13-09-2004 03:31									#61
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quote: 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.

quote: 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.

quote: 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.

	13-09-2004 07:36									#62
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(Edit: I was going to ignore this bait, but it's too tasty....)

quote: 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

	13-09-2004 09:19									#63
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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.

	13-09-2004 12:00									#64
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quote: 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.

quote: 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.

quote: 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.

quote: We need another math voice to speak on this.

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

quote: 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.

quote: 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 .

	13-09-2004 15:40									#65
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quote: shouldn't each odp get sacraficed to stop the pb?

Ari explicitly stated they wouldn't.

That might create interesting results, though.

quote: 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.

quote: 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.

quote: 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.

	13-09-2004 17:10									#66
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quote: 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.

	13-09-2004 17:27									#67
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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...

quote: 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

	13-09-2004 18:17									#68
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quote: 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.

	14-09-2004 01:55									#69
	Re: Zeno's Paradox and Flechette Defense Support Apolyton, buy GURPS/ Alpha Centauri

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

quote: 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.

	14-09-2004 05:46									#70
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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.

	14-09-2004 07:12									#71
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Now it's getting interesting again.

quote: 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!

quote: 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.

quote: 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?

	14-09-2004 08:01									#72
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quote: Originally posted by smacksim quote: 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.

quote: quote: 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.

	14-09-2004 08:19									#73
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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.

	14-09-2004 09:49									#74
	Remove this text

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.

	14-09-2004 13:01									#75
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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.

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

	14-09-2004 13:43									#76
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goomeister, another voice of math. Yeah!

I think it is interesting that everyone, including me, is trying to clarify and narrow the problem after they have solved it! We know that certainly some part of this is an independent event, but what part?

CT brought up the point that (apparently) ODPs are lost after being committed to stopping an inbound ballistic. Its been a long time since I've had a PB war, so I don't remember, but that does ring a bell. Could someone clarify the mechanics of ODPs and flechettes? Yeah, yeah, I know, it's a simple problem and we should be done with it, but several more interesting variations have come up, so it would be good to recall how PBs are handled. Maybe I'll go try it out and report....

Anyways, the premise that:

quote: 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.

can be reversed to state that it is impossible for any single ODP to stop an infinite attack, therefore some attack will get through no matter how many ODPs there are.

The assumption has been that the attack is looked at from the mechanics of a single Planet Buster attempting to poke a hole in a perfect defense. There is no reason why it cannot be framed, or programmed, as one ODP handling x attackers, then moving on to the next ODP.

If you discard this premise, you must come up with some alternate mechanics for the problem. What if we didn't know the answer? What if it were impossible to know before-hand how it would be resolved? If we have n and m quantities we must only accept solutions in which the order of calculations is independent of the result, otherwise we are dictating the mechanics to fit the solution. As CT has earlier shown, quantities below infinity are non-zero, and not surprisingly, approach 1/e for the Planet Busters, while the ODPs have even better odds ...

So we have two kinds of questions:

1. Scenarios in which we start with infinite quantities.
   
2. Scenarios where we start with numerical quantities and build towards infinity, and variations.
   

	14-09-2004 16:23									#77
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quote: Originally posted by Ari Rahikkala ...you can't test it with the methods of physics.

Unfortunately your statement simply means this is all of even less real relevance.

Also, Zeno's Paradox only seems to consider one of the variables involved, distance, as infinitely divisible. I'm not sure whether it compounds matters by dealing with both important parameters (distance and time) as infinitely divisible, or reduces the paradox to, well, not a paradox.

And besides, the uncertainty principle dictates a limit of sorts on the measurement of distance, as it does on time. Thus it is totally meaningless to speak of distances/times smaller than about the value of Planck's constant, which whilst beyond our reach, is in no way infinitesimal.

	14-09-2004 21:32									#78
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quote: Originally posted by 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. code: lim P(x, x) = lim (1-(1-(1/2)^x)^x) = 0 x->+infinity x->+infinity

Here you made an error. You assumed that m = n. With that assumption, your calculations are correct. However, consider the case where m = 2^n. Let both approach infinity, and then work your calculations.

quote: 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.

But it is not impossible for a PB to get through, its odds simply approach zero. Even though your result is valid, given your assumption, your reasoning as to why this is so is flawed. Considering the case m = 2^n should help with this.

	15-09-2004 13:32									#79
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Grrr... problems this simple shouldn't be so difficult!

quote: I think it is interesting that everyone, including me, is trying to clarify and narrow the problem after they have solved it! We know that certainly some part of this is an independent event, but what part? CT brought up the point that (apparently) ODPs are lost after being committed to stopping an inbound ballistic. Its been a long time since I've had a PB war, so I don't remember, but that does ring a bell. Could someone clarify the mechanics of ODPs and flechettes? Yeah, yeah, I know, it's a simple problem and we should be done with it, but several more interesting variations have come up, so it would be good to recall how PBs are handled. Maybe I'll go try it out and report....

Very good points. It appeared to me that the attacks were independent. Consider two PBs and one ODP: whether or not the first PB hits, there's still one ODP for the second PB to contend with. This was my reason for assuming the attacks to be independent. However, if ODPs are lost after stopping an attack (and I have a sneaking suspicion that they are), this cannot be so - if one PB is blocked, the next one has a greater chance of success. Is it possible to separate these events?

quote: Here you made an error. You assumed that m = n. With that assumption, your calculations are correct. However, consider the case where m = 2^n. Let both approach infinity, and then work your calculations.

For the purposes of the original problem, it seemed safe to assume that the production of PBs and ODPs was linear. In that case, the limit is zero.

Actually, I thought about this earlier today - if you want some constant nonzero probability P, what must the relationship be between m and n? For the limit to approach a nonzero value, the number of PBs must increase exponentially with respect to the number of ODPs. My back of the envelope calculations indicate that P approaches 1 when m=e^n.

It is impossible to calculate the limit until you know the relationship between m and n. But, if the ODPs are lost after stopping an attack, there is no simple relation between the two - n varies not only with m, but with the number of failed attacks. How can you calculate that, anyway? It is impossible to know the number of failed attacks beforehand, only the probability that an attack will fail. I wonder, is it even possible to solve this problem?

	15-09-2004 16:34									#80
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My adding of probabilities above is completely wrong, but I won't remove it as CT quoted it. Sorry to confuse the issue for the less stat inclined

It must be done like this dice example

	15-09-2004 18:14									#81
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quote: Originally posted by goomeister It is impossible to calculate the limit until you know the relationship between m and n. But, if the ODPs are lost after stopping an attack, there is no simple relation between the two - n varies not only with m, but with the number of failed attacks. How can you calculate that, anyway? It is impossible to know the number of failed attacks beforehand, only the probability that an attack will fail. I wonder, is it even possible to solve this problem?

I'll assume you're talking about the problem of launching M PBs on a single turn against N ODPs. It's not too difficult, if you know how to solve it:

quote: Originally posted by Chaos Theory 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).

	15-09-2004 20:55									#82
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God, are you nerds still arguing math? Why can't you discuss politics or philosophy like normal rambling posters?

	15-09-2004 22:31									#83
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Sour grapes

	16-09-2004 01:34									#84
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Its so much more satisfying that there might be a solution in mathematics, whereas in politics we can only discover more problems

	16-09-2004 12:09									#85
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Guys, I hate to rain on your boat, but as an old assembler programmer, there is a very simple answer. As your chance PB success is halved, it will eventually reach the point where rounding error exceeds the actual chance. If the smallest number the system can represent is let's say 1%, or .01, you reach that point as you drop below 1%. At this point the processor or the program are going to say that it always equals 1%, or it always zero. If it is zero, the PB never gets through. If it is 1% then eventually it gets through. Obviously it's going to occur at a much smaller quantity, but I am feeling much to lazy to research it and type that many zeroes (to make my point). While mathemiticians could program your calculations into it, trust me for things like Windows and these games, they won't and the rounding issue will prevail. Great discussion though, I've never did get past Calculus 3 (required for Physical Chemistry).

	16-09-2004 12:46									#86
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quote: trust me for things like Windows and these games,

I use the Linux version, ha!

The C language should have a good grasp of infinite values and the such. Well maybe not, but when we get to quantum computing...

	16-09-2004 13:01									#87
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shawnmcc: I'm a young C programmer, and I don't really think the creators of the game bothered to calculate any actual probabilities here. I mean, my computer creates ten million pseudorandom numbers quite happily in half a second (I tested that), so it's not like there's a need to optimise this when no sane person's ever going to build more than a couple of dozen ODPs. Just test if rand() returns a value lower than RAND_MAX>>1 (on little-endian hosts ) as many times as there are defending ODPs.

If we were looking for a really simple answer, we could just declare that playing an infinite amount of turns takes an infinite amount of time...

Last edited by Ari Rahikkala on 16-09-2004 at 17:26

	16-09-2004 13:16									#88
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Oops, you're right. It just tests each time, so it's always 50%. I still tend to think running statistics, my wife is in a doctoral psych program and I've run some of the stats for her. Unfortunately I've never updated my skills, and most of my programming was in Fortran and Cobol which are obsolete for very good reasons. So it's just a simple statistical regression, in which case you're testing for the PB not getting through - the desired result. Which has nothing to do with the infinite case, which I will leave to those very capable people on this thread.

	17-09-2004 08:03									#89
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This brings up infinity:

What exactly IS infinity?

Would it be possible for computers what humans perceive to be infinite as finite?

Is it simply because of the mathematical limitation of our minds?

	17-09-2004 11:49									#90
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quote: Originally posted by Natalinasmpf This brings up infinity: What exactly IS infinity? Would it be possible for computers what humans perceive to be infinite as finite? Is it simply because of the mathematical limitation of our minds?

Infinity is a quantity such that, for any real number n, infinity>n. Also, a set S is infinite if there exists a proper subset of S which has the same cardinality (size) as S.

Infinity is not a real number, so operations that work with real numbers do not work with infinity. This is why we use limits: 1/infinity is not a real number, but we can say that 1/n approaches zero as n approaches infinity.

I'm not sure what you mean by computers "percieving" infinity - computers do not percieve anything. If you mean, "Does infinity exist in reality, or is it just a mathematical tool?", that's a matter of philosophy. Personally, I don't think it matters - if the methods work, there's no reason to doubt the underlying concepts.