Ps, PBs and ODPs

[RedFred]

I have been looking everywhere for my old probability and stats text and I can't seem to find it. So I am hoping somebody can help me out. I seem to have forgotten the basics.

Unless I am sadly mistaken the chance of an PB getting through my ODPs is p to the nth power, where p=.5 (probability of intercept) and n=the number of armed ODPs. Unfortunately I can't figure out the general formula for a situation where there are more than one PBs. Can anyone help me out?

A situation arose in a recent game where another player started building PBs. In this particular game my best option was to build defensive ODPs. With infiltrator status I saw five PBs in production. I built eleven ODPs. I had no clue about the chances that one would get through, but I guessed that about 9 or 10 ODPs would make it even odds. So I built three more ODPs for some extra security.

So I want to be able to plug n=11 and n=14 into a formula where the number of PBs = 5 and see what reduction in chances of a successful strike would be with the three extra ODPs. Beyond that I'd like a general formula so I know how many ODPs to build next time somebody builds PBs to be reasonably assured of stopping them.

Bonus question: Can I treat enemy ODPs as PBs in this formula?In my numerical example above, say the other human also had two ODPs along with his five PBs. Is that the same threat as seven PBs if I have more than seven ODPs?

Since I am asking something I should be offering something as well: My random SMACX factiod of the day is if you pod pop a rush build, but are suddenly forced to change it in the same turn, you lose the entire rush build, not just 50% of the minerals beyond the first row.

Thanks in advance for any help.

[Flubber]

I am no math whiz but it would seem to me that the chance of 5 PBs getting through against 14 ODPs would be the sum of the chances of their individual success. I seem to recall that the ODP is destroyed in killing a PB ( or is it just unavailable as against further strikes ??) so would that not be 0.5 to the 14th plus 0.5 to the 13th plus . . .down to 0.5 to the 10 th.


As for his two ODPs, I would do a probability for the worst case contingency-- both ODPs kill one of yours (probability of this is 25%)

Lets see, worst case is he reduces you to 12 ODPs and therefore using the probabilities of a single PB getting through
against

.5 1 ODP
.25
.125
.0625
.03125 5 ODP
.01562
.00781 7 ODP
.00391
.00195 9ODP
.00097
.00049 11ODP
.00025
.00012
.00006 14 ODP


Its easy to see that a single PB has less than 4 10ths of 1% of getting past 8 ODPs even after the destruction/unavailability of 4 ODPs in killing the first 4 PBs and 2 assumed killed by his ODPs . The aggregate chance is still a good bit less than 1%.

To compare, if you can conceive that he could get you down to 5 ODPs and could then fling 3 PBs at you, he would have have odds of about 21% that ONE of them would succeed. It appears that just 3-4 ODPs are pretty darn good against a single PB but you may want more when the opponent starts stockpiling PBs
.

Right now he has slim odds and I for one would not fire my PBs in that circumstance. But If he has the capability to start an ODP war, that is what I would watch. 5 PBs are not a huge threat now, but if he could match your ODP total then you could end up undefended.


I hope my assumptions were correct and await correction from better mathematicians

[Mongoose]

Been a long time since I fought a PB war, but I don't remember ODPs working quite as you have modeled them. As I recall, each ODP has a 50% of destroying an incoming PB. Succeed or fail, that ODP is set to 'used' or 'fired' status for the remainder of the turn. If the PB is not destroyed by the ODPs in this process, the defender gets an option to destroy the incoming PB by sacrificing a 'fired' ODP.

So, unless my memory is faulty (and that's certainly possible! Seems to be more so with every passing week. ) An attacker cannot penetrate a defense of as many ODPs as he has PBs, as long as the defender is willing to accept the sacrifice of his ODPs as a measure of last resort. (That would usually be the case, I think.)

A mathematical model is only applicable, I think, when the attacker has ODPs of his own with which to reduce the defender's number of ODPs. The relevant question isn't how many ODPs does the defender need to stop X PBs; it's how many ODPs does the attacker need to deliver X PBs. (Hope that makes sense!) Part of the model also must be the level of ODPs each wishes to hold in reserve against either a counterstrike or against a potential third (or fourth!!) participant in the PB orgy!

Nobody has mentioned the inclusion of the flechette defense system. (SMAX only) I'm not even sure they work, as I never got into a PB brawl in SMAX. But, if they DO work as advertised, that adds an additional layer of calculation to the requisite ODP/PB combination for a strike.

Of course, the fun REALLY starts when solar flares destroy everybody's ODP system!

[RedFred]

Mongoose, is it a conceptual problem I am having? I agree that a PB is out of the picture for the whole turn after it has either intercepted or failed to intercept. But I don't recall seeing any options in PBEM to sacrafice my ODP. The strike is going to happen in my opponent's turn. I agree that according to the manual I should have that option to sacrifice my ODP; does it just happen automatically?

Are you suggesting that I could never stop all PBs if my number of ODPs is less and I would always stop all of them if my ODPs are equal or more to his PBs? That wasn't how I thought it worked.

After thinking about it again I also agree that an opponent's ODP cannot be treated the same way as an additional PB. So the answer to my bonus question appears to be "no". Which leads to another perhaps more complex question, what model or formula would work in that instance. Flubber seems to be suggesting that I multiply the probability of each of the conditions where my ODP is defeated (in this case the probability of zero ODP defeats, one ODP defeat as well as the probability of two ODP defeats) with the chance that it will get through under the number of ODPs I have remaining.

We might be getting a bit ahead of ourselves adding in Flechettes as we seem to be in disagreement about how things would work prior to their introduction, but...I tend to have crawlers or other bases close enough to my key bases, yet outside the two square FDS limit, that an opponent could damage or destroy the whole zone. That, combined with their ability to only activate in the base it is built means I view the FDS as fairly useless except as a stopgap defence until ODPs.

I admit I know little about this type of war. I so rarely experience space wars in SP, but I am starting to play a few MP games now. The chances of PBs being launched seems much higher in MP.

[Flubber]

I never thought a single ODP could be an absolute defence against a single PB as suggested by mongoose but I am very inexperienced in PB warfare and was merely commenting earlier on the math.

and Redfred wrt your opponents ODPs if he has 2 of them and attacks with both, I am simply saying there is a 25% chance that both of his will win , 50% chance of a win each and 25% chance that both yours will win in any given 2 battle combat. I just thought for the sake of safety you should assume the worst case.

With respect to the math I provided earlier I think it is an accurate reflection of the odds based on the 50% shootdown ratio. The odds of a specific PB getting through are 0.5 to the power of n where n is the number of unused ODPs remaining to take a shot at that PB. IF there is some option to self-destruct to knock down the PB then your defence will always be absolute for so long as your ODPs outnumber their PBs.

[Mongoose]

RedFred: I don't know how (if?) the sacrifice mechanic works in pbem. No pbem I played in ever used ODPs. Usually, the PB fireworks are in that (overlengthy) interval between Orbital Spaceflight and Self Aware Machines (is that the right tech, even?)

I think you get the sacrifice option in IP Multiplay. I know you do in SP.

[Chowlett]

It should work out (if I'm judging the problem correctly) to be a geometric sum at some point. However, it's too late here to work it out. Maybe tomorrow.

However - the addition thing is definitely wrong, since the events aren't mutually exclusive - ie if PB 1 gets through, that doesn't stop PB 2 getting through. We'll need to take 1 minus the probability for each individual buster, multiply them, and take 1 minus the result. If someone wants to take a crack at it before me then go ahead.

[MariOne]

Correct approach, by one sporting a klein bottle in his avatar you could not expect less.
Without having ANY direct exp in PB-ODP battles, one catch strikes tho my mind at first sight.
You say "probability for each individual buster"
For 1st PB, you can apply a simple formula, as all you care is hit/miss.
To calculate 2nd PB's hit/miss chance tho, you NEED the actual breakdown of HOW the 1st one missed, as each of the single event chances summing up into "1st PB miss" leaves a different chance for blocking 2nd PB.
What I mean offhand is that the formula calculating the hit/miss chance of 2nd PB will not be the same you can use for 1st pb.
Off the top, I can't even think of a synthetic formula, could it be the usual binomial? I'd have to recur to drawing the probability tree, as of now.

Let's take a simple example.

If you have to block 1 PB, here's how it goes.

50% 1st ODP success: PB destroyed, 1 ODP used and STOP
of the remaining 50% in which 1st ODP misses, the surviving PB has 50% of getting hit by 2nd ODP, and 50% of still surviving to face the next ODP
and so on till PB is eventually destroyed or you run out of ODPs

so, to summarise in short, let's assume you have 3 ODP.

50% PB destroyed, 1 ODP used
25% PB destroyed, 2 ODP used
12.5% PB destroyed, 3 ODP used
12.5% PB hits (or 1 used ODP sacrificed), 3 ODP used (or 2 left for next turn)

that's indeed (1 - 0.5)^n where n=3

How would you now calculate the hit/miss chance when you're facing a 2nd PPB in the same turn?
At which value will you set n for the 2nd PB? You can't.

You have
A) 50% chances of being left with 2 available ODP after 1st PB launch
B) 25% chances of being left with 1 available ODP
C) 25% chances of being left with 0 available ODP (regardless of having blocked 1st PB)

A) With 2 available ODP the 2nd PB hit chances are 25%
B) With 1 available ODP the 2nd PB hit chances are 50%
C) With 0 available ODP the 2nd PB hit chances are 100%

This has to be weighed against the integter (100% = certitude = total experiment options) with the above percentages.
That is, the chances BEFORE you launch the 1st PB, for the 2nd PB to hit are:
A) 25% of 50% = 12.5%
PLUS
B) 50% of 25% = 12.5%
PLUS
C) 100% of 25% = 25%
TOTAL 50%

Could THIS chance for the 2nd PB have been easier calculated with a straightforward formula for any number of ODPs? I really can't figure it off the top while I'm posting form work.

And for the 3rd PB you'll have to eventually furtherly split case A) to discriminate whether you're left with 1 or 0 ODP

This tells you when each subsequent PB will hit.
As you first pointed out, these are not mutually exclusive events, INDEED we could have splitted case C), in half of it BOTH PBs had hit.

Actually, and here comes the peculiarity of this problem, ONCE a PB has hit, this means that ALL the ODPs have been fired! This means that ALL the subsequent PBs will find NO ODP to intercept them, thus they will ALL hit.
Of course here I continue talking of PB hit not considering that you can sacrifice a used ODP.

If you want (and IF Mongoose model holds true, I rely on it offhand!), N ODPs can only block N incoming PBs.
If you're lucky, you'll only USE 1 ODP per PB, hitting them all at 1st attempt.
If you're unlucky, you'll fire them all for 1st PB, missing (even if you have a zillion ODPs), and then you'll have to SACRIFICE an ODP for each incoming PB.

So, the matter reduces to:
if they have more PBs, the excess PBs WILL HIT
if you have more or equal ODPs you will surely stop al the PBs; you want to know how many will you have to probably sacrifice, and how many will you probably be able to save for next turn.

Let's split the above A) case
A1) 25% 3 ODP, 2 used 1 available
A2) 12.5% 3 ODP, all used
A3) 12.5% 2 ODP (2nd PB hit), all used

B1) 12.5% 3 ODP, all used
B2) 12.5% 2 ODP, all used

C) 12.5% 2 ODP, all used (12.5 % 1st PB killed by 3rd ODP, 100% 2nd PB hit)

D) 12.5% 1 ODP, used (12.5% 1st PB hit, 100% 2nd PB hit)

If you sum this up:
3 ODP facing 2 PB, before battle starts

25% no PBhit/ODPloss, and 1 av.ODP (for 3rd ev.PB)
25% no PBhit/ODPloss, no defenses left for 3rd ev.PB (A2+B1)
37.5% 1 PBhit/ODPlost (and also of course no further defense left)
12.5% 2 PBhits/ODPlost "

It will "jump to your eye" (italian way of telling!) that when 1st PB hits, 2nd PB will ALSO, CERTAINLY hit.
Thus the hit % of 1st PB is INCLUDED in 2nd PB's 50% chance to hit.
That is, not only the PBhit events are not mutuallly exclusive, they're... subsequently *inclusive*, as for sure the subsequent PBs will have to face (at least) 1 less ODP than their predecessor.

This makes a bit hard to apply the common, comfortable, no-brainer probabilistic patterns...

___


About the "ODP sacrifice - PBEM" issue, I have not the slightest idea about hiw it was resolved.
But I can tell for sure that this is the SAME problem pattern of PBEM implementation, and I've been hammering on this concept for so long.

SMAC is designed to be a *simultaneously* played Single Player game.
This reflects in Elections, in withdrawal requests, in probing reactions, to name few that come to my mind.
When a 2nd faction is *involved in resolving* an action of the player, that faction IS online although it's not its turn to play, because all the other factions are played by the AI, which is always online by definition.

Pbem is SEQUENTIAL, and I mean in the physical sense of how the flow of play is implemented.
When you probe another human pbem player, he is NOT online to decide his reaction on the fly.
When you ask another human pbem player to withdraw using the commlink menu, he is NOT online to decide his reaction on the fly.
When you ask or offer a bribe to another human pbem player during YOUR election vote, he is NOT online to decide his reaction on the fly.

So, when you launch a PB against another human pbem player, he is NOT online to decide whether he wants to sacrifice one of his ODPs or he prefers being hit by the PB.

Considering how FurXs approached the cases above they might have:
- left your PB hanging in the air till the turn gets to the target????? I doubt even FurXs could choose this! A PB launch MUST be resolved on the launch turn, or not?
- prompted the PB launcher if he wanted to let the target proctect himself against his PB or not (!)
- always sacrifice the ODP
- always save the ODP and let the PB hit
- let the game engine decide, applying the AI algorythms (considering for instance the worth of the target???)
- take a *random* decision

I have never needed to launch a PB against the AI (except in tests), and as Mongoose said all my pbem PB experiences were long before any ODP could be around, so I can't say which of the above applies.
I'd be indeed very interested if some player could report such firsthand pbem experience.

__
EDIT: corrected "has been weighed" in "has to be weighed"!

[Chowlett]

It looks possible to analyse, I might do so later on. I'm "working" at the moment.

[MariOne]

yeah, a little more thought seems yielding some promise.

For now I can tell RedFred that the probability of the 2nd PB being destroyed by the Nth ODP overall (that is summing also those deployed to *succesfully* destroy 1st PB TOO) is:

(N-1) p^N

I mean, *exactly* by the Nth ODP overall.
If you want to know *within* the Nth ODP overall you'll have to cumulate (from N=2, of course).

I'm "working" at work on elaborating it for 3, 4, X PBs

__

I'm heavly relying in these formulas on our special case condition that p=(1-p), being it 0.5
It would be a true PIA otherwise reducing all in such a slim form.
I would dare to tell that YES, you can use the binomial for the chance above

that is, the chance of the 3rd PB to be destroyed exactly by the 5th OdP, is

(3 5) p^5

where with (3 5) I here intend to symbolise the 3rd number in the 5th row of this famous table (Tartaglia's triangle)

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...

that is, 6

I will check it further

[Chowlett]

Interesting - we know it as Pascal's triangle.

[MariOne]

Actually, it's a reminiscence from italian school

If you search on the web, you find some interesting links.
Mainly biographical like these:

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Tartaglia.html

http://es.rice.edu/ES/humsoc/Galileo/Catalog/Files/tartalia.html

He's more famous for inventing the soultion of cubic equations, but one page cites also his interest in combinatorial analysis.
A site in italian quoted "Pascal-Tartaglia triangle", and it can be found here wiht only Tartaglia's name

http://www.ticalc.org/archives/files/fileinfo/179/17909.html

I can't say wheter the triangle was disputed beween the two, or if Tartaglia gave some contribution or particualr application to it and italian math schooltext elected to enhance tha patriotic reference.
Of course I know that it's (also) Pascal's and as such is internationally reknown, but when I think to it, I don't know why I never recall the french philosopher's name, but only the more picturesque one of my compatriot

Anyway, I don't know the english name, but the "combinatorial binomial" notation (3 5), where the numbers are vertically stacked on the books, actually gives 10. That is indeed the # of possible combinations of 3 elements out of 5 (ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE).
So, in the tabe rows and colums shouls be numnered starting from 0...

For our purpose, the probability of the Xth PB being hit exactly by the Nth ODP overall must then be more accurately expressed:

( (X-1) (N-1) ) 0.5^N

That "combination" function should be found in Excel, since older versions too, I presume.
So it should be easy for anyone (provided with Excel or as powerful spreadsheet) build his own table.

[MariOne]

If the above holds true, then...

against 5 PBs
the cumulated probability of hitting them all with

11 ODPs is 72.56%
12 ODPs is 80.62%
13 ODPs is 86.66%
14 ODPs is 91.02%

... just to refer to RedFred's example

You see that even building 11 ODPs, you still have 27.44% chance of being hit by *one or more* PBs, that's pretty high considering the investment done.
But add just 3 more, and that chance drops down to 8.98%

Ah, all that brain for something that I will 99.9999999% NEVER have the ocasion to apply in any pbem!!!

[Capt Dizle]

Yes, Mr. PlanetBuster, I think it ironic that you would be discussing PB defense.

[RedFred]

Thanks for everyone's insight. I could mostly follow everything until MariOne's second to last post. Sadly, this seems to be the key one.

If 'combinatorial binomial' is what I think it is, I seem to have a hazy memory of a formula for it which involves factorials. No?

BTW, in my PBEM, I had a bit of a tech lead so the other player did not have ODPs. The PBs were built as a last ditch attempt to thwart my Transcendance. He wound up launching two PBs at me without success. Two PBs were 'melted down' to build the Transcendance SP just after me just in case I messed up and my base went into drone riot and failed to build the SP that turn. One PB was apparently retained.

All 14 of my ODPs showed up as 'online' the following turn, contrary to my expectations. Now I really regret wiping the game from my PC. It would have been most interesting to continue to play after transcendance and allow him to build 15+ PBs to launch at me as an experiment.