Originally posted by royalchickenYes, but this is all consistent with our original premise that all potential states are equilikely. For example, let E_x be the event
Even better, since we proved at FW that conditional probability is a probability function, we can conclude that since all probabilities are 1/2, P(E|E) = 1/2, so we have 00 = 1/2 = 2. Thus 1 = 4, so, 1/4 = 4 etc 😛.
E_x: the sum of 1+2 = x
Since all events are equally likely, E_3 is just as likely to occur as E_17.
So, the various equalities you present are acutally consistent rather than paradoxical. Not that we needed extra proof of this, but consistency checks are always good to make sure we're not wandering off into the land of nonsense.
The question is has been changed that is why both answers are correct.
Prisoner has and always will have a 33.3% chance of surviving provided 2 of 3 prisoners die.
However, when isolating that he versus ONE the the other two prisoners will die, he obviously has a 50% chance, because the question has been altered from 1 out of 3, to 1 out of 2.
... i like the concept though. I think if i were one of those prisoners i would argue forever that my chances HAD increased, while i lose my mind waiting!
Originally posted by DoctorScribblesI'm not saying they're paradoxical at all. What I am saying is that:
Yes, but this is all consistent with our original premise that all potential states are equilikely. For example, let E_x be the event
E_x: the sum of 1+2 = x
Since all events are equally likely, E_3 is just as likely to occur as E_17.
So, the various equalities you present are acutally consistent rather than paradoxical. Not that we needed ...[text shortened]... sistency checks are always good to make sure we're not wandering off into the land of nonsense.
4 = 1 = 1/4 = 1/16 = 4^-n for all n.
0 = 2 = 1/2 = 1/4 = 1/8 = 2^-n for all n. This looks worrying, but it is consistent with everything else. However, each real number can be represented as:
ak*2^k + a(k-1)*2^k-1 + ... + a0*1 + a(-1)*(1/2) + ...
Where ai = 0 (since by the above two equations, 0=1) for all i. Thus all real numbers are equal. (Alternatively, let ai = 1 for all i. Then the powers of 2 we need to get the integer part of the real are all 1/0 = 4 = 1, as are the powers of 1/2. Thus the real is k*1 + 1 + 1 + 1 +... = k+2 . This means that there is one real for each k, so the reals are countable!)
Thus either our Riemann sphere friend is a single point or a ball of spacious Swiss cheese.
Chances are said to be the number of "good" outcomes divided by the total number of outcomes.
At first the total numer of outcomes is three (A lives, B lives or C lives)
For Prisoner A there is only one "good" outcome; A lives.
His chances of survival are thus 1/3
Now when the guard tells him that B will die, the total number of outcomes changes from three to two (A lives or C lives).
For Prisoner A there is still only one "good" outcome; A lives.
His chances of survival are thus 1/2.
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Chances are said to be the number of "good" outcomes divided by the total number of outcomes.
At first the total numer of outcomes is three (A and B die, B and C die, A and C die)
For Prisoner A there is only one "good" outcome; B and C die.
His chances of survival are thus 1/3
Now when the guard tells him that B will die, the total number of outcomes changes from three to four (A and B die, B and C die while guard mentions B).
For Prisoner A there is still only one "good" outcome; B and C die.
His chances of survival are thus 1/2.
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Last time:
Say prisoner A gets to choose wich prisoners place he wants to take. At this point, it makes no difference, he chooses right with chance 1/3. He chooses A, himself.
Now the guard will say wich of the others is certain to die, he will mention only one name.
In the case A was going to live (chance 1/3), the guard has a choice wich name he's going to say (B with chance 1/2 and C with chance 1/2).
In the case B was going to live (chance 1/3), the guard will say C.
In the case C was going to live (chance 1/3), the guard will say B.
Now the guard say B is going to die. With chance 1/6 we have the first case and with chance 1/3 we have the last case. The other cases are not possible anymore! B's chances to live are 0, and C's chances to live are twice as much as A's! So the chance for A to live is 1/3 and for C 2/3.
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My question to you; Wich reasoning is the correct one, and give a step by step explenation why the other(s) are wrong.
Originally posted by TheMaster37The first two are identical.
Chances are said to be the number of "good" outcomes divided by the total number of outcomes.
At first the total numer of outcomes is three (A lives, B lives or C lives)
For Prisoner A there is only one "good" outcome; A lives ...[text shortened]... e, and give a step by step explenation why the other(s) are wrong.
The problem with these two seems to be this - correct me if I am wrong, as I find this hard to understand:
From A's perspective, there are three possibilities: A lives, B lives, C lives. Whoever does not live dies.
A knows that at least one of the other two will die. When he asks the guard to point out which one of the other two will die, he recieves no new information; he already knew one of the other two would die. The guard did not randomly choose one of the other two, he intentionally chose the one who would die.
After the guard says B will die, there are two possibilities; A will live, or C will live. These two possibilities are not evenly weighted however. If the guard had randomly chosen B and then said B would die, the situation would be different. The guard may have had a chance to choose the one who wouldn't die. Since he did not, the chance that A was that person would increase a little bit. This chance did not exist in the actual scenario though.
It's like me saying - I am either red haired or blonde. I will roll a six sided die. If the die rolls a 1, I won't tell you what I rolled, but I will tell you my hair color. If the die doesn't roll a 1, then I will say blonde no matter what. In this case there are 12 possibilities; whether I have blonde or red hair times the number of possible numbers I can roll. 11/12 times I will tell you blonde. If I do tell you blonde, you know that I didn't both have red hair and roll a 1. Now it's like the possibility of me being blonde has gone from 1/2 = 6/12 to 6/11. The randomness has to be there though so that one can eliminate a possibility; randomness is what allows one to lessen the total number of possibilities.
I'm still kind of confused, but I think that's the kind of reasoning that is used for this kind of problem.
No matter how you look at the problem, prisoner A's chances of survival are 1/3.
If the guard picks the condemned prisoners one at a time by lottery, the chances that prisoner A will not be named on the first pick is 2/3, and the chance that he will not be named on the second pick is 1/2. Taken together, his chances of survival are (2/3)*(1/2) = 1/3.
If the guard has already picked which pair will die before being asked any questions, prisoner A's chances of survival are 1/3, and therefore his chances of not surviving are 2/3. After the guard says "prisoner B will die", prisoner A knows that it's impossible that both he and prisoner C have been picked, but the probability that he wasn't picked in the first place remains 1/3.
Just to clarify, in the lottery situation the guard's statement that the first pick was prisoner B reduces the number of possibilities for the second pick. In the other situation, both prisoners were picked beforehand and so the knowledge that prisoner B will die doesn't impact on the original selection.
Originally posted by AThousandYoungI understand what you're trying to say 🙂 That's more or less what I hoped what it would be. Bonus points for you for being able to explain!
The first two are identical.
The problem with these two seems to be this - correct me if I am wrong, as I find this hard to understand:
From A's perspective, there are three possibilities: A lives, B lives, C lives. Whoever does not live dies.
A knows that at least one of the other two will die. When he asks the guard to point out which ...[text shortened]... nd of confused, but I think that's the kind of reasoning that is used for this kind of problem.
Originally posted by AThousandYoungI'm still confused ...
A knows that at least one of the other two will die. When he asks the guard to point out which one of the other two will die, he recieves no new information; he already knew one of the other two would die.
.....
Assuming the guard tells the truth, then the statement "B will die" is
equivalent to the following situation:
A doesn't ask at all but A WATCHES that B is killed - and only A and C
are still alive and one of them is going to be free ... what now?
It's different because when A asks the question, he knows he can't be the answer while if he just watched B die then he could have been the chosen one to die first.
If the guards said to him: "If you die, you'll be the last one to go" then when he watched B die it wouldn't improve his chances.
Originally posted by EdSchwartzEd - maybe this will help.
I'm still confused ...
Assuming the guard tells the truth, then the statement "B will die" is
equivalent to the following situation:
A doesn't ask at all but A WATCHES that B is killed - and only A and C
are still alive and one of them is going to be free ... what now?
Initially, the chance of each prisoner surviving is 1/3. Let's look at the following outcomes, before the guard gets a chance to speak:-
(1) B lives (The guard MUST tell A that C dies). P=1/3.
(2) C lives (The guard MUST tell A that B dies). P=1/3.
Looking at the scenario if A lives (P=1/3), there are still 2 possible "sub"-outcomes...
(3) A lives - and the Guard tells A that B dies. P=1/6
(4) A lives - and the Guard tells A that C dies. P=1/6
The prob for each of options (3) and (4) must be 1/6, as each is equally likely, and the total must add up to 1/3.
So we have four possible scenarios, each with different probabilities of occurring. Notice that the total probabilities is 1/3 + 1/3 + 1/6 + 1/6 = 1, and that the prob of each person living is 1/3.
Now, our big lumbering guard has opened his mouth and tells A that B is for the chop. So...
Option (1) above is immediately redundant (B can't live, the guard told us so)....
and Option (4) is also redundant - the Guard didn't tell A that C dies, he only told him B dies.
Which leaves us with... options (2) and (3) are the only possible outcomes.
Now, we all know that probabilities of all possible outcomes must add up to 1, so the probabilities aren't 1/3 and 1/6 anymore. The ratios of the probs is still the same, so now:-
(2) C lives - The guard MUST tell A that B dies. P=2/3.
(3) A lives - and the Guard tells A that B dies. P=1/3.
Hence, the chance of A living even though the guard told him B dies, is still 1/3.
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FOOTNOTE, in lay-persons language
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(If you're unsure of why the probs of 1/3 and 1/6 double up to 2/3 and 1/3, think of the inital problem this way:
There are 2 chances that option (1) might happen.
There are 2 chances that option (2) might happen.
There is 1 chance that option (3) might happen.
There is 1 chance that option (4) might happen.
6 "chances" altogether, so 2 chances is the same as P=1/3, 1 chance P=1/6
As above, Options 1 and 4 are ruled out when the guard spills the beans on B.
So, there are 2 chances that option (2) might happen (C lives)
And, there is 1 chance that option (3) might happen (A lives).
So, A has 1 chance in 3 that he gets off scot-free.
QED?
Well, folks, thanks a lot for dealing with my twisted brain.
In fact, I DO know, how to calculate probabilities, and I DO know,
that you are correct (e.g. PawnCurry). What confuses me is the fact,
that each practical scenario that I may think of seems to contradict
my (and your)calculations.
Here's another one:
A, B and C are standing in front of exactly three guns: Ga, Gb and
Gc. Again, A knows that B will be shot, thus there must be a bullit in
gun Gb. There must be another bullit EITHER in Ga OR in Gc (not in
both of them). Now, hasn't A a chance of .5 that the bullit is in gun Gc
?????????
No! There isn't a 50% chance that the bullet is in Gc. Just because there are two remaining options doesn't make the chance 50%.
Extreme scenario: if I jump out of a plane without a parachute, there are 2 outcomes - I live, or I die. Don't want to test this one, but I'd be willing to wager that most of the time I would die!!! Hence, not a 50% chance.
So in the prisoner/gun example, why isn't it 50%? Seems logical without analysing it that it should. We need to ditch our assumptions, and go through all the possible outcomes, looking at what we know (as above).
The reason it's not 50/50, is that the guard's choice is LIMITED if Ga is true. He can ONLY respond Gb. If Gc is true, well he could have said either Gb or Gc. Hence we have the various "options" mentioned in my previous post.
Good problem though!
Unlike the prisioners problem, in the problem EdSchwartz describes there is a 50% chance because it is never said that two of three of the guns have bullets, only that b has a bullet and a and c another.
By his words, you cannot conclude that the bullet in Gb has conditioned choice between either Ga or Gc having the bullet.
It's like the monty hall problem except that one door has been opened before the first choice.
Originally posted by Palynka
.... because it is never said that two of three of the guns have bullets,
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 ....
Originally posted by PawnCurry
The reason it's not 50/50, is that the guard's choice is LIMITED if Ga is true. He can ONLY respond Gb. If Gc is true, well he could have said either Gb or Gc. Hence we have the various "options" mentioned in my previous post.
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 ....