Originally posted by EdSchwartzIf A just finds the information that B dies then A has a 50% chance of survival.
Assume, there's no guard at all; "somehow" A gets the information
that B will be shot ... and then the same scenario that I described ....
BUT if he already knew he wouldn't find information about himself, then he would still have 1/3 chance of survival.
That's the key in the prisioners problem. A is excluded from the first answer and that's why his probabilities remain the same.
Originally posted by EdSchwartzNot exactly.
Of course it's assumed that there's exactly one gun aiming at one
prisoner - and every gun is only fired once. Since two prisoners are
going to die, two of the three guns must have one bullet each ....
If I said to you:
There are 3 inverted cups on the table, the first one has a ball under it and one of the other two also has a ball under it. Which one of those two is it?
Here you have a 50/50 chance of guessing but there were 3 cups with 2 balls on the table.
In the bullets' problem you said B has a bullet and either A or C has a bullet too. Said like that it's 50/50 chance.
Subtly different problem.
Your problem's starting conditions:-
Cup "1" has a ball, and one of the other 2 cups has a ball. What are the chances of it being in a particular cup. (As you've stated, chances are 50😵.
In the original problem, the initial starting conditions were:
3 prisoners, 2 will die, and the guard cannot tell a prisoner if he dies or not.
If you changed the wording of your problem to...
I have three cups, 2 of them contain a ball. (Your cup is "cup 1" for arguments sake). I cannot tell you if your has a ball in it. I CAN tell you that "cup b" contains a ball.
...then this problem is essentially the same as 3 Prisoners.
And... chance of your cup containg a ball? 2 in 3.
........ Chance of cup3 containg a ball? 1 in 3.
For an explanation, see one of my earlier posts.
:-)
Originally posted by PawnCurryPawnCurry,
Subtly different problem.
Your problem's starting conditions:-
Cup "1" has a ball, and one of the other 2 cups has a ball. What are the chances of it being in a particular cup. (As you've stated, chances are 50😵.
In the original problem, the initial starting conditions were:
3 prisoners, 2 will die, and the guard cannot tell a prisoner if ...[text shortened]... ce of cup3 containg a ball? 1 in 3.
For an explanation, see one of my earlier posts.
:-)
That's what I've said. His problem of the guns is NOT like the prisioners but like the one with the cups.
Originally posted by Palynka
If A just finds the information that B dies then A has a 50% chance of survival.
BUT if he already knew he wouldn't find information about himself, then he would still have 1/3 chance of survival.
That's the key in the prisioners problem. A is excluded from the first answer and that's why his probabilities remain the same.
Well, let's see:
While A speaks with the guard, the old lurker C has his ear at the
wall and he just hears "B will die". This would give him a .5 chance,
while A's chance ist still 1/3. Possible?
Or:
We said that after A gets his answer A's chance is still 1/3 while
C's chance has increased to 2/3. What, if C (without knowing anything
about A's conversation with the guard) has the same idea 5 minutes
later and C asks the guard and gets the answer "B will die". What do
we have now?
Since I found this problem several weeks ago in Gigerenzer's book
"Calcualted Risks" it really drives me crazy
Continued:
Chances (=probabilities) are the number of "positive" outcomes divided by the number of "possible" outcomes.
Could it be, that all of us are mixing up two different entities of
"possibles outcomes"? The number of possible combinations who will
die is one entity and the number of allowed answers is the other.
Not sure about that. But:
(I)
There are three possible outcomes of who will die:
AB
AC
BC
After A knows that B will die there are only two possible outcomes left,
giving him a chance of .5
(II)
When A asks the guard, there are four possible(=allowed) answers:
(1) if AB will die, the guard MUST say B
(2) if AC will die, the guard MUST say C
(3) if BC will die, the guard MAY say B
(4) if BC will die, the guard MAY say C
Two out of these four are positive for A; chance: .5
Originally posted by EdSchwartzBefore asking any questions, every prisioners believes his probabilities of living are 1/3. However, the guard, knows which two prisioners have 0% probability of living and which one has a 100% probability of living.
Originally posted by Palynka
If A just finds the information that B dies then A has a 50% chance of survival.
BUT if he already knew he wouldn't find information about himself, then he would still have 1/3 chance of survival.
...[text shortened]... Gigerenzer's book
"Calcualted Risks" it really drives me crazy
This is not incompatible because we are dealing with different information.
If C had heard only "B will die", it is possible that he ASSUMED that the question had no restrictions. Under those assumptions C could calculate his chances to be 50%, however he had made that calculation under the wrong assumptions. In his calculations, A would also have 50% chances of survival. If he knew what A had asked he would know his probabilities were 2/3 now.
The important is that probabilities are consistent under the same information, not with different informations.
-----
Example:
Imagine A asked the question he did, with the guard saying "B will die" and C had simply asked the question, "will B survive?" to which the guard answered "no". B asked no question and none of them listened to other prisioners' conversations with the guard.
According to A:
P(A living)=1/3
P(B living)=0/3
P(C living)=2/3
According to B:
P(A living)=1/3
P(B living)=0/3
P(C living)=2/3
According to C:
P(A living)=1/2
P(B living)=0/3
P(C living)=1/2
According to the guard (2 possibilities):
P(A living)=1
P(B living)=0
P(C living)=0
or
P(A living)=0
P(B living)=0
P(C living)=1
Edits:spelling and formatting
Originally posted by EdSchwartzIn this problem:
Originally posted by Palynka
Or:
We said that after A gets his answer A's chance is still 1/3 while
C's chance has increased to 2/3. What, if C (without knowing anything
about A's conversation with the guard) has the same idea 5 minutes
later and C asks the guard and gets the answer "B will die". What do
we have now?
Since I found th ...[text shortened]... problem several weeks ago in Gigerenzer's book
"Calcualted Risks" it really drives me crazy
For A:
P(A living)=1/3
P(B living)=0
P(C living)=2/3
For C:
P(A living)=2/3
P(B living)=0
P(C living)=1/3
If they discuss the results among themselves, then they could conclude:
P(A living)=1/2
P(B living)=0
P(C living)=1/2
(Note: we are assuming that when the prisioner that will survive asks the question to the guard, the guard has a 50% chance of saying B or the other prisioner who will die)
Originally posted by EdSchwartzFor the answer to this post, please look at iamatiger's post in the first page of this thread.
Continued:
Chances (=probabilities) are the number of "positive" outcomes divided by the number of "possible" outcomes.
Could it be, that all of us are mixing up two different entities of
"possibles outcomes"? The number of possible combinations who will
die is one entity and the number of allowed answers is the other.
Not sure about that. ...[text shortened]... if BC will die, the guard MAY say C
Two out of these four are positive for A; chance: .5
Originally posted by THUDandBLUNDERok, are you going to make some moves now? 😕
There are 3 prisoners in a prison. Let’s call them A, B, and C. Tomorrow, 2 of them will be executed, but the prisoners don’t know which of them have been chosen. Prisoner A reasons that his chances of survival are 1/3 (a third). He ...[text shortened]... nces of survival have increased from 1/3 to 1/2. Is he right?
.
Originally posted by BLReidduh, you said the answer! 50/50
You have 3 ice cream cones.
You have 2 kids.
You give 1 cone to 1 child.
You let the other child pick between the 2 remaining, say chocoloate and vanilla.
Waht are your odds of being left with chocolate? (assuming the child has no pre-defined preference)
50/50
BLR