That's what i was thinking. Since A already knows that one of the other two will be executed, he knows that his chance of survival is 1/2;
Case 1; both of them die
Case 2; one of them lives
But then again, i suspect this might be the same reasoning as this;
Case 1; i win the lottery
Case 2; i don't win the lottery
Therefor my chances of winning the lottery are 1/2.
But who says that this kind of reasoning is wrong? Who can actually prove that?
Originally posted by TheMaster37Lots of INFORMed people. Probability is based upon INFORMation, and you are not using all of the available INFORMation..
Therefor my chances of winning the lottery are 1/2.
But who says that this kind of reasoning is wrong?
Instead, you should be working out the conditional probability -
something like 'the probability that I will win the lottery GIVEN THAT there is a total of 1 zillion tickets, only about 1,000 of which will win anything.'
The bookies rely upon punters' ill-INFORMed choices in order to stay in business.
.
I that case, the prisoners chance of survival is still the same after the guard told him something. In this type of reasoning the chance of something to happen is either 1 or 0. That i don't know that something will happen, doesn't mean it's less likely to happen.
The chances of getting shot by someone who hates me on this forum is les then 1 percent. But if one of you out there is getting on the plane this night, comes here and shoot me, that chance was 1, i simply didn't know enough to conclude that.
Originally posted by TheMaster37Excuse my candour, but the above two statements are incompatible.
As a mathematician....
Originally posted by TheMaster37
That's what i was thinking. Since A already knows that one of the other two will be executed, he knows that his chance of survival is 1/2;
.
*grins*
They are 🙂
From A's point of view there are two cases;
B or C will most certainly die. If it's B then A has chance 1/2 of surviving, if it's C as well. So A's chances of survival are 1/2*1/2=1/4.
all the nonsense above is fun, though it will all be cleared when someone gives a good definition of chance. I have one, i'm just trying to provoke someone else to try a definition (btw, A's chances have indeed increased from 1/3 to 1/2, as is the deeper math in that door-switch-game. Hard to believe that the enitire mathworld was in chaos because of such a simple thing).
Originally posted by TheMaster371) Are you serious?
Before the prisoner asks the guard, there are still three possibilities;
A lives, B lives or C lives.
When the guard says B dies, that eliminates one possibility. That leaves only two;
A lives or C lives.
If yes
2) Do you believe it is better to switch in the Monty Hall puzzle?
If yes
3) Can you see that this puzzle is isomorphic to the Monty Hall puzzle?
If yes
4) Why do you still believe that the answer to this puzzle is 1/2?
.
Yes, No, Yes. It is not perfectly isomorphic to the game. The same type of reasoning applies though. The prisoners have no operation 'switch' available, wich in my opinion, makes it hard to create an isomorphism. If the prisoner could say "Ok, then i want to trade places with C", then it was completely isomorphic.
The Monty Hall game is solved likewise to this problem, say the prize is behind Door 2;
Three possibilities at first:
Door 1, Door 2 or Door 3.
You pick a door.
-If you pick Door 2, the host opens either door 1 or 3, and it's not good to switch.
-If you pick Door 1, the host will open Door 3, and it's good to switch.
-If you pick Door 3, the host will open Door 1, and it's good to switch.
Two of three cases in wich it's good to switch, so the chance of getting the prize when switching is 2/3.
The solution to this problem is either 1/2 or 1/3. The problem statement must be clarified in order to determine which is the correct solution. In particular, the guard's knowledge must be specified more clearly.
Consider this scenario, which is consistent with the problem statement.
There are three pieces of paper, labeled on one side with A, B, or C, and on the
other side, with Live or Die. They are laid on the table before the guard with
the letters facing up. The guard has no knowledge of their undersides. To answer
A's question, he chooses the paper marked either B or C. There are now three
cases to consider:
1. Guard chose paper B and saw Die.
2. Guard chose paper C and saw Live.
3. Guard chose paper C and saw Die, then chose paper B and saw Die.
In case 1, papers A and C remain undetermined, thus A's survival chance is 1/2.
In case 2, papers A and B remain undetermined, thus A's survival chance is 1/2.
In case 3, A's survival chance is is known to be 1
Thus in all cases, A's survival chance is at least 1/2.
In order to transform to the Monty Hall problem, it must be assumed that
the guard already has complete knowledge, just as Monty Hall does. Under
that assumption, the solution is obviously 1/3.
So, either 1/2 or 1/3 is correct depending on the meaning of the problem
statement.
Cribs
But A doesn't know what the guard knows, so for A, the chances are different.
Time for a definition of chance then (not the official one);
For a given observer, the chance on A is the number of events in wich A occurs, divided by the total number of events possible.
Chance depends on the one calculating the chance. If i don't know that i'm going to win the lottery, my chances are tiny. If i know beforehand that i am going to win the lottery, i know my chance to be 1.
BTW, i still don't see how the prisoners problem is the Monty Hall problem; the prisoners can't switch! Since B dies, one of the possibilities is cancelled, so there are still two possibilities left or am i missing one?
EDIT;
1) C lives (guard automatically names B)
2) A lives, guard names B
3) A lives, guard names C
4) B lives (guard automatically names C)
This is known to A before the guard names a prisoner to die (this because the guard will name one of the two that dies, wich is not A).
Two out of Four makes 1/2 chance of survival. The guard names B, wich leaves 1) and 2), still 1/2 chance of survival for A.
Now don't simply say "this is monty hall problem". Indicate where i'm wrong, and how EXACTLY this is the monty hall problem, since the prisoners can't switch as is a possibility in the Monty Hall game.
Originally posted by TheMaster37Consider the first scenario I present, in which I conclude that A's chances of survival are at least 1/2. It suffices for A to know the guard's process, and it is not necessary for him to know which of cases 1, 2, or 3 occurred. He can come to a valid logical conclusion that his chances of survival are no worse than 1/2 immediately upon the guard's answer of "B will die."
But A doesn't know what the guard knows, so for A, the chances are different.
So I disagree that A's knowledge has to be the same as the guard's in order
to reach my conclusion from my first scenario. I do agree that A's knowledge about the process governing the guard's knowledge is key; that in fact was the point of my post.
Cribs
Now don't simply say "this is monty hall problem". Indicate where i'm wrong, and how EXACTLY this is the monty hall problem, since the prisoners can't switch as is a possibility in the Monty Hall game.Substitute "Live" for "Prize", "Die" for "No Prize", and "Would you rather be Prisoner C?" for "Would you like to switch to the remaining closed door?". Further, specify that the guard knows from the outset everybody's fate, just as Monty knows the contents of every door. Prisoner A asking to reveal the death of B or C is analogous to the contestant asking to have a door with no prize revealed.
That's really all there is to this problem and the Monty Hall problem.
Is the transformation clear now?
If not, think about physically playing the prisoner problem out on the Monty Hall stage. One of Doors B or C is revealed to Prisoner A with the grim reaper behind it. 2/3 of the time the other reaper will be behind Door A, and 1/3 of the time it will be behind Door C. Upon seeing this, Prisoner A would wish that he was really prisoner C. (Similarly, if Prisoner C were the one playing the game, he would end up wishing he were Prisoner A.)
Cribs