in 3, to 1 in 2.

But that's not correct. He's not factoring in the variable that the answer the guard was going to give him was already at 50/50 since the guard couldn't tell him his fate the information given to him only showed that Bill and Harry each have a 1 in 3 chance of dying and he already new that. It doesn't matter (for our mathematics purposes here for Ted POV) whether it's Bill or Harry that die, he knows at least one of them is going to and no new information has been added to his perspective that really changes anything.
I think he is correct in determining his chances improved. Basically, the guard says Bill does not go free. That means it is either Ted or Harry. Fiddy fiddy.

eta forgot: that the guard eliminates a chance altering variable, so it must affect the odds from Ted's pov
__________________
"Half of what he said meant something else, and the other half didn't mean anything at all" -Rosencrantz, on Hamlet

Last edited by Thermal; 21st February 2019 at 09:48 AM.
Thermal is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 10:27 AM   #49
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Thermal View Post
I think he is correct in determining his chances improved. Basically, the guard says Bill does not go free. That means it is either Ted or Harry. Fiddy fiddy.

eta forgot: that the guard eliminates a chance altering variable, so it must affect the odds from Ted's pov
Isn't that the same mistake people make with the monty hall problem?

Ted's odds haven't changed. All that happened was that the guard showed him a goat, Ted's odds of being a car (going free) are still 1 in 3.
His odds of being the other goat (executed) are still 2 in 3.

Last edited by p0lka; 21st February 2019 at 10:30 AM.
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 10:41 AM   #50
JoeMorgue
Self Employed
Remittance Man
 
JoeMorgue's Avatar
 
Join Date: Nov 2009
Location: Florida
Posts: 18,774
Exactly. No new information has been given to Ted. Because Harry dying or Bill dying are the same thing from Ted's perspective. It's the number of people who aren't Ted who are going to die, not the identity of any one person.

Ted already knew that either Bill or Harry was gonna die before the guard said anything and which one didn't matter.

Now he knows that Bill is going to die. He has not gotten any further information that matters to his fate.

Before the guard's info he knew that at least one of two people (Harry and Bill) are definitely going to die.

After the guard's info he still only knows that... one of two people are definitely going to die. He has no concrete evidence about his or Harry's fate so the exact same variables are there. 1 person is going free, two are dying. That's all he knows now and all he knew before. Assigning value to Bill's specific fate is a Texas Sharpshooter-esque fallacy since his fate didn't have a particular value before.

No probability has been altered.

And yes this is massively counter-intuitive, even more so then the Monty Hall problem since as noted people tend to get hung up on the "honesty" of Monty Hall's new information in the Monty Hall Problem even though that's not really a factor.
__________________
- "Ernest Hemingway once wrote that the world is a fine place and worth fighting for. I agree with the second part." - Detective Sommerset
- "Stupidity does not cancel out stupidity to yield genius. It breeds like a bucket-full of coked out hamsters." - The Oatmeal
- "To the best of my knowledge the only thing philosophy has ever proven is that Descartes could think." - SMBC

Last edited by JoeMorgue; 21st February 2019 at 10:47 AM.
JoeMorgue is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 10:51 AM   #51
Thermal
Philosopher
 
Thermal's Avatar
 
Join Date: Aug 2016
Location: Currently Dismembered
Posts: 7,622
But it has. Bill is no longer possible to be the one going free. Your logic only works if the guard says an unknown one of the other prisoners dies.

From Ted's POV, one fate is no longer left to chance and not part of the guesswork any longer.

By your reasoning, if the guard said Bill and Harry will die, Ted still has only a one in three chance of freedom.
__________________
"Half of what he said meant something else, and the other half didn't mean anything at all" -Rosencrantz, on Hamlet
Thermal is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:03 AM   #52
jrhowell
Muse
 
Join Date: Jun 2009
Posts: 607
Originally Posted by JoeMorgue View Post
Ted, Bill, and Harry are arrested. They are told that tomorrow morning, two of them are going to be executed and one is going to be set free, but none of them are told who it is going to be. They are then put in separate cells to await the morning.

As the night goes on Ted becomes more and more nervous. He asks, begs, pleads with the guard to please, please, tell him if he's one of the ones who is going to die.

The guard thinks for a moment and says, and we will assume a truthful guard for our purposes, that he can't tell Ted if he is going to die or not. But he will tell him this. Bill is definitely one of the two who is going to be executed.

Ted is pleased. ...
At this point we discover that Ted is actually one of the X-men with the mutant ability to transfer his consciousness into others. He uses that to become Harry, increasing his chances of surviving.
jrhowell is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:10 AM   #53
JoeMorgue
Self Employed
Remittance Man
 
JoeMorgue's Avatar
 
Join Date: Nov 2009
Location: Florida
Posts: 18,774
Because "Tell me which of us are going to die" and "Tell me which us are going to die but you can't tell me specifically if I am going to die" are not the same question, so the "missing" probability that on the surface looks like a jump from 1/3 to 1/2 is hidden in the fact that the guard can't tell Ted directly if he is going to live so one possibility has been taken off the table, so it has to be assumed.

Look at it this way. In all scenarios the guard will not actually lie about anything and in all scenarios he is forbidden from directly telling Ted whether or not he will live or die. He is only allowed to give a definitive statement about 1 of the 2 prisoners who isn't Ted.

Scenario 1: Ted is the one who has been selected to be released in the morning. That means Harry or Bill are going to die. Here's all the possible scenarios.

1. Bill and Harry are going to die, the guard tells Ted that Harry will be executed. Bill and Harry are executed. Ted goes free.
1. Bill and Harry are going to die, the guard tells Ted that Bill will be executed. Bill and Harry are executed. Ted goes free.

Scenario 2. Ted is one of the two scheduled to be executed. That means either Harry or Bill is going to go free. Here's all the possible scenarios.

1. Bill and Ted are going to die, the guard tells Ted that Bill is going to die. Harry goes free.
2. Ted and Harry are going to die, the guard tells Ted that Harry is going to die. Bill goes free.

That's it. That's all possibilities. Any other combination of executions/information given by the guard would require the guard to either lie, directly tell Ted if he is going to live or die, or give information about both the other two prisoners, known of which he does.

If I'm wrong walk me through scenario with a different outcome that doesn't break the rules.
__________________
- "Ernest Hemingway once wrote that the world is a fine place and worth fighting for. I agree with the second part." - Detective Sommerset
- "Stupidity does not cancel out stupidity to yield genius. It breeds like a bucket-full of coked out hamsters." - The Oatmeal
- "To the best of my knowledge the only thing philosophy has ever proven is that Descartes could think." - SMBC

Last edited by JoeMorgue; 21st February 2019 at 11:12 AM.
JoeMorgue is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:23 AM   #54
Thermal
Philosopher
 
Thermal's Avatar
 
Join Date: Aug 2016
Location: Currently Dismembered
Posts: 7,622
I get the Monte Hall reasoning, but I see it as torturing logic to leap frog the math. Kind of like when you say something is a twenty five percent decrease or a thirty percent increase, depending on which way you look at it.

The odds are to determine which of the three goes free. Up front, each has a one in three chance, because there are three of them. By eliminating one possibility of going free, a variable has been removed from the equation. Three fates were no longer at play, just two.
__________________
"Half of what he said meant something else, and the other half didn't mean anything at all" -Rosencrantz, on Hamlet
Thermal is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:32 AM   #55
RecoveringYuppy
Philosopher
 
Join Date: Nov 2006
Posts: 8,858
Originally Posted by JoeMorgue View Post
Exactly. No new information has been given to Ted. Because Harry dying or Bill dying are the same thing from Ted's perspective. It's the number of people who aren't Ted who are going to die, not the identity of any one person.
Let me state this a different way. Ted has new information, it's just information about himself. His odds haven't changed. But the odds of the one of the other prisoners just jumped to 2 in 3 because it's that other prisoner that is analogous to the remaining door in the Monty Hall problem.
__________________
REJ (Robert E Jones) posting anonymously under my real name for 30 years.

Make a fire for a man and you keep him warm for a day. Set him on fire and you keep him warm for the rest of his life.
RecoveringYuppy is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:45 AM   #56
The Great Zaganza
Maledictorian
 
The Great Zaganza's Avatar
 
Join Date: Aug 2016
Posts: 8,650
The point is that Ted doesn't get to make a choice, based on the new data.
Therefore, his odds haven't changed.
__________________
Careful! That tree's bark is worse than its bite.
The Great Zaganza is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:52 AM   #57
Thermal
Philosopher
 
Thermal's Avatar
 
Join Date: Aug 2016
Location: Currently Dismembered
Posts: 7,622
Originally Posted by The Great Zaganza View Post
The point is that Ted doesn't get to make a choice, based on the new data.
Therefore, his odds haven't changed.
But if you are considering the fates of the unknowns, its 50/50.

Say there were 100 prisoners in the original dillema. Your odds of freedom are 1 in 100. You learn the fate of 98 of them. Now, between you and the last prisoner, are your odds of freedom 1 in 100?
__________________
"Half of what he said meant something else, and the other half didn't mean anything at all" -Rosencrantz, on Hamlet
Thermal is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 11:56 AM   #58
RecoveringYuppy
Philosopher
 
Join Date: Nov 2006
Posts: 8,858
Originally Posted by Thermal View Post
But if you are considering the fates of the unknowns, its 50/50.

Say there were 100 prisoners in the original dillema. Your odds of freedom are 1 in 100. You learn the fate of 98 of them. Now, between you and the last prisoner, are your odds of freedom 1 in 100?
You knew that 98 other people were going to be executed the minute the situation was first explained to you.
__________________
REJ (Robert E Jones) posting anonymously under my real name for 30 years.

Make a fire for a man and you keep him warm for a day. Set him on fire and you keep him warm for the rest of his life.
RecoveringYuppy is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 12:03 PM   #59
Thermal
Philosopher
 
Thermal's Avatar
 
Join Date: Aug 2016
Location: Currently Dismembered
Posts: 7,622
Originally Posted by RecoveringYuppy View Post
You knew that 98 other people were going to be executed the minute the situation was first explained to you.
Yes, but they have been eliminated as variables, and it is down to two, of which one walks free. One in one hundred only applies when no fates are known. As each fate is eliminated as a possibility to be the one to go free, your personal odds improve, till it is .5 when two of you and 1 if his fate is to die. Your odds of one in a hundred only apply when there are 100 variables. Each possibility eliminated change your personal odds
__________________
"Half of what he said meant something else, and the other half didn't mean anything at all" -Rosencrantz, on Hamlet
Thermal is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 12:11 PM   #60
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Thermal View Post
But if you are considering the fates of the unknowns, its 50/50.

Say there were 100 prisoners in the original dillema. Your odds of freedom are 1 in 100. You learn the fate of 98 of them. Now, between you and the last prisoner, are your odds of freedom 1 in 100?
Yes.
The decision on who is going free was made at the beginning, you knew 99 people were going to die at the start, now there is just you and another prisoner.
The odds that you are going free is still 1 in 100, assuming the decision was made at the beginning. (You are on the car)

It's the same as the monty hall problem, when a goat is revealed (guard gives you a name), the odds do not change.

Imagine 99 goats and 1 car, 98 goats are then revealed, what are your odds of being on a car? 1 in 100.

On a goat? 99 in 100. Your odds haven't changed.

Last edited by p0lka; 21st February 2019 at 12:22 PM.
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 12:28 PM   #61
Thermal
Philosopher
 
Thermal's Avatar
 
Join Date: Aug 2016
Location: Currently Dismembered
Posts: 7,622
Originally Posted by p0lka View Post
Yes.
The decision on who is going free was made at the beginning, you knew 99 people were going to die at the start, now there is just you and another prisoner.
The odds that you are going free is still 1 in 100, assuming the decision was made at the beginning. (You are on the car)

It's the same as the monty hall problem, when a goat is revealed (guard gives you a name), the odds do not change.

Imagine 99 goats and 1 car, 98 goats are then revealed, what are your odds of being on a car? 1 in 100.
On a goat? 99 in 100.
That is the odds of any one specific goat in the begginning.

Take it to the extreme: you know the fates of 99 prisoners. Your odds of freedom are not one in one hundred, but they were at the beginning.

Actually, if the prisoner was chosen for freedom in advance, the odds were zero or one when the choice was made, no 2 in 3 or whatever in play.

Put another way, if a fair coin is flipped, the odds are .5 of it coming heads. Doesn't matter how some other coins landed. There is a choice of only two outcomes with two prisoners: one or the other goes free. Whether there were initially three or a hundred or ten thousand no longer matters. It only mattered at the beginning. Its down to a coin flip
__________________
"Half of what he said meant something else, and the other half didn't mean anything at all" -Rosencrantz, on Hamlet
Thermal is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 12:33 PM   #62
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Thermal View Post
That is the odds of any one specific goat in the begginning.

Take it to the extreme: you know the fates of 99 prisoners. Your odds of freedom are not one in one hundred, but they were at the beginning.

Actually, if the prisoner was chosen for freedom in advance, the odds were zero or one when the choice was made, no 2 in 3 or whatever in play.

Put another way, if a fair coin is flipped, the odds are .5 of it coming heads. Doesn't matter how some other coins landed. There is a choice of only two outcomes with two prisoners: one or the other goes free. Whether there were initially three or a hundred or ten thousand no longer matters. It only mattered at the beginning. Its down to a coin flip
Once the prisoner goes free, and there are 99 prisoners dead, what were the odds of the prisoner being the one to go free?

Last edited by p0lka; 21st February 2019 at 12:35 PM.
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 12:35 PM   #63
Hellbound
Merchant of Doom
 
Hellbound's Avatar
 
Join Date: Sep 2002
Location: Not in Hell, but I can see it from here on a clear day...
Posts: 13,630
Originally Posted by p0lka View Post
Once the prisoner goes free, and there are 99 people dead, what were the odds of the prisoner being the one to go free?
1. As is true for any event in the past.

ETA: Just to go back to an earlier post:

There are originally three possibilities with the three prisoners, who I'll call Al, Bob, and Chuck (just cause).

1. Al and Bob are killed.
2. Al and Chuck are killed.
3. Bob and Chuck are killed.

Al is worried about his chances and asks the guard. The guard says "Bob will die". THis means the possible outcomes are now:

1. Al and Bob are dead
2. Bob and Chuck are dead.

One of the outcomes that included Al dying has been eliminated by the new info.

Last edited by Hellbound; 21st February 2019 at 12:39 PM.
Hellbound is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 12:38 PM   #64
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Hellbound View Post
1. As is true for any event in the past.
Oops.
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 01:20 PM   #65
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Hellbound View Post
1. As is true for any event in the past.

ETA: Just to go back to an earlier post:

There are originally three possibilities with the three prisoners, who I'll call Al, Bob, and Chuck (just cause).

1. Al and Bob are killed.
2. Al and Chuck are killed.
3. Bob and Chuck are killed.

Al is worried about his chances and asks the guard. The guard says "Bob will die". THis means the possible outcomes are now:

1. Al and Bob are dead
2. Bob and Chuck are dead.

One of the outcomes that included Al dying has been eliminated by the new info.
Referring back to the monty hall problem again,
doors A,B and C.

1. A and B are goats.
2. A and C are goats.
3. B and C are goats.

A is shown that B is a goat. This means that the possible outcomes are now:

1. A and B are goats.
2. B and C are goats.

One of the outcomes that included A being a goat, A and C, has now been eliminated by the new info.
Does it change the odds in the monty hall case?
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 01:21 PM   #66
Hellbound
Merchant of Doom
 
Hellbound's Avatar
 
Join Date: Sep 2002
Location: Not in Hell, but I can see it from here on a clear day...
Posts: 13,630
Originally Posted by p0lka View Post
Referring back to the monty hall problem again,
doors A,B and C.

1. A and B are goats.
2. A and C are goats.
3. B and C are goats.

A is shown that B is a goat. This means that the possible outcomes are now:

1. A and B are goats.
2. B and C are goats.

One of the outcomes that included A being a goat, A and C, has now been eliminated by the new info.
Does it change the odds in the monty hall case?
I think it does in both cases. Changes the odds from 1 in 3 to 1 in 2.
Hellbound is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 01:36 PM   #67
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Hellbound View Post
I think it does in both cases. Changes the odds from 1 in 3 to 1 in 2.
Oh,
well if you disagree with the conclusions of the monty hall problem, then you can literally just grab a pencil and some paper and write down the permutations then count it up for yourself.
You don't have to take anyone's word for it.
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 01:41 PM   #68
Hellbound
Merchant of Doom
 
Hellbound's Avatar
 
Join Date: Sep 2002
Location: Not in Hell, but I can see it from here on a clear day...
Posts: 13,630
Originally Posted by p0lka View Post
Oh,
well if you disagree with the conclusions of the monty hall problem, then you can literally just grab a pencil and some paper and write down the permutations then count it up for yourself.
You don't have to take anyone's word for it.
Eh, I'd have to go back over the monty hall thing. Been too long. But what you gave didn't specify any door was already chosen, so yeah, the odds go from choosing 1 out of 3 doors to 1 out of 2. But assuming a door is already chosen, then a goat is shown from the two unchosen doors (and assuming an honest monty, always shows a door, etc, etc) then your odds are 2 in 3 for switching.

I'm not seeing a similar thing for the prisoner problem, but maybe I'm looking at it wrong.
Hellbound is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 02:20 PM   #69
p0lka
Graduate Poster
 
Join Date: Sep 2012
Posts: 1,531
Originally Posted by Hellbound View Post
Eh, I'd have to go back over the monty hall thing. Been too long. But what you gave didn't specify any door was already chosen, so yeah, the odds go from choosing 1 out of 3 doors to 1 out of 2. But assuming a door is already chosen, then a goat is shown from the two unchosen doors (and assuming an honest monty, always shows a door, etc, etc) then your odds are 2 in 3 for switching.

I'm not seeing a similar thing for the prisoner problem, but maybe I'm looking at it wrong.
I think in both cases, referring back to what you said about an outcome being removed,
the outcome that is removed by revealing the goat, or the name of the prisoner to be executed, is the outcome of both 'you and the other prisoner who is going free' being executed, or in the monty hall case both 'your door and the person with the car door', being goats.

There's nothing that changes by that outcome being removed.
p0lka is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 02:28 PM   #70
Hellbound
Merchant of Doom
 
Hellbound's Avatar
 
Join Date: Sep 2002
Location: Not in Hell, but I can see it from here on a clear day...
Posts: 13,630
Originally Posted by p0lka View Post
I think in both cases, referring back to what you said about an outcome being removed,
the outcome that is removed by revealing the goat, or the name of the prisoner to be executed, is the outcome of both 'you and the other prisoner who is going free' being executed, or in the monty hall case both 'your door and the person with the car door', being goats.

There's nothing that changes by that outcome being removed.
Except trials on the monty hall problem show that it does. Assuming:
1. Monty always shows a goat, no matter what
The switching leads to a higher percentage of wins; 2/3 over 1/3 of staying. Tested by Mythbusters (among others):
https://mythresults.com/wheel-of-mythfortune

So, staying with a door means a 1/3 chance of winning, switching is 2/3. Nothing fundamental is changed, but your knowledge of the situation changes.

Likewise in the prisoner situation. Probability of this type is based on incomplete knowledge. If your knowledge improves, you can better calculate the odds. IF the guard can tell you definitely that (for example) Bob will be executed, then that means one of two things:
1. It was never a 1 in 3 chance, as Bob was always going to be executed and they're just going to randomly choose one other (a 1 in 2 chance)
or
2. the two executees were pre-chosen, and by knowing that one of the possible outcomes is eliminated you've increased your knowledge of the probabilities, leaving a one in two chance that you'r the second one chosen to die (rather than the 2 in 3 chance of being one of two chosen to die that existed before you gained knowledge).


ETA: or to try and explain it differently, if 2 of the 3 prisoners has to die, there has to be some selection process. There is a 1 in 3 chance of being the first chosen (assuming it's random). Once that first is chosen, there is a one in 2 chance of being the second one chosen to die. The guard's information is the same as knowing the first prisoner chosen.

ETA2: And the problems are different because in the prisoner scenario, there's no option to switch. It's the same as the monty hall problem if you're shown a goat door before you choose; there are two options left, your odds change to 1 in 2 to pick the prize.

Last edited by Hellbound; 21st February 2019 at 02:33 PM.
Hellbound is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 02:32 PM   #71
JoeMorgue
Self Employed
Remittance Man
 
JoeMorgue's Avatar
 
Join Date: Nov 2009
Location: Florida
Posts: 18,774
Hey here's a good one along these lines, to step away from the Monty Hall archetype for a moment.

You are standing before the King, facing the possibility of execution. (All great philosophical math problems have to include threat of death.)

Two jars sit in front of you. Each jar contains 50 white marbles and 50 black marbles. You have to reach, sight unseen, into one of the two jars and pick out a marble. White marble, you go free. Black marble, you meet the ax man.

You ask the king if you can rearrange the marbles to increase your odds. The King, on the advice of his scheming Vicar, agrees. The number of jars and marbles is fixed, you cannot increase your odds.

You dump all the marbles into one jar so it has 100 black and 100 white. You then take one white marble and put it in the empty jar.

Boom. You have now increased your odds from 50% to 75% (well 74.99% or whatever if you want to get technical.)

You now have a 50/50 chance of picking a jar with only white marbles, and if you by chance pick the jar with both marbles you have a (nearly) 50/50 chance of picking a white marble.
__________________
- "Ernest Hemingway once wrote that the world is a fine place and worth fighting for. I agree with the second part." - Detective Sommerset
- "Stupidity does not cancel out stupidity to yield genius. It breeds like a bucket-full of coked out hamsters." - The Oatmeal
- "To the best of my knowledge the only thing philosophy has ever proven is that Descartes could think." - SMBC
JoeMorgue is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 02:39 PM   #72
casebro
Penultimate Amazing
 
casebro's Avatar
 
Join Date: Jun 2005
Posts: 16,627
But isn't the question in Monte Hall "after Monte shows one goat, does the contestant changing his choice change his odds? "
__________________
Great minds discuss ideas.
Medium minds discuss events.
Small minds spend all their time on U-Tube and Facebook.
casebro is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 02:40 PM   #73
Hellbound
Merchant of Doom
 
Hellbound's Avatar
 
Join Date: Sep 2002
Location: Not in Hell, but I can see it from here on a clear day...
Posts: 13,630
Originally Posted by casebro View Post
But isn't the question in Monte Hall "after Monte shows one goat, does the contestant changing his choice change his odds? "
Yes, it is, and yes, it does.

ETA3: (referring back to previous): Or, to see how just information changes your odds, if the guard in the prisoner problem tells Al that Bob and Chuck are both going to be executed, then your odds (from Al's point of view) are obviously NOT still 2 in 3 chance to die. It's a 0 chance for death and a 1 for living. The added information changes the odds.

Last edited by Hellbound; 21st February 2019 at 02:48 PM.
Hellbound is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:03 PM   #74
Myriad
Hyperthetical
 
Myriad's Avatar
 
Join Date: Nov 2006
Location: A pocket paradise between the sewage treatment plant and the railroad
Posts: 15,043
Originally Posted by Hellbound View Post
Yes, it is, and yes, it does.

In the Monty Hall problem, switching changes your odds exactly because NOT switching does NOT change your odds. That is, the revelation of one goat doesn't alter your odds of winning with the door you initially chose, which remain at 1/3. Which means switching improves your odds to 2/3.

In the Prisoner problem, you have no option to switch, so your odds of survival don't change. They remain at 1/3. (The odds of survival of the third person, who is not you and not the doomed person the guard names, improves to 2/3.)
__________________
A zÝmbie once bit my sister...
Myriad is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:10 PM   #75
Myriad
Hyperthetical
 
Myriad's Avatar
 
Join Date: Nov 2006
Location: A pocket paradise between the sewage treatment plant and the railroad
Posts: 15,043
Originally Posted by JoeMorgue View Post
Ted, Bill, and Harry are arrested. They are told that tomorrow morning, two of them are going to be executed and one is going to be set free, but none of them are told who it is going to be. They are then put in separate cells to await the morning.

As the night goes on Ted becomes more and more nervous. He asks, begs, pleads with the guard to please, please, tell him if he's one of the ones who is going to die.

The guard thinks for a moment and says, and we will assume a truthful guard for our purposes, that he can't tell Ted if he is going to die or not. But he will tell him this. Bill is definitely one of the two who is going to be executed.

Ted is pleased. From his POV his chances of survival have raised from 1 in 3, to 1 in 2.

But that's not correct...

Here's a variation where Ted's chances of survival do improve: after pleading with the guard to tell him whether he's going to be executed, without getting any answers, he finally says "Okay, then, just tell me whether that double-crossing bastard Bill is going to be the one released." The guard says, "No, Bill is going to be executed." (As usual, assume he's telling the truth.)

In that case, Ted's chances of survival improve to 1 in 2.
__________________
A zÝmbie once bit my sister...
Myriad is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:13 PM   #76
RecoveringYuppy
Philosopher
 
Join Date: Nov 2006
Posts: 8,858
Originally Posted by Myriad View Post
Here's a variation where Ted's chances of survival do improve: after pleading with the guard to tell him whether he's going to be executed, without getting any answers, he finally says "Okay, then, just tell me whether that double-crossing bastard Bill is going to be the one released." The guard says, "No, Bill is going to be executed." (As usual, assume he's telling the truth.)

In that case, Ted's chances of survival improve to 1 in 2.
I'm not seeing the variation here. What's different from what you quoted?
__________________
REJ (Robert E Jones) posting anonymously under my real name for 30 years.

Make a fire for a man and you keep him warm for a day. Set him on fire and you keep him warm for the rest of his life.
RecoveringYuppy is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:20 PM   #77
Myriad
Hyperthetical
 
Myriad's Avatar
 
Join Date: Nov 2006
Location: A pocket paradise between the sewage treatment plant and the railroad
Posts: 15,043
Originally Posted by RecoveringYuppy View Post
I'm not seeing the variation here. What's different from what you quoted?

The difference is that Ted asks about a specific one of the other prisoners, Bill, and the guard answers that question.

Since the guard is honest, if Bill were the one to be freed, the guard would have said so and Ted would know he was doomed. (Or perhaps the guard would have refused to answer.) Since that's not what happened, Ted's chances of survival have improved.
__________________
A zÝmbie once bit my sister...
Myriad is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:25 PM   #78
RecoveringYuppy
Philosopher
 
Join Date: Nov 2006
Posts: 8,858
Ah, got it. And that corresponds to a frequent misunderstanding of the Monty Hall problem where people think Monty doesn't know what's behind the doors and occasionally reveals the grand prize.
__________________
REJ (Robert E Jones) posting anonymously under my real name for 30 years.

Make a fire for a man and you keep him warm for a day. Set him on fire and you keep him warm for the rest of his life.
RecoveringYuppy is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:34 PM   #79
Myriad
Hyperthetical
 
Myriad's Avatar
 
Join Date: Nov 2006
Location: A pocket paradise between the sewage treatment plant and the railroad
Posts: 15,043
Originally Posted by RecoveringYuppy View Post
Ah, got it. And that corresponds to a frequent misunderstanding of the Monty Hall problem where people think Monty doesn't know what's behind the doors and occasionally reveals the grand prize.

Exactly so.
__________________
A zÝmbie once bit my sister...
Myriad is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Old 21st February 2019, 03:36 PM   #80
baron
Guest
 
Join Date: Dec 2006
Posts: 8,627
The easiest way to grasp the prisoner one is to imagine 100 prisoners from which only 1 will be freed; the rest will be executed. A's chance of being freed is therefore 1 in 100.

The guard, being a nice guy, agrees with A that he will shoot 98 of the prisoners who are destined for death right there and then, but he will leave A and one other alive. This he does.

It's much easier to see that A's chances are not suddenly 1 in 2 that he'll be set free, they're still 1 in 100.

The other prisoner is doing cartwheels.
baron is offline   Quote this post in a PM   Nominate this post for this month's language award Copy a direct link to this post Reply With Quote Back to Top
Reply

International Skeptics Forum » General Topics » Science, Mathematics, Medicine, and Technology

Bookmarks

Thread Tools

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump


All times are GMT -7. The time now is 03:16 PM.
Powered by vBulletin. Copyright ©2000 - 2019, Jelsoft Enterprises Ltd.

This forum began as part of the James Randi Education Foundation (JREF). However, the forum now exists as
an independent entity with no affiliation with or endorsement by the JREF, including the section in reference to "JREF" topics.

Disclaimer: Messages posted in the Forum are solely the opinion of their authors.