Resource preservation exploit

[punkbass2000]

Originally posted by Dominae
I figured there was something tricky going on. But I still maintain (stubbornly, I know) that the question is not well-posed: by saying "you know one of them is male", in ordinary language, this means "that this or that one is male", and not "either A is male or B is male". The problem, however, is still a good one, it just doesn't have much to do with probability, but with how we represent and label things.

Anyone know the "3 prisoners dilemma"?


Dominae

In 'ordinary language' I agree, but not in probability, which was my point. It may sound 'tricky', but it is exactly the type of problem one encounters when dealing with probability.

[One_Brow]

Originally posted by punkbass2000 If you know there are two people walking down a hall, and you know that one of them is male, what are the odds they are both male?

punkbass2000:
As Dominae mentioned, your riddle is ambiguous. His answer was correct for his understanding of the riddle, which was different than the intended question.

Better would be:
If you know there are two people walking down a hall, and you know that at least one of them is male, what are the odds they are both male? The "at least" mitigates the specificity of the "one".

Dominae:
I'm aware of two-prisoner dilemma. How does the three-prisoner differ, besides in count?

[One_Brow]

Originally posted by punkbass2000 In 'ordinary language' I agree, but not in probability, which was my point. It may sound 'tricky', but it is exactly the type of problem one encounters when dealing with probability.

The trouble is that the "ordinary language" must first be translated into the language of probability, i.e., abstract mathematical symbols. The English sentence could be translated as either of (where m(x) means x is a male):

[P(m(b))|{a,b} ^ m(a)='True']

or

[P(m(b))|{a,b} ^ Ex(x å {a,b} ^ m(x) = 'True')]

The first translation leads to one-half, the second to one third.

[Dominae]

punkbass2000, I'm not putting down your question in any way. I'm simply commenting that much of the difficulty in "probability" problems isn't the actual stuff of probability, but our (sometimes faulty) representation of them. As One_Brow is pointed out, you can have a very good grasp of probability and still get "probability" questions wrong, because you're answering what you interpreted to be the right question. The fact that people like Catt and DaveMcW get the answer right is not so much their expertise with probability (however good it may be), but their experience with those types of questions.

One_Brow, I don't want to go too off-topic here, but here's a short version of the 3-prisoners dilemma:

Out of 3 prisoners, one will be sentenced to die the next day, and the two others will go free. The jailer knows which one will die, but is not allowed to tell the prisoners until the next day. Prisoner A asks the jailer to give a letter to either B or C, but it must be one that will go free (this does not give A any information because at least one of them will go free). The jailer does so. After, A asks the jailer which one the letter was given to (again, this gives him no information). The jailer says: "B".

Now, the chances that A would die were 1/3 before he sent the letter out, and are 1/2 after. What happened?


Dominae

[DaveMcW]

I haven't heard this one before, but it can be solved in a very similar way as the first. (I did hear the male/female problem before.)

"Out of 3 prisoners, one will be sentenced to die the next day, and the two others will go free."
There are 3 combinations:

PrisonerA dies, PrisonerB free, PrisonerC free
PrisonerA free, PrisonerB dies, PrisonerC free
PrisonerA free, PrisonerB free, PrisonerC dies

"After, A asks the jailer which one the letter was given to (again, this gives him no information). The jailer says: "B".
I disagree with this statement. The jailer does not violate his orders to keep the dying prisoner secret, but he does give new information that affects probabilities (obviously B now has 0% chance of dying). So we delete the combination in which B dies.

PrisonerA dies, PrisonerB free, PrisonerC free
PrisonerA free, PrisonerB free, PrisonerC dies

And A's probability of dying has gone up to 50%.

This would not be as confusing a problem if not for the misleading comments about "gives no information".

Of course, the probabilities are meaningless to A himself and should not have prevented him from writing the letter. He has already bet his life on the outcome of the problem and has no way to exploit his knowledge of the changed odds.

[Dominae]

This thread is now way off-topic, but there's no turning back!

DaveMcW, somehow I'm not surprised you'd be the first to try your hand at this one! Your analysis looks okay, but you've only got half the story (or is it a third? )

Consider: Had the guard said that C was going to be set free, the chances would have been 50% too (by symmetry). Therefore it appears prisoner A's chances to die were 1 in 2 to begin with. But then didn't the problem indicate that each prisoner had an equal chance of dying (1 in 3)?

So there's a problem. By the way, it's not the "3 Prisoners Dilemma" but the "3 Prisoners Paradox" (which makes more sense). If no one comes up with the solution I'll post it tomorrow sometime.

Getting back on topic, has anyone ever seen a resource get depleted when it wasn't "roaded"?


Dominae

[JohnM2433]

A's chance of dying is still 1 in 3, but C's is now 2 in 3.

See, if the guard gave the letter to B, that meant that either A or C will be killed. But if A is to be killed, then he just as likely would have given the letter to C. If C was to be killed, however, he would have to give it to B. So it's twice as likely that he gave it to B because C would be killed.

To put it another way, to start off there are the following possibilities, with the associated probabilities:

1. A to die, guard decides to give letter to B. (1/6)
2. A to die, guard decides to give letter to C. (1/6)
3. B to die, guard must give letter to C. (1/3)
4. C to die, guard must give letter to B. (1/3)

Once the guard gives the letter to B, only 1 and 4 are possible. Since 1 is still twice as likely as 4, that means C now has a 2/3 chance of dying.

[One_Brow]

Actually, this whole dilemma is a misconceived exercise. Probability is a discussion of unknown future events. Here, there is no unknown event. One of the prisoners has already been selected, he will be executed. The others will not.

Thus, prisoner A's "probability" is either 1 or 0, and we don't know which one. This does not change simply because B has probability 0.

What has changed is the rate at which random guesses will accurately identify the person who will die. Obviously, a reduction of the sample space (by eliminating B) will change that probability, *if* we adjust our sample space to reflect the jailer's actions.

[Dominae]

Originally posted by One_Brow
Actually, this whole dilemma is a misconceived exercise. Probability is a discussion of unknown future events. Here, there is no unknown event. One of the prisoners has already been selected, he will be executed. The others will not.

So, if I roll a die and cover it, you would argue that the probability of any given number was different from 1/6 (assuming a fair die)? I think we'll both agree that (othe than the realm of quantum mechanics) things either are or they aren't, but there is still something to say for probability, no?

I had completely forgotten about this poser that I posed, so sorry to anyone who was eagerly awaiting the answer (probably no one, but I feel bad not posting anyways).

Here we go:

The paradox can be resolved by noting that 'the guard said B is to be released' implies 'B is to be released', but not vice versa; if 'B is to be released' is true, then differing probabilities can be assigned to 'the guard said B is to be released'. Another way to look at it is that the fact that 'the guard said B is to be released' could indicate two very different state of affairs, namely, that C is to be killed (where the guard has no choice but to say 'B'), and that A is to be killed (where the guard does have a choice).

Here's the dirty math:

---

Let:

IB; 'B will be declared innocent'
GA; 'A will be declared guilty'

Then:

P(GA|IB) = P(IB|GA)*P(GA)/P(IB)
= P(GA)/P(IB)
= (1/3)/(2/3)
= 1/2

This is the straightforward analysis that leads to the paradox, since it's stipulated that all three prisoners have an equal chance of being declared guilty.

---

Let:

IB2: 'The guard said B will be declared innocent'

Then:

P(GA|IB2) = P(IB2|GA)*P(GA)/P(IB2)
= (1/2)*(1/3)/(1/2)
= 1/3

This resolves the paradox.


I'm now not speaking off-topic (on any thread) for another month.


Dominae

[One_Brow]

Originally posted by Dominae

So, if I roll a die and cover it, you would argue that the probability of any given number was different from 1/6 (assuming a fair die)?

Yes, that is exactly correct. We may not know which value the probabilities is for any individual number, but the probability for any number is either 1 or 0 at that point. This is different from the probability of my guessing the number correctly (again, this only applies before I guess).

I think we'll both agree that (othe than the realm of quantum mechanics) things either are or they aren't, but there is still something to say for probability, no?

There is plenty to say for it in future event, but you can't discuss the probability of a past event any more than you can "predict" who will win the 2000 Presidential election.

[DaveMcW]

One_Brow, I propose a game for you.

You pay me $1 and I roll a fair 6-sided die and cover it. If you guess correctly what number comes up I will pay you $3. You may play as many times as you want, as long as you keep paying me $1 per try. Would you play?

Let's increase the payoff to $10 for a correct guess. Now would you play?

I hope this shows that probability is still useful on past events if you don't already know the outcome.

[One_Brow]

Originally posted by DaveMcW
One_Brow, I propose a game for you.

You pay me $1 and I roll a fair 6-sided die and cover it. If you guess correctly what number comes up I will pay you $3. You may play as many times as you want, as long as you keep paying me $1 per try. Would you play?

Let's increase the payoff to $10 for a correct guess. Now would you play?

I hope this shows that probability is still useful on past events if you don't already know the outcome.

You have the order of events wrong.

ou have already rolled the dice, and covered it. I give yo a dollar. I have already written the number on a piece of paper. You now make this offer: if the numbers match, I'll pay you $3/$10. Either I'll get the money, or I won't. With no decision to make, there is no probability to9 consider.

Getting back to the origianl "paradox", knowledge that B will live does not affect the fate of A, as this has already been decided. Probability is not a meaningful theory in this case.

[DaveMcW]

Even with your piece of paper and my covered dice, there is still a decision to make: will you pay me $1 for a chance at $3? You need probability to determine whether it is a smart bet.

Comparing it to the prisonner problem... suppose I peek at the dice and say "It's not 1, 2, 3, or 4." If you have written 5 on your paper you should take the 3:1 bet, because your odds have gone up to 1 in 2.

But suppose I take your piece of paper, peek at it AND the dice, and say "It's not 2, 3, 4, or 6." You should not take the bet even though you have more information, because I will never tell you that the number on the paper is wrong. Just like the jailer will never tell A that he will go free.

[One_Brow]

You refer to a decision to make, and thus to a future event, in the example of the dice and written number.

What is the decision that must be made in the 3-prisoner "paradox"?

Once we have such a decision, or some other not-yet-determined event, than probability will become meaningful. I don't see where any decision is required.

[nbarclay]

Originally posted by One_Brow
Actually, this whole dilemma is a misconceived exercise. Probability is a discussion of unknown future events. Here, there is no unknown event. One of the prisoners has already been selected, he will be executed. The others will not.

From the prisoners' perspective, who will live and who will die is certainly an unknown future event. Similarly, from the time a die is rolled or a coin is flipped until the time someone looks at it, what value will be seen when someone looks is an unknown future event.

As any science fiction fan knows, human language is not always well suited to dealing with temporal (time-related) phenomena. In this case, terms such as "probability" are used in what might be called a retroactive tense: even though the event has already been decided, we treat it as undecided in our speech until we actually know the outcome. And my impression is that such usages are more than sufficiently common to be regarded as "correct" in terms of common usage of the language.

Nathan