Zeno's Paradox and Flechette Defense

[Chaos Theory]

Originally posted by fender
To keep cutting the odds of the PB getting through in half, is IMO flawed logic.

Each failure or successfully take down of a PB must be looked at separately. Each attempt is must be looked at on it's own. The odds are always 50/50.

But if a PB gets through a single ODP, you still don't know whether the Drones survive or die. You need to know if it gets through all the ODPs. The odds are not always 50/50 for this decision.

For example. If you could have such a thing as a perfectly balanced roulette wheel and an infinite number of tries at the wheel. If red was success and black failure, you will hit red at some point.

Are you trying you tell me that you could never hit red? You know I will, and then down goes the PB, and the next and then next.

We're working with time-varying probabilities. An analogy would be if every time you spun the roulette wheel half the red tiles became black. You would have a chance to hit red eventually, and a chance to never hit it.

In this situation the PB, being the attacker, set the whole thing in motion. I only have to deal with one PB at a time with an infinite number of tries to do it. The playing field is NOT level, so to speak. I will always be able to shoot it down so long as time itself is infinite.

This doesn't make much sense to me. Each PB must make it through all the remaining, undeployed ODPs. If only one per turn is being launched, then it must simply make it through all the ODPs. For many in a single turn, I've actually made an error in my previous posts.

When launching M PBs against N ODPs, each ODP that deploys, either successfully or unsuccessfully, is unavailable to stop the next PB. My calculations were all based on having to go against the full set of ODPs each turn.

To accurately represent M PBs against N ODPs, first assume that M <= N, otherwise a PB will get through every time. Now, consider X, the number of ODPs that a PB will penetrate before being stopped. X is a random variable with an exponential distribution across the integers, such that P(X) = .5^X (my notation here might be screwy, but I hope it gets the idea across). The expected value (or mean) of X is the sum from 1 to infinity of X*P(X) = x/(2^x). This series has a limit of 2. Therefore, the mean of X is 2. By other calculations, the standard deviation of X is also 2.

The number of ODPs that M PBs will penetrate is represented by M samplings of X. As M -> infinity, this will be a Gaussian distribution with mean 2*M and standard deviation 2*sqrt(M). To find out the odds of whether at least 1 of M PBs will penetrate N ODPs, calculate

(2M - N) / (2*sqrt(M))

and compare the value to a normal distribution chart (or use a function I don't care to find).


Whoha:
These calculations are based on the assumption that ODPs are never sacrificed. Of course, a sensible player would almost always sacrifice an ODP if needed, but we're (or maybe it's just me) having fun with math!


smacksim:
It may be crazy to you, but not to me nor anyone experienced with math. Not to be mean, but try thinking about it, rather than flailing. Consider a finite case, and see what happens in the limit as the finite parameters approach infinity.

[smacksim]

(Edit: I was going to ignore this bait, but it's too tasty....)

smacksim:
It may be crazy to you, but not to me nor anyone experienced with math. Not to be mean, but try thinking about it, rather than flailing. Consider a finite case, and see what happens in the limit as the finite parameters approach infinity.

Oh comeon Chaos Theory, come down off your pedastal for a little bit, would ya?

As (statement) --> ("not to be mean, but...."), 1/("not to be mean") --> true connotation

To be perfectly plain, in the hopes that it may be useful to you, your non-math statements are insulting. I don't see a need for that. I respect you and your intelligent posts. I find that I always respect what you have to say and that you bring a lot to any discussion. It is sad to find out that the reverse is not true, but I can accept that. However, I would ask that you refrain from this kind of inflamatory remark, if only because we are teammates in the ACDG and must get along.

If your line of reasoning is correct, it will speak for itself I should think. I can sympathise with the need to be right, but there is no need to insult anyone, is there?

We need another math voice to speak on this. Until then, or until I'm convinced of some real number solution, I'll stick with my own line of reasoning which seems like more of a solution to me, obviously.

I do not understand your premise that this is a time-varying series. We never approach infinity, we have infinite weapons/defenses. Please explain how and why our solutions differ. If I am really so far beneath you that my argument should be ignored, I will be surprised and delighted to learn the truth, though growth be painful. So I beg you, enlightened one, speak one more time to the thick-headed crowd, if you have the time in nirvanna for such things.

-Smack

[MrWhereItsAt]

Well I'm a physicist, and although I understand most of what has gone on here, I would tell you that this is physically unrealisable and thus not worth worrying about. In any case, whatever your conclusion, someone will find something in testing that goes against it, and you'll then find your mistake, and make another prediction.... ad infinitum



Mathematicians.

[Chaos Theory]

Originally posted by smacksim
(Edit: I was going to ignore this bait, but it's too tasty....)





Oh comeon Chaos Theory, come down off your pedastal for a little bit, would ya?

As (statement) --> ("not to be mean, but...."), 1/("not to be mean") --> true connotation

To be perfectly plain, in the hopes that it may be useful to you, your non-math statements are insulting. I don't see a need for that. I respect you and your intelligent posts. I find that I always respect what you have to say and that you bring a lot to any discussion. It is sad to find out that the reverse is not true, but I can accept that. However, I would ask that you refrain from this kind of inflamatory remark, if only because we are teammates in the ACDG and must get along.

Okay, but I don't mean to be insulting - I mean to give you a metaphorical kick in the butt to cause you to stop repeating yourself, and broaden your horizons. When I read your posts, I don't see that I've had much effect, and you still hang up on an infinity. I probably have had more math than you (how much have you had, anyway?) and ignorance is not a flaw.

If your line of reasoning is correct, it will speak for itself I should think.

I had hoped so, but I'm not sure why I'm not being understood. With just a few of us arguing over this, I can't tell if I'm just being esoteric, or if you're being stubborn.

I can sympathise with the need to be right, but there is no need to insult anyone, is there?

I argued on Usenet. I figured out a while ago that it's pointless to insult people over the internet. Pointing out something you're doing that I don't like isn't insulting. At worst, it's humiliating, but I don't expect anyone to be humiliated at not knowing complicated math, especially when it's being applied to a trivial purpose.

We need another math voice to speak on this.

That would be quite welcome, but I wouldn't know who on Apolyton to fetch.

Until then, or until I'm convinced of some real number solution, I'll stick with my own line of reasoning which seems like more of a solution to me, obviously.

But the catch is your solution is that a solution is impossible, whereas I can come up with a concrete solution. If our results differed, either could be correct, but this situation is asymmetric. You should be able to poke a hole in my reasoning if you are correct. Reinforcing your point is insufficient.

I do not understand your premise that this is a time-varying series. We never approach infinity, we have infinite weapons/defenses. Please explain how and why our solutions differ. If I am really so far beneath you that my argument should be ignored, I will be surprised and delighted to learn the truth, though growth be painful. So I beg you, enlightened one, speak one more time to the thick-headed crowd, if you have the time in nirvanna for such things.
-Smack

Well, for starters let's make sure we're dealing with the same problem. How about you phrase the problem? It doesn't help that we've dealt with several in this thread.

As far as dealing with the thick-headed, I'm currently a graduate teaching assistant. No head is too thick for me .

[Natalinasmpf]

shouldn't each odp get sacraficed to stop the pb?

Ari explicitly stated they wouldn't.

That might create interesting results, though.

Each failure or successfully take down of a PB must be looked at separately. Each attempt is must be looked at on it's own. The odds are always 50/50.

Yes, but then you have to SUM up the entire thing. Each attempt is looked as its own, before being summed up in a series: hence formulas, rather than going through all of them at once.

It will NOT be 50/50, since there is more than 1 ODP, for example.

For example. If you could have such a thing as a perfectly balanced roulette wheel and an infinite number of tries at the wheel. If red was success and black failure, you will hit red at some point.

Are you trying you tell me that you could never hit red? You know I will, and then down goes the PB, and the next and then next.

Ah, but thats different. An ODP only halves the chance. The thing is, in your analogy, include an infinite amount of roulette wheels. What is the chance of hitting red constantly, for all eternity? Is there a chance it will eventually reach convergence and stop? Thus no chance?

Well not an infinite amount of roulette, wheels, I meant spin that infinite times, anyway.

One roulette wheels only represent one ODP. Lets have a roulette wheel with say, huge odds, 50% is red, 50% is black.

The playing field is NOT level, so to speak. I will always be able to shoot it down so long as time itself is infinite.

What do you mean by the "playing field is not level"? The resolution of Zeno's paradox requires that an infinite terms can be resolved to create a finite result: ie. reduction of percentage to ZERO.

Say, can you people start stressing important points (ie. by bolding or italics) when you post the mathematical arguments? Its easier to follow, then.

[Ari Rahikkala]

Originally posted by MrWhereItsAt
Well I'm a physicist, and although I understand most of what has gone on here, I would tell you that this is physically unrealisable and thus not worth worrying about. In any case, whatever your conclusion, someone will find something in testing that goes against it, and you'll then find your mistake, and make another prediction.... ad infinitum



Mathematicians.

Wrong. We've defined the problem to be in the field of mathematics, therefore you can't test it with the methods of physics.

Physicists.

[Ari Rahikkala]

Another single quotation that I wanted to criticise separately.. I'll read the rest of the thread after I'm done with this and quote the rest of this stuff in a lump post... if I've got anything interesting to say anymore...

Originally posted by Chaos Theory
We're working with time-varying probabilities. An analogy would be if every time you spun the roulette wheel half the red tiles became black. You would have a chance to hit red eventually, and a chance to never hit it.

It's a bit more complicated than that. Yes, the situation is pretty much that half of the red tiles would become black every time you spin the roulette. The problem is that you need either an infinite number of both kinds of roulette tiles, or a continuum of tiles so that you could colour fractions of them. Otherwise you'd eventually arrive at a situation where you have either zero or one red tiles left for all eternity, depending on how you define the result of division that does not yield an integer for integer numbers.

Hitting a red tile <-> a PB getting through
Hitting a black tile <-> a PB getting destroyed
Spinning the roulette <-> launching a PB
Turning half of all red tiles to black <-> building an ODP

[Chaos Theory]

Originally posted by Ari Rahikkala

It's a bit more complicated than that. Yes, the situation is pretty much that half of the red tiles would become black every time you spin the roulette. The problem is that you need either an infinite number of both kinds of roulette tiles, or a continuum of tiles so that you could colour fractions of them. Otherwise you'd eventually arrive at a situation where you have either zero or one red tiles left for all eternity, depending on how you define the result of division that does not yield an integer for integer numbers.

Hitting a red tile <-> a PB getting through
Hitting a black tile <-> a PB getting destroyed
Spinning the roulette <-> launching a PB
Turning half of all red tiles to black <-> building an ODP

Yes, that is correct. We are dealing with continuous probability here. Silly real discrete roulette wheels.

[smacksim]

Re: Zeno's Paradox and Flechette Defense

Chaos Theory has asked for a restatement of the problem. Here's the core of the setup from Natalina's first post:

Originally posted by Natalinasmpf
Thus, suppose I have an infinite amount of planet busters (or just so much, it has crossed the threshold), a huge amount of RAM and such, and a critical city, lets say Free Drone Central, since with planned and eudaimonic, they can halve satellite costs, and with space elevator, get it done in 25% of the time, say one turn. Thats not really important though. Now, lets say there was no turn limit, and suppose (anyway there's 1 million+ mission years in one of the interludes ) the base now has an infinite amount of Flechette Defenses, or enough to cross the threshold of infinity into the surreal number category (where adding +1 no longer matters, same effect).

The key is: "suppose I have an infinite amount of planet busters", and "suppose (...) the base now has an infinite amount of Flechette Defenses"

ODPs or Flechette, no matter. We are focused on a defense that is 50% effective. With 2 such defenses the probability of failure against a single missle is 25%, etc..

We are dealing with infinite weapons vs. infinite defenses. Natalina suggests that we have inifinite turns to resolve the issue, should we not choose to launch all the weapons at once. To rephrase as simply as possible:

Suppose we have 2 factions, one with infinite planet buster weapons, and one with infinite defenses. A wepon has a 50/50 chance of getting past a single defense. A defense has a 50/50 chance of stopping an inbound weapon. Multiple weapons can act on each defense, and each defense can attempt to block multiple weapons (perhaps until it is successful, when it is destroyed?). The question is: Will any Planet Buster be successful and make it through to destroy the enemy faction, or will infinite defenses guarantee that no PB will ever get through, or neither?

I suggest that because we can prove both that all the PBs will get stopped and, on the other hand, that at least one will get through, that the answer, in combination, is undefined. Further, I suggest that the actual percentages do not matter at all, so long as they are between 0 and 1. Others disagree.

[Chaos Theory]

Okay, I was working with a related problem:

Hive against Drones
Hive builds and uses 1 PB/turn
Drones build 1 ODP/turn and never sacrifice any
What are the odds the Drones will survive forever?

I also handled:

Hive against Drones
Hive has M PBs
Drones have N ODPs
The Hive launches all its PBs on 1 turn - what are the odds the Drones will survive, as M and N -> infinity?

and

Hive against Drones
Hive has M PBs
Drones have N ODPs
The Hive launches 1 PB/turn and production for both sides halts completely - what are the odds the Drones will survive, as M and N -> infinity?


If all you have is infinity PBs vs infinity ODPs, with no further information, the odds of survival are undefined and the problem is insoluble, except perhaps in surreal mathematics, which I don't know.

[smacksim]

Now it's getting interesting again.

Originally posted by Chaos Theory
Okay, I was working with a related problem:

Hive against Drones
Hive builds and uses 1 PB/turn
Drones build 1 ODP/turn and never sacrifice any
What are the odds the Drones will survive forever?

Edit: Whoops, I didn't take into account the accumulating ODPs. That is interesting!

Originally posted by Chaos Theory
I also handled:

Hive against Drones
Hive has M PBs
Drones have N ODPs
The Hive launches all its PBs on 1 turn - what are the odds the Drones will survive, as M and N -> infinity?

I think I see how this is interesting....I see how you could look at examples along the way with varying probabilities, but, as soon as we introduce infinity the problem explodes, so to speak. What I'm arguing is that the case must be examined from the point of view of survival and the point of view of destruction. As each probability converges on a value, they diverge completely as the series approaches infinity. Survival + Destruction == not solvable.

One can look at the case of 100 PBs and 100 ODPs, for example. It looks like there's very little chance for any PB to make it through that screen if we take the point of view of a single PB going through 100 ODPs, 100 times. Its a very small number. But if we look at the obverse, the situation changes: What are the chances that a single ODP could stop 100 PBs, and repeat for 100 ODPs? The same tiny number.

So I suppose the real issue is mechanical: How does the battle get resolved: all at once or sequentially? Sequentially from the point of view of the PB passing through the ODP screen, or from the point of view of each ODP handling all the PBs?

If we mash all the battles together into one big moment, we have 100 PBs making 100 survival checks each, or do we have 100 ODPs making 100 'Did I kill it?' checks each?

-----------------------
Edit: The following is totally incorrect! One can't add probabilities like that.
-----------------------

From the point of view of the Planet Buster's Survival
Reducing the numbers to 3 PBs and 3 ODPs...
PB#1 has .5*.5*.5 = .125 chance of survival
PB#2 has .5*.5*.5 = .125 chance of survival
PB#3 has .5*.5*.5 = .125 chance of survival

Summed = .375 chance of a PB getting through. This number decreases to something like 1/e over as we approach infinity.

From the point of view of the ODP's being successful
ODP#1 has .5*.5*.5 = .125 chance of knocking down all the PBs
ODP#2 has .5*.5*.5 = .125 chance of knocking down all the PBs
ODP#3 has .5*.5*.5 = .125 chance of knocking down all the PBs

Summed = .375 chance of any ODP taking care of the situation by itself. This number decreases to something like 1/e over time. However, we can see that there is a problem with this line of thought.....

On the other hand, there are several ways to set up the battle more sequentially, which is how it is probably done by the code.

Originally posted by Chaos Theory
Hive against Drones
Hive has M PBs
Drones have N ODPs
The Hive launches 1 PB/turn and production for both sides halts completely - what are the odds the Drones will survive, as M and N -> infinity?

One.

Any one PB will be stopped by the ODPs, as I've argued before. Because you are launching a single PB in a given turn, the odds of it making it are exactly zero. Or is there another way to go about this that you had in mind?

[Chaos Theory]

Originally posted by smacksim

Hive against Drones
Hive has M PBs
Drones have N ODPs
The Hive launches all its PBs on 1 turn - what are the odds the Drones will survive, as M and N -> infinity?

I think I see how this is interesting....I see how you could look at examples along the way with varying probabilities, but, as soon as we introduce infinity the problem explodes, so to speak. What I'm arguing is that the case must be examined from the point of view of survival and the point of view of destruction. As each probability converges on a value, they diverge completely as the series approaches infinity. Survival + Destruction == not solvable.

One can look at the case of 100 PBs and 100 ODPs, for example. It looks like there's very little chance for any PB to make it through that screen if we take the point of view of a single PB going through 100 ODPs, 100 times. Its a very small number. But if we look at the obverse, the situation changes: What are the chances that a single ODP could stop 100 PBs, and repeat for 100 ODPs? The same tiny number.

So I suppose the real issue is mechanical: How does the battle get resolved: all at once or sequentially? Sequentially from the point of view of the PB passing through the ODP screen, or from the point of view of each ODP handling all the PBs?

If we mash all the battles together into one big moment, we have 100 PBs making 100 survival checks each, or do we have 100 ODPs making 100 'Did I kill it?' checks each?

From the point of view of the Planet Buster's Survival
Reducing the numbers to 3 PBs and 3 ODPs...
PB#1 has .5*.5*.5 = .125 chance of survival
PB#2 has .5*.5*.5 = .125 chance of survival
PB#3 has .5*.5*.5 = .125 chance of survival

Summed = .375 chance of a PB getting through. This number decreases to something like 1/e over as we approach infinity.

From the point of view of the ODP's being successful
ODP#1 has .5*.5*.5 = .125 chance of knocking down all the PBs
ODP#2 has .5*.5*.5 = .125 chance of knocking down all the PBs
ODP#3 has .5*.5*.5 = .125 chance of knocking down all the PBs

Summed = .375 chance of any ODP taking care of the situation by itself. This number decreases to something like 1/e over time. However, we can see that there is a problem with this line of thought.....

On the other hand, there are several ways to set up the battle more sequentially, which is how it is probably done by the code.

Your second analysis isn't accurate, and here's why:
One ODP doesn't have to stop all the PBs. They can cooperate. If ODP #1 stops 2 PBs and ODP #2 stops 1 PB, then none get through. The odds of all 3 being stopped are therefore higher than you suggest.

The other problem with this is that an ODP deploys itself and becomes unavailable as soon as it is used, whether or not it is successful. However, if you apply this analysis to the next problem, that's not the case, as the PBs are launched over time. This would also work if flechette defenses were satellites.

Hive against Drones
Hive has M PBs
Drones have N ODPs
The Hive launches 1 PB/turn and production for both sides halts completely - what are the odds the Drones will survive, as M and N -> infinity?

One.

Any one PB will be stopped by the ODPs, as I've argued before. Because you are launching a single PB in a given turn, the odds of it making it are exactly zero. Or is there another way to go about this that you had in mind?

If M rises much faster than N, the odds the Drones will survive will be less than 1. For example, if M = 2^N, the odds of survival are 1/e, as I demonstrated low on page 2. The odds of a single PB penetrating are not exactly zero, they simply approach zero. For a common example, consider what happens to sin(x) / x as x -> 0. The top goes to 0, but so does the bottom. The quotient approaches 1, in this case. You can of course replace x with 1/x and have it tend towards infinity.

[smacksim]

Which is why I said " However, we can see that there is a problem with this line of thought....."

But there are myriad problems with it unless we know how the mechanics of resolution work out. Clearly, infinite comparisons cannot be done simultaneously, and as you point out, perhaps we should consider the defenses being 'used up' if they block a PB.

I'm still not grasping the last example, but that's alright. I'll take another look.

[Chaos Theory]

The comparisons cannot be done simultaneously, but if they're done correctly (which I believe they aren't, but that's beside the point), they should be equivalent to the simultaneous case. PB are launched one at a time and so must be checked this way. This doesn't mean you can't calculate statistics for how likely M are to penetrate. The ODPs also function one at a time and should be checked that way. However, a computer program might instead generate a random number that represents how many ODPs it takes to stop a particular PB. This would have the same result as checking them one at a time, if the random number has the right distribution (exponential, discrete).

One nice thing about math is if you're doing something legit, correctly, it doesn't matter how you arrive at your answer, it must always be the correct answer. Conflicting answers indicate that either what you're measuring is undefined, or that you are making a mistake.

[goomeister]

Hm, interesting problem. I can't resist a good mathematical workout, so let's take a crack at it, shall we? The problem boils down to "What is the probability of m PBs successfully attacking a base with n ODPs, as m and n approach infinity?" So, we need to find the success formula in terms of m and n.

So, let's consider the special case where m=1. Against one ODP, the probability of success are 1/2. Against two ODPs, the probability is 1/4. Against three ODPs, the probability is 1/8. A PB has a 1/2 probability of passing each PB, so the probability that it will succeed against n ODPs is (1/2)^n.

Now, we can derive the formula for m PBs. The chance that a PB will fail against n ODPs is 1-(1/2)^n. Thus, the chance that m PBs will fail is (1-(1/2)^n)^m. The chance that they will succeed is one minus the chance of failure, or 1-(1-(1/2)^n)^m.

P(m, n)=1-(1-(1/2)^n)^m.

With this formula in hand, we can solve the original problem, by finding the limit of P as m and n approach infinity.

lim P(x, x)    =    lim (1-(1-(1/2)^x)^x) = 0
x->+infinity        x->+infinity

Therefore, the probability that infinitely many PBs will penetrate infinitely many ODPs is zero.

Think about it this way: the probability that a PB will pass by an ODP is 50/50, a coin toss - heads, the PB survives, tails, it is caught. You have to flip a coin for every ODP, and get heads each time to successfully attack. Now consider infinitely many ODPs: you'd have to flip infinitely many coins and never get tails. But, if you flip a coin infinitely many times, eventually, after a finite number of flips, one will land tails! Thus, one PB can never penetrate infinitely many ODPs. And if the probability of success for one PB is zero, the probability for two PBs is also zero, and for three, and four, and so on up to infinity. If it is impossible for one PB to attack, it is impossible for any number to attack, because each PB attack is an independent event.