Zeno's Paradox and Flechette Defense

[smacksim]

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.

[Chaos Theory]

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.

[Natalinasmpf]

*scratches head*

0.29?

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

[Ari Rahikkala]

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.

[Natalinasmpf]

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?

[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.

[Natalinasmpf]

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)

[Dry]

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.

[#endgame]

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.

[Chaos Theory]

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.

[GeoModder]

You guys still feel well in those Ivory Towers?

[Ari Rahikkala]

More like embarrassed, I'd say. These problems shouldn't be this difficult...

[Chaos Theory]

I like the view.

[smacksim]

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.

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.

[Natalinasmpf]

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.