09-09-2004 23:32									#31
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Ari, I'm no mathematician, but could one not also prove the opposite and have the equations cancel? I mean, there's a 50% chance that an ODP will fail, but there is a 50% chance that it'll succeed. As we approach infinity we can prove that of infinite ODPs, one will succeed in preventing a PB, just as we can prove that one PB will always succeed, no?

If there are infinite chances of failure and infinite chances of success, the event will always succeed and fail, and thus a conclusion is not meaningful. Or can you demonstrate that this line of thought is false?

Chaos, thanks for the F4 buttons. I'd always thought they just showed the 'best base' in each category, and now see that the whole list is re-ordered. Very nice. Smac is laden with buttons hidden everywhere. The overlay/report system from RRTycoon 2 or Tropico would be nice though, and as complex and changeable as the governor system is, its not quite good enough....I was thinking that for SMAC 2 what we really need is a scripting language to sit on top of the game, allowing control of everything from what is displayed to AI behavior, to governorships. Then it would be easy to write bots for the AI, not to mention produce better build orders and governors and reports.

	10-09-2004 11:52									#32
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Simple way:

A PB passing through an ODP happens .5 of the time.
A PB passing through n ODPs happens .5^n of the time, since each event is independent. That is, the probability that a PB will pass through an ODP is .5 regardless of what happened before.

Now the Hive vs Drone thing:

Surviving a given turn happens with probability P(t).
P(t) = 1 - .5^t

Surviving all the turns from 1 through n happens with probability equal to the product of P(1) through P(n).

Multiplying out, this becomes
(2^(2n) - 2^(2n-1) - 2^(2n-2) + 2^(2n-3) - 2^(2n-4) + ...) / 2^(2n)
though I am somewhat suspicious of this expansion. However, I am too tired to figure out what might be wrong with it.

Each power of 2 is represented in the numerator only once, but with a + or - depending on which terms needed to be multiplied to produce it. This fraction is therefore greater than

(2^(2n) - 2^(2n-1) - 2^(2n-2) + 2^(2n-3) - 2^(2n-4) - 2^(2n-4)) / 2^(2n)

and is therefore some number greater than 1/4. Tests on a calculator show that it is smaller than .29.

I'm not sure how to actually sum this series, so I've settled for this.

	10-09-2004 12:12									#33
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*scratches head*

0.29?

Hmm, I thought the probability would either be 1 or zero.

	10-09-2004 12:51									#34
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Thanks, Chaos. That's the part that I couldn't figure out... how to calculate the probability of surviving all the turns from 1 through n when n approaches infinity..

Natalinasmpf: It wouldn't make sense for the probability of survival to be 1, because it drops to 0.5 on the very first turn. It would also sound strange for it to be 0, because the series of probabilities of hitting converges.

	10-09-2004 12:59									#35
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Well I mean 1 out n. Slight chance, very very slim.

Or 0 out of n, no chance.

Isn't the whole point is whether there's a chance of surviving or not?

	10-09-2004 13:25									#36
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Well I don't see how the probability of a single PB considered infinite times could have anything but zero chance of survival. It is a limit and y-->0. If the denominator isn't increasing to infinity, there is something off with the equation methinks.

The interesting thing to me though is that there are infinite PB's in Natalina's proposition. Thus, as n-->infinity, survival-->zero AND as n-->infinity, success-->1, which can be represented as:

1/2^n (chance of failure)
1/2^n (chance of success)

These produce the result 0==1, or NaN, or merely 0/0, depending on how you like to have an equation commit seppuku.

To summarize: Given infinite Planet Busters and infinite ODPs over infinite time, one Planet Buster will eventually get through the defenses, and the ODP screen will never fail, always blocking all Planet Busters.

	10-09-2004 14:49									#37
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Well, there's a CHANCE the ODP screen will never fail, saving Free Drone Central for all eternity (while the remaining garrison at Free Drone Central will have enough time as things progress on to equip themselves with string disruptors and stasis generators to finally attack Plex Anthill ) - but this chance has to be calculated.

There's also a chance a single Planet Buster might break through all ODP defenses (if there are infinite), as well....but the scenario Ari gave us, Free Drone Central doesn't really have infinite ODP's...and having infinite something of matter (PB's) is more complex to imagine then of time....(mission years)

	10-09-2004 15:15									#38
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If your problem is simply an infinite (not determined) number of busters vs an infinite (not dtermined) number of ODP, then the result is
zero (prob.) times infinite (#bust) = undetermined.

Now, a more interesting problem would be the SAME infinite number of busters vs the SAME infinite number of ODPs.
Like each base would have been building exactly 1 improvement per turn, resulting each turn N busters for Drone City, N ODPs for Banana City.
Nothing is launched, until turn Z. Drone City launch his Z busters against Banana City protected by Z ODPs.

Probabilty of Banana's survival is (if I'm not mistaken):
((2^z - 1) / (2^z))^z
So the question is: is this converging? To what?
Looks to me that it is growing with z, but even that, I have no proof.
My math are soooo far away.

	10-09-2004 16:44									#39
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I think the designers would know what they were doing and have realised that you can't have 1/2 of an ODP, or -1 ODPs thus making the amount of ODPs you can have most likely an unsigned int. now is it short, normal or long? either way you will hit a point where x+1<x.

Last edited by #endgame on 10-09-2004 at 16:58

	10-09-2004 18:20									#40
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quote: Originally posted by smacksim Well I don't see how the probability of a single PB considered infinite times could have anything but zero chance of survival. It is a limit and y-->0. If the denominator isn't increasing to infinity, there is something off with the equation methinks. The interesting thing to me though is that there are infinite PB's in Natalina's proposition. Thus, as n-->infinity, survival-->zero AND as n-->infinity, success-->1, which can be represented as: 1/2^n (chance of failure) 1/2^n (chance of success) These produce the result 0==1, or NaN, or merely 0/0, depending on how you like to have an equation commit seppuku. To summarize: Given infinite Planet Busters and infinite ODPs over infinite time, one Planet Buster will eventually get through the defenses, and the ODP screen will never fail, always blocking all Planet Busters.

Absolutely not! You fail if any one PB gets through, not if all of them get through! Furthermore, failure and success are mutually exclusive, and there is no third option. Therefore, whatever the chance of success (equal to the odds that every PB is blocked; relatively easy to compute), the chance of failure is 1 - that.

The catch when considering an infinite number of PBs is that the odds a PB will get through decrease dramatically with time, so that the product of all the chances of survival probably converges, to some number around .288. If you had an infinite string of PBs against a fixed number of ODPs, failure *would* be inevitable, but that's not the case.


As far as x + 1 < x, that's an implementation detail, a flaw, that has nothing to do with the math we're fiddling with.


N PBs vs N ODPs on a single turn, N -> infinity:

Odds that 1 is blocked = 1 - .5^N
Odds that all N are blocked = (1 - .5^N)^N

That reminds me a lot of the expression for e, (1 + 1/N)^N. 1/e = (1 - 1/N)^N. However, .5^N decreases much, much faster than 1/N. Therefore, I suspect the odds that N are blocked converges to 1.

	10-09-2004 21:09									#41
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You guys still feel well in those Ivory Towers?

	10-09-2004 21:26									#42
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More like embarrassed, I'd say. These problems shouldn't be this difficult...

	10-09-2004 21:58									#43
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I like the view.

	11-09-2004 00:46									#44
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quote: Originally posted by Ari Rahikkala More like embarrassed, I'd say. These problems shouldn't be this difficult...

I agree. This is basic calculus. The problem is that we are dealing with the qualities of two infinities. We are always told to treat infinity in certain ways in math. I always pressed my professors on that. What's the point of studying a system of logic if at some point you are told to take something as true without logical proof?

Anyways....I can definately see your point CT. If any one PB gets through, boom, and it doesn't work in reverse, ie, if any ODP blocks, then fizzle............or does it?

I think, if I'm correct, the difficulty framing the problem is conceptual. Yes, any PBuster that gets through wins. But do look at the reverse of that situation: For any PB there are infinite ODPs to block it. The chance of it being blocked is 1/1.

quote: Furthermore, failure and success are mutually exclusive, and there is no third option. Therefore, whatever the chance of success (equal to the odds that every PB is blocked; relatively easy to compute), the chance of failure is 1 - that.

Can you agree that there are some problems, even in statistics, that produce undefined answers? Like x/0, or infinity/infinity? Failure and success are mutually exclusive in a system where the results are already known and in the past. But consider quantuum physics for a second. The chance of a particle being at X location and Y speed is meaningless because X and Y are probabilites that must interact. This is a similar case. Failure and success are not only mutually exclusive, they are indeterminate. This is because you must solve for two infinities and have them agree. That is simply not possible.

This is one of them! You must consider both infinities to solve the problem. You cannot walk away from it because consideration of one leads to a real number, and consideration of the other leads to a real number. These must 'add up', which they do not.

Now I'm sure I'm missing something.

	11-09-2004 01:39									#45
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Hmm, I thought there were three different scenarios, but I began to see the point (of only there being one):

I had actually thought that after a PB was launched and failed, the corresponding ODP would go down. Oops....thus, I was thinking, is there a slim chance the Drones can survive every attack? Ie. the chance of the Drones surviving the first attack is 50%, if they do not sacrifice a pod (if they did this would become much more complicated...hmm), second attack 25%....third, 12.5%, till I realised that successful ODP's accumulate, so I could see the probability eventually curve and decelerate (hyperbolar?)...hmm!

Hmm, could we consider hyperbolars here?

There of course is the possibility of a PB working on turn 1 against infinite ODP's, and 1 PB working against infinite ODP's on turn 1.

I wonder if one of us should engage in a TCP/IP and try this out.

	11-09-2004 03:46									#46
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quote: Originally posted by smacksim I agree. This is basic calculus. The problem is that we are dealing with the qualities of two infinities. We are always told to treat infinity in certain ways in math. I always pressed my professors on that. What's the point of studying a system of logic if at some point you are told to take something as true without logical proof? Anyways....I can definately see your point CT. If any one PB gets through, boom, and it doesn't work in reverse, ie, if any ODP blocks, then fizzle............or does it? I think, if I'm correct, the difficulty framing the problem is conceptual. Yes, any PBuster that gets through wins. But do look at the reverse of that situation: For any PB there are infinite ODPs to block it. The chance of it being blocked is 1/1.

Yes, the odds that 1 PB will get through infinity ODPs is 0, and the odds that infinity PBs will get through 1 ODP is 1. However, infinity is not a real number and does not obey the same laws of math real numbers do. If you want to know whether infinity PBs will get through infinity ODPs, you need more information. How many PBs do you have relative to your ODP count? Are you launching them all at once or over infinity turns? You need to replace infinity with a real number that tends towards infinity according to some formula to make any sense out of this situation.

quote: Can you agree that there are some problems, even in statistics, that produce undefined answers? Like x/0, or infinity/infinity?

Division by a literal zero is undefined. Division by a number that tends towards zero makes perfect sense. Same for all operations performed on infinity.

quote: Failure and success are mutually exclusive in a system where the results are already known and in the past. But consider quantuum physics for a second. The chance of a particle being at X location and Y speed is meaningless because X and Y are probabilites that must interact. This is a similar case. Failure and success are not only mutually exclusive, they are indeterminate. This is because you must solve for two infinities and have them agree. That is simply not possible.

We aren't working quantum physics, but even there the wave function integrates to 1 over all space - if you want to think of it as representing the probability that a particle is at a particular place, then all points in space are mutually exclusive. If you really want to work quantum mechanics, know that conventional concepts of location and speed (or momentum) are insufficient.

quote: This is one of them! You must consider both infinities to solve the problem. You cannot walk away from it because consideration of one leads to a real number, and consideration of the other leads to a real number. These must 'add up', which they do not.

Not at all! This is a simple mathematical limit. The limit may be 0, infinity, .578, or may not exist at all (which is provably not the case for the problems we've looked at in this thread) but you can solve the problem by extricating infinities. Even when a limit doesn't exist, you can characterize the limiting behavior of the series.

quote: Now I'm sure I'm missing something.

Probably a good helping of graduate-level math!

	11-09-2004 05:56									#47
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I disagree, obviously. We need another math voice in here....

Another interesting thing is that it doesn't matter what the probability of an ODP working might be, so long as it is between zero and one. As long as there are infinite ODPs and PBs, the result is the same: At least one PB will get through, and at least one ODP will manage to stop an inbound PB........== nonsense.

For instance, if the chance of an ODP stopping a nuke is 99%, then 1/100 PBs will get through the first one. 1/100 of those survivors will get through the next one. The actual chances don't matter at all, so long as we have infinite supplies on both sides.

--------------------------------

To respond to some of your statements Chaos:

quote: However, infinity is not a real number and does not obey the same laws of math real numbers do.

Infinity is extremely useful in limits, obviously. This is how we can accept results of zero or one for an infinite series. It might not obey the same laws as other numbers, but that makes it more useful. I don't think I'm using it incorrectly, so what are you saying?

quote: If you want to know whether infinity PBs will get through infinity ODPs, you need more information. How many PBs do you have relative to your ODP count? Are you launching them all at once or over infinity turns? You need to replace infinity with a real number that tends towards infinity according to some formula to make any sense out of this situation.

When dealing with infinity you need exactly zero additional information about quantities. If you have 10*infinity PBs and just infinity ODPs, it's the same problem. The calculation may happen in a series if you like, but those initial 10 PBs don't end up in virgin space just because some make it through the first ODP. There are infinite ODPs.

Now if we're talking about the build-up to infinity, say when one side has 500 ODPs and the other has 500 PBs, then yes, it does matter if it's actually 501 to 499 or something, because if they launch that turn, then those are the odds. But as long as we assume no launch until there is an infinite supply (which would take infinite turns to produce, but assuming we are gifted infinite weapons), then it doesn't matter at all how many ODPs we had last turn. If we have infinite now, then we have achieved the impossibly non-sensical situation of having perfect defense against guaranteed offese. There can be no result of that war.

We don't need to replace infinity with a real number to make sense of the situation. There is no way to make sense of it, no matter how we sneak up on it.

quote: If you really want to work quantum mechanics, know that conventional concepts of location and speed (or momentum) are insufficient.

Which is why I brought it up. Infinite series can behave in a similar fashion, as we've demonstrated. Common sense says that there must be a winner and a loser in this coin-toss. But we can't know if there is a winner and a loser, only if there is a winner. Since we must determine if the ODPs and PBs both 'win' in the same formula, we discover that that is not possible. That's the pain of infinity on the noggin. Well, at least it makes me crosseyed to try and imagine I recognize that you disagree, but I don't see proof. Zeno was right: it is a paradox.

quote: Probably a good helping of graduate-level math!

Please demonstrate.

	11-09-2004 22:41									#48
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If you simply have infinity ODPs and PBs, then your odds of survival are indeterminate. Nothing more can be said.

You *cannot* use infinity in mathematical operations as if it were a real number. Trying to do so leads to some of the garbage results you see. Replace it with a parameter that you can vary, and observe the limit as that parameter approaches infinity. Surreal math can deal with infinities, but I don't know surreal math.

If you build up to infinity ODPs and PBs at the rate of 1 per turn, then fire all the PBs, then your odds of survival are well-defined and can, at least in theory, be calculated. To calculate them, replace infinity by N, and let N go to infinity. As I showed in an earlier post, the odds of survival become

(1 - .5^N)^N

I speculated that, based on similarities to the equation for 1/e, that this number tends towards 1 as N tends towards infinity. That's your chance of survival; the result of that war.

quote: We don't need to replace infinity with a real number to make sense of the situation. There is no way to make sense of it, no matter how we sneak up on it.

What about my result? I made sense of the situation by replacing the infinities.

For graduate-level math, I'd say start with advanced calculus. You go back and work out calculus and why it works, dealing with infinite sequences and series at the same time.

	11-09-2004 23:38									#49
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quote: If you simply have infinity ODPs and PBs, then your odds of survival are indeterminate. Nothing more can be said.

The thing is, how many PB's there are at one moment, can affect how many ODP's there are, hence probability.

The more PB's fail, the more ODP's are built. The moment a PB is sucessful, ODP production stops. Thus, the more PB's don't become successful, the higher chance of it failing the next turn. The thing is: where exactly does this chance stop advancing?

Of course I may be repeating the obvious, but sometimes that helps - reiteration.

	12-09-2004 07:30									#50
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To me this is not a paradox at all although it seems like one at first read. The math on this page is very impressive, though I don’t understand most of it, I don’t think it’s really a math problem. It’s a logic problem in my opinion. I see no need for long or complicated formulas at all to answer this question.

A PB coming in to any base comes in ONE at a time. If you have an infinite amount of ODP’s, there is no way the PB will ever get through. How could it be anything else? Each time it would be 1 PB against an endless supply of ODP’s. No matter how many PB’s you have, they only have to be dealt with one at a time.

Anyway, that the way I think, right or wrong. I must say that I’m impressed with all the math, though most of it’s over my head

	12-09-2004 11:48									#51
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It is possible to have sufficiently many more PBs than ODPs, even as both approach infinity. Here's how to find out how many more you need:

Odds of landing a PB against N ODPs: .5^N
Odds of landing 1 or more of M PBs against N ODPs: 1 - (1 - .5^N)^M

The idea is to find N and M in terms of some other variable, t, such that both go to infinity, but (1 - .5^N)^M does not tend towards 1.

Define N = (log t)/(log 2)

(1 - .5^N)^M = (1 - 1/t)^M

Now define M = t

= (1 - 1/t)^t

The limit of this expression as t -> infinity is exactly 1/e. This would work just as well for the following definitions of N and M:

N = t
M = 2^t

For N = t^p for any fixed, positive p, the odds of getting at least 1 PB through will still tend towards a non-zero number.

Therefore, to maintain a non-zero probability of getting a PB through ODPs, given infinite numbers of each, the number of PBs must rise exponentially and the number of ODPs must rise slower than exponentially (such as polynomially). Still, given infinite PBs and infinite ODPs, PBs can still get through with non-zero probability.

	12-09-2004 13:19									#52
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quote: A PB coming in to any base comes in ONE at a time. If you have an infinite amount of ODP’s, there is no way the PB will ever get through. How could it be anything else? Each time it would be 1 PB against an endless supply of ODP’s. No matter how many PB’s you have, they only have to be dealt with one at a time.

Zeno's paradox of ODP's:

First ODP halves PB's success to 25%. Second ODP halves a PB's success to 50%, 25%, 12.5%, 6.25%, and so on....but it will NEVER halve a PB's success to zero, because an ODP only divides it by half. Halve infinitely, there is no end. Therefore at least 1 PB out of infinity will be successful.



In order to prove this not true, you need "complicated math", which isn't really complicated IMO.

Anybody thought about e? 2.71828183? In relation to the probability of the drones surviving.....I thought it curious how the probability falls between 25 and 29 or something. A very wild link though. I just thought of it...I haven't even used e in school yet.

quote: It’s a logic problem in my opinion. I see no need for long or complicated formulas at all to answer this question.

Even logical statements need formulas. p

http://en.wikipedia.org/wiki/Falsifiability

	12-09-2004 20:52									#53
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quote: First ODP halves PB's success to 25%. Second ODP halves a PB's success to 50%, 25%, 12.5%, 6.25%, and so on....but it will NEVER halve a PB's success to zero, because an ODP only divides it by half. Halve infinitely, there is no end. Therefore at least 1 PB out of infinity will be successful.

I disagree. Each ODP has a 50% chance of success. If the first ODP fails, the next ODP still has a 50/50 chance of success and so on. Sooner or later the PB is stopped.

Further more. If your really talking about a infinite number of ODP's it doesn't matter what the percent to hit ratio is. It could be anything. If it only had a 1% chance of success, sooner or later you're going to get that 1% and down goes the PB.

More like the laws of probability than a paradox.

	12-09-2004 21:09									#54
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quote: I disagree. Each ODP has a 50% chance of success. If the first ODP fails, the next ODP still has a 50/50 chance of success and so on. Sooner or later the PB is stopped.

Yes each ODP has a 50% chance of success, but a PB has a 25% chance of avoiding both.

It has a 0.5^100000000% chance (which is very very small, but still a chance) to pass through 1,000,000,000 ODP's.

If you have infinite PB's, eventually one will get through, even if it takes googleplex years.

quote: If it only had a 1% chance of success, sooner or later you're going to get that 1% and down goes the PB.

Yes but you have infinite PB's.

And ODP's have a way higher chance, it will OFTEN take it down, but it will NEVER take all of it down, or is it? It has to be resolved it won't. Or because the chance has to be calculated as well.

quote: More like the laws of probability than a paradox.

See title post. If you walk 1 foot, firstly, you have to cross half a foot, then a half of that half, then cross half of the half of that half, and this gos on forever. You will only achieve 1/2 + 1/4 + 1/8 + 1/16, but never 1 foot!

Says Zeno paradox!

Now, here is percentage. To achieve 0% chance of a PB hitting, firstly you have to build one ODP, then two, than three, but this is only 100% - 50% - 25% - 12.5%, but you will never have enough ODP's to reach 0%.

There's NO chance the Drones will survive forever, for all eternity. Free Drone Central will eventually be destroyed, even if they build a googleplex ODP's, or it takes a googleplex mission years.

Or is it?

	12-09-2004 21:38									#55
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Ah, but there is a chance the Drones will survive forever. I showed in an earlier post that, if the Hive launches 1 PB/turn foreve while the Drones build 1 ODP/turn, the Drones have a ~.28 chance of surviving forever. You can feel free to construct other scenarios and I can examine the odds for those.

	12-09-2004 23:06									#56
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quote: Originally posted by Chaos Theory Ah, but there is a chance the Drones will survive forever. I showed in an earlier post that, if the Hive launches 1 PB/turn foreve while the Drones build 1 ODP/turn, the Drones have a ~.28 chance of surviving forever. You can feel free to construct other scenarios and I can examine the odds for those.

I know, but it requires disproving Zeno's paradox (that infinite ODP's will eventually get probability of any PB (after infinite has been launched) hitting down to zero)....which fender doesn't think is necessary....

That 0.27+% of the time in this scenario, by the time say, infinite PB's have been launched, the probability of them hitting is so low, then it buys the Drones time to eventually reach infinite ODP's and clinch 0% probability of the next PB hitting (after say, a lot have been launched).

	12-09-2004 23:50									#57
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Nit: .28, not .28%. I dislike percents and generally express probabilities as numbers, and never mean percent when I don't use a % sign.

	12-09-2004 23:57									#58
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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.

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.

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.

	13-09-2004 00:57									#59
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quote: Originally posted by Chaos Theory Ah, but there is a chance the Drones will survive forever. I showed in an earlier post that, if the Hive launches 1 PB/turn foreve while the Drones build 1 ODP/turn, the Drones have a ~.28 chance of surviving forever. You can feel free to construct other scenarios and I can examine the odds for those.

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

	13-09-2004 02:54									#60
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quote: Originally posted by fender 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.

And at the same time, with infinite PBs, one will get through, no matter what. To look at the obverse: As infinite PBs deal with each ODP, one is guaranteed to make it past that ODP. Applied to the set of all ODPs, and one PB will make it through. That's what makes the concept of infinity so crazy in this case. One PB will make it, and no PBs will make it.