Originally posted by nemesioI did post this, probably while you were writing this.
I'm sorry to waste your time. Could you please, as simply as you did with the Monty Hall problem, explain what you mean by zero and complete information; that is, why one analysis applies to one situation and a different one to another? I think this is where the (substantial) gap in my understanding/knowledge must lie.
I thank you all again for all the help.
Nemesio
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That all depends. You have a 1/3 chance of getting the red marble with your first pick. That means there is a 2/3 chance that it is still in the bag.
Now, if your friend randomly reaches in the bag and pulls out a marble there is a:
1/3 chance that he will pick a red marble and the game ends
2/3 chance that he will pick a yellow or blue marble and the game continues
If you know he didn't get the red marble, then this fact raises the likelihood that you have the red one yourself. In fact your chances increase from 1/3 to 1/2
However, what your friend might do is look in the bag and carefully pull out a yellow or blue marble. In this case you know he's not going to pull out the red one whether its in there or not, and so the fact that as you predict, he does pull out a yellow or blue marble doesn't alter your chances of having the red marble and they stay at 1/3. He's just made sure that if the red one was left in the bag then its the only one left. The red marble had a 2/3 chance of being in the bag before he pulled one out and it still does.
Originally posted by nemesio
Let's say I have a bag with three marbles in it, colored red, yellow and blue, and the goal is to get the red one. I reach in and clasp a marble and pull it out without looking. Now, someone reaches in the bag and pulls out a marble that isn't red from the bag. Given the choice, how is it better for me to switch. Either the one I have or the one I don't have is red. Seems like 1/2 chance.
ow ow ow ow
Originally posted by iamatigerI was about to write something similar myself. The point is that the host doesn't just pick a door at random, he deliberately avoids the winning door (if you didn't pick it initially), and this condition further increases the probability that the door he doesn't open is a winner. Many people, on hearing the problem, imagine that the host randomly opens another door and happens to reveal a goat, which isn't the case at all.
That all depends. You have a 1/3 chance of getting the red marble with your first pick. That means there is a 2/3 chance that it is still in the bag.
Now, if your friend randomly reaches in the bag and pulls out a marble there is a:
1 ...[text shortened]... I don't have is red. Seems like 1/2 chance.
ow ow ow ow[/b]
Another common (and more fundamental) fallacy is that two superficially similar events which are both possible are equally probable. Believers in this might wrongly assume that a pair of dice is as likely to come up '2' as '7', for example, even though the latter is six times more likely.
Originally posted by CribsThis is a controversial matter. There are two schools of thought on this - the frequentists and the Bayesians.
probability is a measure
of information; it is not a description of reality
Frequentists define the probability of an event as the proportion of times it would happen if you carried out the random process repeatedly. Probability is thus an objective property of the random process.
Bayesians define the probability of an event as the odds you'd need to be offered to want to bet on it (expressed in suitable terms, eg '6 to 1' corresponds to 1/7). Probability is thus a property of how much you know and what you believe, so its connection to the random process (which may in fact not be random at all) is subjective.
IMO, frequentism is fine in the abstract, but in real life you never know enough to draw objective conclusion; so I find the Bayesian approach more honest by embedding the subjectivity in the system, rather than having to make the somewhat arbitrary choices you see in frequentist statistics. For example, why choose a significance level of 5% in a (frequentist) hypothesis test? The disadvantage of Bayesian statistics is that you need a prior distribution that reflects a lack of information, and while in the Monty Hall case there's an obvious theoretical starting point, for other things it's not so simple. Suppose the mean length of a piece of string is x. If you have no idea about lengths of pieces of string, how are you supposed to construct a prior for x? You can't just make it uniform over the positive reals!
Originally posted by AcolyteThis is my favorite example to illustrate what I mean when I say that
This is a controversial matter. There are two schools of thought on this - the frequentists and the Bayesians.
Frequentists define the probability of an event as the proportion of times it would happen if you carried out the random proc ...[text shortened]... r for x? You can't just make it uniform over the positive reals!
probability is a measure of information rather than a description of reality.
Consider that you are holding a deck of cards, and you don't know
where any particular cards are. What is probability that the top card
is red? 1/2, because you have zero information. But in reality, that physical
top card in that physical deck is not a random process (it was before the shuffle; now its value has been realized, since someone could peek at it). Now it
really is either red or black, and that won't change until the next shuffle.
Now suppose that holding that same deck, you peel off the bottom 13 cards and they happen to all be red. Again, what is the probability that that same top
card is red? This time, 1/3.
The physical card on top has not changed, so probability can't describe
reality, since the probability changed from 1/2 to 1/3. What did change?
Your information. So that is what probability measures.
Cribs
Originally posted by CribsExactly; this is the Bayesian position. (The frequentist position is that the top card isn't random at all, once the shuffling is complete.) The subjectivity comes into play, though, when you assume as a prior that the cards have been permuted uniformly at random. You can't actually prove that your shuffling method really does randomise the cards uniformly - the best you can do is to list all the ways in which your shuffle is 'likely' to be biased and test for them in the hope that the posterior probability of each bias occurring is low enough to be negligible.
This is my favorite example to illustrate what I mean when I say that
probability is a measure of information rather than a description of reality.
Consider that you are holding a deck of cards, and you don't know
where any particular cards are. What is probability that the top card
is red? 1/2, because you have zero information. But in reality, that ...[text shortened]... 2 to 1/3. What did change?
Your information. So that is what probability measures.
Cribs
Originally posted by CribsThe card being red or black, even before you look at is is only when you don't give any value to quantumtheory.
This is my favorite example to illustrate what I mean when I say that
probability is a measure of information rather than a description of reality.
Consider that you are holding a deck of cards, and you don't know
where any particular cards are. What is probability that the top card
is red? 1/2, because you have zero information. But in reality, that ...[text shortened]... 2 to 1/3. What did change?
Your information. So that is what probability measures.
Cribs
Your situation; deck of cards, you don't know the distribution of the cards, The top one is either red of black, because you don't have information suggesting otherwise. Now in the quantumtheory the card is in a state of flux, between black and red. One fraction of a second it is red, another black. Only when you flip the card does it's colour shift to a set state, either black or red...
Oh, Cribbs, iamatiger and Acolyte, i do get the Monty Hall problem perfectly now. Especially due to Cribbs explenation of the chance increasing from 1/2. Thank you very much, i was wrecking my brain over it :p
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Originally posted by CribsJust read this point and I beg to differ. In the prisoner problem the execution is due to happen in the future and the guard is telling the prisoner sometime before. This means that the prisoner's chances don't increase at all.
Finally, it is because of this subtle difference that the prisoner problem
must be carefully specified with respect to the guard's information. In that
thread people presented valid arguments for a solution of both 1/3 and 1/2,
but they failed to recognize that both solutions are correct, depending on
what the guard knows. If the guard has zero knowledg ...[text shortened]... 50/50 after hearing B's fate; if the guard has perfect
knowldege, they remain at 1/3.
Cribs
Let me illustrate:
The guard says to A, "B is going to die" but then admits "That's just a guess, I have zero knowledge". A thinks thinks "Hooray! - my chances of survival have increased to 1/2"
The day of execution dawns. C is selected for execution. A thinks "Excellent! I am going to survive". Then A is selected for execution. The guard says "Sorry, looks like I was wrong about B after all!"
In fact, the guards randomly selected information will only help A if he knows whether it is wrong or right. In the case of the guard/prisoner problem A has no idea whether the guard's information is wrong or right so it won't help him.
And we have already shown that if the guard knows who is to be executed his information doesn't increase A's chances. So there is no need to state in the problem whether the guards information is correct or is a guess, it won't help A either way!
Originally posted by iamatigerThis is incorrect. If you read my full discussion, you will see why.
And we have already shown that if the guard knows who is to be executed his information doesn't increase A's chances. So there is no need to state in the problem whether the guards information is correct or is a guess, it won't help A either way!
It is essential to specify the guard's knowledge. Previously, I
said you have to distinguish between the guard knowing everything
a priori and the guard choosing a pair to die at random once the
question is asked. Now you introduce a third case, in which the
guard just guesses or lies.
Your third case in inconsistent with the statement and the spirit of
the problem. Why would the problem include the guard at all
if he was going to lie? You must assume that the guard is telling
the truth: in other words, everything in the problem statement
is axiomatic. If the guard says B will die, B will die.
Perhaps we are not seeing eye to eye on the timeline. I assert
that there are two cases. In the first, the guard knows eveything
before the question is asked. In the second, the guard knows
nothing before the question, and at least knows that B will
die after the question. Both of these cases are completely
consistent with the problem statement. Your third case is that
the guard still knows nothing after the question, and I maintain
that this is not consistent with the problem statement, since
the guard answers that B will die. Since this case is not
consistent with the problem, it need not be considered.
So that addresses your third case. Do you really not believe that
the problem must distinguish between the first two cases? In other
words, do you really believe that the solution to the Monty Hall
problem holds even if Monty does not have complete knowledge?
Surely you don't. The very same analysis applies to the prisoner
problem.
Cribs
Regarding the Monty Hall problem: Thanks Cribbs. I've finallyt wrapped my mind around it, with three doors or more.
I guess the problem I had was pounding in Monty's absolute knowledge. Although contrary to the way that the puzzle is stated, I had in my mind that the removal of the door was somehow totally random -- that the prize could be eliminated or that the originally chosen door could be eliminated -- in which case the person holding onto the door (unless it were removed) would not be given partial information at all.
I am a bit lost, though, with the prisoner problem still. I think I share Iamatiger's confusion. It seems that the guard imparts no useful information by telling Prisoner A that either B or C is going to be executed, irrespective of the guard's accuracy:
If the guard says B, and is correct:
B %100 executed.
C %50 chance;
A %50 chance;
and if he's incorrect:
B %0 executed;
C %50 chance;
A %50 chance.
And similarly if the guard guesses C correctly or incorrectly.
I know the reasoning is flawed, because it suggests that A always has a 50% chance, not a 33% chance.
On a related note, assuming the guard tells the truth, does it make a difference whether the guard has complete knowledge or partial knowledge?
That is, if does it make a difference if the King uses three slips of paper (AB, BC, AC) and shows the guard or if he has three slips (A, B, C) and only shows the guard "B?"
Still swimming a bit here with this problem.
Nemesio
Originally posted by nemesioFirst, let's agree that the guard always tells the truth; the problem statement
Regarding the Monty Hall problem: Thanks Cribbs. I've finallyt wrapped my mind around it, with three doors or more.
I guess the problem I had was pounding in Monty's absolute knowledge. Although contrary to the way that the puzzle i ...[text shortened]... "B?"
Still swimming a bit here with this problem.
Nemesio
doesn't really allow for him to guess or to lie.
I also think we all agree that if the guard has complete, a priori knowledge
about everybody's fate, then the prisoner's survival chances reamin at 1/3.
(This is identical to the Monty Hall problem, in which the contestant's
chances remain at 1/3 by not switching doors.)
So the remaining case is when the guard doesn't have complete knowledge,
but only sufficient knowledge to answer "B will die". The question you seem
to ask is: are there different processes that the guard can use to get his
information, such that he imparts different information to the prisoner?
The answer is no. Whatever method is used will result in the prisoner having
a 1/2 survival chance.
Sure, there are differnt processes that the guard could use to get his
information. But in the end, he is only passing that same one fact along
to the prisoner: B will die. So if we can prove for one such process that
the prisoner's chances improve to 1/2, we have proved it for all such
processes. I gave a proof of one scenario in the prisoner thread, and a
similar analysis could be used for whatever other process you could
construct. The idea here is that the guard and the prisoner are essentially
the same party (they start with the same knowledge), and someone else (the king) is the puppet master revealing that knowledge. (In the complete
knowledge case above, the guard and the king are essentially the same party.)
The king may reveal a little more to the guard, but only some gets passed
along to the prisoner.
Cribs
Originally posted by nemesioLet's be sure we understand the difference between "the guard imparts no
I am a bit lost, though, with the prisoner problem still. I think I share Iamatiger's confusion. It seems that the guard imparts no useful information by telling Prisoner A that either B or C is going to be executed, irrespective of th ...[text shortened]... ause it suggests that A always has a 50% chance, not a 33% chance.
useful information" and "the prisoner's chances of survival don't improve."
In the prisoner problem, as in the Monty Hall problem, these are exact opposites.
When one occurs, the other doesn't. Consider the version of Monty Hall
in which Monty has no knowledge. We saw the solution was that the contestant's
chances improved to 1/2. Likewise, in the standard Monty Hall, the contestant's
chances don't improve: they remain at 1/3 for his original choice.
The same is true in the prisoner problem. Consider playing the prisoner
problem out on a physical Monty Hall stage. Let the guard act as Monty.
If the guard has complete knowledge, he knows which of two doors are
safe to open, and the prisoner's chances don't improve. If the guard does
not have complete knowledge, he opened B at random and saw "Die",
and since it was random the prisoner now has a 1/2 chance of survival.
It is 1/2 for the very reason that you mistakenly thought the solution to the
standard Monty Hall problem was 1/2: namely, there are two remaining
doors, both equally likely to contain "Live." The proof is a simple proof
by contradiction: it you believe the remaining doors are not equally likely,
then you have information that one is more likley than another. But neither
the prisoner nor the guard have information, so we have a contradiction,
and it must be that the doors are equally likely.
Cribs
P.S. In your example, if the guard is lying, A's survival chances are 0. But
I still maintain that that is an irrelevant situation to analyze. For if you are
going to allow for the possibility that the guard lies, there's no reason not
to allow for the possibility that the king is lying about the executions in the
first place. Maybe nobody dies! Then A's chances are more than 1/3 if you're
going to allow that. That's why the whole idea of the guard lying is nonsense.
here is my take on this, and I may be way off. The concept of prior knowledge is said to be disclosed to monty and not to the contestant, but I think this is wrong. The contestant does'nt start with zero knowledge. we have 3 doors and we pick one, but we also know prior to our choice that one of the other doors will be eliminated. so I think we have a two door problem the one we pick or the one remaining.the third to me is a distractionary tactic, as it will always be disclosed as wrong. we will always have the one we pick, or the one left. to me this seems a 50/50 chance regardless as to which one we originally pick, because the third was never an option.
um like I tried anyway