Resource preservation exploit

[One_Brow]

I don't recall seeing this one before:

You have a small supply of a resource that often disappears (1 iron, two oil, etc.). Station a military unit and a few workers on top of the resource, and do not road it until you need it. When you need the resource to build something, create the road, set your city(ies) build queue, and pillage the road. It shoudl never disappear.

[Dominae]

Yeah, that works. But there are a few "complications".

First, you need to time the production of all your cities to take maximum effect of the window where you've connected your Iron. If city A is almost finished a Swordsman and city B is ready to build another one, you're going to need to build a road twice to get them both to produce another Swordsman. Multiply this by the number of cities you have (or the number of Iron-requiring units you want to build) and you've got some micromanagement to do.

Second, you need quite a few Workers to build a road in one turn, especially before Democracy and Replaceable Parts. I don't remember off-hand how many it is (forget Industrious civs or "difficult" terrain for now), but it's definitely more than I can usually afford at the time in the game when I'm using Iron.

Third, the exploit (although possible) isn't really needed. If you really want to mass-produce units requiring a strategic resource, chances are you're going to do some conquering with them, meaning you'll find yourself another source of that resource. This may seem like a weak argument, but practically it makes a lot of sense.

However, I do see the advantage of this "exploit" for Modern-era resources (Uranium and Aluminum) where the problems I discussed above are easily worked around.


Dominae

[Catt]

Warning: I do not know that the following is true -- I have never seen the Firaxian comment -- but it so often repeated here (with reference to Firaxian comment) that I have taken it to be true (partly becuase whether true or false it doesn't affect my gameplay in the least ).

Suppossedly, resources won't disappear unless connected to a trade route (i.e., roaded). But, if ever roaded, even for one turn, then the chance for depletion and disappearance exists. Pillaging your one roads may very well do nothing in terms of preserving the resource but may simply deny your civ the presence of the resource.

Catt

[punkbass2000]

I think this is useful, if taken past this application and to its logical conclusion. It has been said before, but simply don't connect any of your extra resources.

[ducki]

Also, once a unit is under construction, I don't think you need the resource.

Example: My first Regent game, I got lucky enough to have Iron and Coal in the same city radius.

I started the Ironworks, which requires both in the radius. 5 turns before completion, the source of Coal depleted. But the Ironworks didn't change out, like GWs do when someone else builds them.

So I let it ride, and sure enough, it finished, even without the coal.

I would venture to guess that requirements for producing a unit or improvement or Small Wonder are only checked at the beginning of construction, so for many resource needs, just don't connect it until you are ready to _start_ using it.

[punkbass2000]

Originally posted by Catt
Warning: I do not know that the following is true -- I have never seen the Firaxian comment -- but it so often repeated here (with reference to Firaxian comment) that I have taken it to be true (partly becuase whether true or false it doesn't affect my gameplay in the least ).

Suppossedly, resources won't disappear unless connected to a trade route (i.e., roaded). But, if ever roaded, even for one turn, then the chance for depletion and disappearance exists. Pillaging your one roads may very well do nothing in terms of preserving the resource but may simply deny your civ the presence of the resource.

Catt

I see what you're saying, but your argument against his proposal falls apart under probability. Although it does not matter how long your resource is hooked up, each turn that it is hooked up is one more turn that it's possible for it to be lost. In theory, if you followed One_Brow's idea and it resulted that, on average, every other turn your resource was disconnected, you would be half as likely, in the long run, to lose that resource. I hope that ddn't sound condescending, or anything, but I've recentlyt discovered how difficult it is to determine probabilities. Here's an (IMO) interesting and tough one:

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?

The answer wil either be posted today by 5ET or sometime after the weekend. (Bonus marks if anyone can accurately determine the probability of it being posted today )

[One_Brow]

Originally posted by ducki
Also, once a unit is under construction, I don't think you need the resource.

Yes, that was the point. You can connect, change your build chouice, and disconnect within a single turn, and your build queue remains.

[One_Brow]

Originally posted by Catt
Warning: I do not know that the following is true -- I have never seen the Firaxian comment -- but it so often repeated here (with reference to Firaxian comment) that I have taken it to be true (partly becuase whether true or false it doesn't affect my gameplay in the least ).

Suppossedly, resources won't disappear unless connected to a trade route (i.e., roaded). But, if ever roaded, even for one turn, then the chance for depletion and disappearance exists. Pillaging your one roads may very well do nothing in terms of preserving the resource but may simply deny your civ the presence of the resource.

Catt

If that is true, there is also this question: does the computer notice that the resource has been connected within a turn, or only that it was connected during the between-turn period. If the latter, then you might not ever be oficially connected tot he resource you're using, so it still won't expire.

BTW, I've never tried this, as I don't use exploits. I would be interested in hearing from someone who has tired it, though.

[Dominae]

punkbass2000, I think you misunderstood Catt's comment. Basically the claim is that if ever you connect up a Resource, a flag fires in the game that gives a probability that the resource will be depleted sometime in the future, regardless of whether it's connected to a road at that point or not. In my experience I've never seen a Resource disappear unless it was connected to my trade network, but then again I've rarely connected a resource, then disconnected it on purpose.

As for the answer to your question (assuming I'm reading it correctly), the chances that both people are men is the same as the chances that the other person is a man, which is 50% (I presume).


Dominae

[punkbass2000]

Originally posted by Dominae
punkbass2000, I think you misunderstood Catt's comment. Basically the claim is that if ever you connect up a Resource, a flag fires in the game that gives a probability that the resource will be depleted sometime in the future, regardless of whether it's connected to a road at that point or not. In my experience I've never seen a Resource disappear unless it was connected to my trade network, but then again I've rarely connected a resource, then disconnected it on purpose.

Hmm, upon rereading Catt's post, I can see it being interpreted in either way. If your interpretation is correct, I withdraw my objection.

As for the answer to your question (assuming I'm reading it correctly), the chances that both people are men is the same as the chances that the other person is a man, which is 50% (I presume).


Dominae

Given your answer, I think you are reading it correctly. Unfortunately, it is not the right answer. It is, however, the 'obvious' answer that virtually everyone would give the first time they hear it, which is why I posted it to demonstrate the difficulty of probability. It's what I answered

[DaveMcW]

Originally posted by punkbass2000

I've recentlyt discovered how difficult it is to determine probabilities. Here's an (IMO) interesting and tough one:

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?

The answer wil either be posted today by 5ET or sometime after the weekend. (Bonus marks if anyone can accurately determine the probability of it being posted today )

"If you know there are two people walking down a hall..."
There are 4 possible permutations of 2 people:

Person1:male Person2:male
Person1:male Person2:female
Person1:female Person2:male
Person1:female Person2:female

"...and you know that one of them is male..."
This narrows down our set to only 3:

Person1:male Person2:male
Person1:male Person2:female
Person1:female Person2:male

"...what are the odds they are both male?"

1 in 3, or 33%

-DaveMcW

[punkbass2000]

BTW, the answer is not based upon the fact that there are technically more women in the world than men or any other tricks like that. For the purposes of this question, if there was one person walking down the hall, the odds of them being male would be one in two.

[punkbass2000]

Originally posted by DaveMcW


"If you know there are two people walking down a hall..."
There are 4 possible permutations of 2 people:

Person1:male Person2:male
Person1:male Person2:female
Person1:female Person2:male
Person1:female Person2:female

"...and you know that one of them is male..."
This narrows down our set to only 3:

Person1:male Person2:male
Person1:male Person2:female
Person1:female Person2:male

"...what are the odds they are both male?"

1 in 3, or 33%

-DaveMcW

Correct! I'm going to guess that this is not first question of this nature that you have dealt with. If it is, then I am simply shocked. I think this illustrates how probability is not as straightforward as it can seem. This example shows how you can't assign positions to the objects in the probability question (ie. Person1 is male, and there is a 50% chance of Person2 being male, so there is a 50% chance of them both being male).

[Catt]

Originally posted by punkbass2000


I see what you're saying, but your argument against his proposal falls apart under probability.

Even though my comments were less than crystal clear, I think that your interpretation of them was somewhat strained.

But Dominae is correct in that I intended to share a widespread and oft-repeated belief on the forums that resources won't disappear unless connected to a trade route, but that once roaded, a resource has a chance of depleting every turn thereafter, regardless of whether or not the road is subsequently pillaged.

I knew the answer to the "two men in a hall" paradox, but only becuase I've seen similar mathematical riddles before (most people will immediately say 1/2).

Catt

[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

[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