
28 Dec 2005, 21:06  #1 (permalink) 
Kroot Shaper
Join Date: Dec 2005
Posts: 82

forks guide to Theoryhammer (v2.0)
[size=16pt]fork’s guide to Theoryhammer v2.0[/size]
or common mistakes involving statistics There are a lot of misconceptions considering the use of statistics as a tool to measure the efficiency of a unit or configuration. This little guide will try to sort out the worst of these delusions. And yeah, it’s pretty basic and this is not my native language – bear with me. [size=12pt]1. The common error[/size] I think where most mistakes are made is when we separate the probability of succeeding with a single event and the joint probability of several events. Say for an example that your Fire Warriors have a 60% chance to kill a target with its Pulse Rifle. They question then would be how many times that warrior must shoot in order to guarantee a kill. It might be tempting to add probabilities together and say that a second shot would yield a 120% chance (60% + 60%) to kill the target, thus ensuring a kill. Two things should bother you with that conclusion. The first is that all probabilities lies between 01 (0100%) where 0 means that the event always fail/never occur and 1 means that the event never fails/always occur. 120% is clearly more than 100% and I can make no sense out of the concept that 120% would mean that the action occurs more often than always. The second thing is that since we know that there is a chance to fail each roll there is a chance that we roll horribly and fail to wound with both of them. Since we know that there is a chance to fail, what then does a 120% chance to succeed really mean? The answer is pretty much nothing, but I will try to sort this mess out. F= Failure S= Success Take a moment to look at the table above. It shows the earlier described situation. First you fire a shot and have a probability of 0.6 to succeed. After that you fire a second shot with the same probability of success. Given this layout we have four possible outcomes: 1. F first shot and F second shot = Chance to receive zero wounds = chance for target to survive 2. F first shot and S second shot = Chance to receive one wound from the second shot 3. S first shot and F second shot = Chance to receive one wound from the first shot 4. S first shot and S second shot = Chance to receive one wound from each shot (overkill... but its a success nevertheless) I will use a simple form of notation during this text. “The probability of succeeding with a shot is 60%” would be written as P(S)=0.6. “The probability of succeeding to wound on the first and then failing to wound on the second” would be written as P(SF). As we already seen above P(SF) is not the same as P(S)+P(F)=0.6+0.4=1. This is just wrong just like our earlier attempt. Instead the golden rule is to multiply all joint probabilities. P(SF)=P(S)*P(F)=0.6*0.4=0.24. We can now calculate the probabilities for all of our possible outcomes: P(FF) = 0.16 = P(target survives) P(FS) = 0.24 = P(inflicting 1 W on first shot) P(SF) = 0.24 = P(inflicting 1 W on second shot) P(SS) = 0.36 = P(inflicting 2 W) Since these events are mutually exclusive (meaning that one of them have to occur and no two of them can occur at the same time) the sum of the probabilities of all possible outcomes must be 1. 0.16+0.24+0.24+0.36=1 so we are all fine so far. Joint probabilities should be multiplied and not added together. [size=12pt]2 A bit of elaboration[/size] A second common error is one of picking the wrong probabilities to multiply. In our case we wanted to find out how great our chances are to kill the target. Our chance of success with each shot was 60%, P(S)=0.6 and we figured two shots would do then. P(S) * P(S) = P(SS) = 0.36. This then seems like we actually have less chance to kill the target with two shots rather than one since P(S)>P(SS). But P(SS) is of course the probability to inflict one wound with each weapon and we only need to inflict one. Take a look again at our probability tree above again. If the chance if success is 0.6 the chance of failure must be 0.4 (10.6=0.4). Since we only need one wound we can use all outcomes with at least one S in them together. So each of P(FS), P(SF) and P(SS) will be a success and only P(FF) is a failure. Since both the middle branches of the tree above have the same effect we combine them into one and get: P(FF) = 0.16 = P(target survives) P(FS) + P(SF) = 0.24 + 0.24 = 0.48 = P(inflicting 1 W) P(SS) = 0.36 = P(inflicting 2 W) Since we only have one target we are only interested if we wound at all: P(FF) = 0.16 = P(target survives) P(FS) + P(SF) + P(SS) = 0.24 + 0.24 + 0.36 = 0.84 = P(inflicting at least 1 W) So in other words: P(target survives) = 16% P(target dies) = 84% Our total chance of success is then the sum of all the individual branches in the tree that inflicts at least one wound = P(FS) + P(SF) + P(SS) = 0.84 = 84%, so still a 16% chance of failure. The good thing with understanding this process is that it could be simplified a lot. As your calculations become more and more entangled, perhaps involving entire units of Warriors you may end up with several hundreds of branches. You don’t want to calculate all of them. In the example above we can use the fact that all of the branches are mutually exclusive. Since P(FF) + P(FS) + P(SF) + P(SS) = 1, then P(FS) + P(SF) + P(SS) = 1 – P(FF). This way we only need to calculate the risk of us not inflicting a single hit and then subtract that probability from 1 to get the probability to inflict at least one hit. The beautiful part is that this could be used for any number of shots. Say that we want to calculate the chance of a twelve man strong unit of Fire Warriors killing a single target, each having a 60% chance of inflicting a wound. All we need to do then is to calculate the risk of them all failing. S(FFFFFFFFFF)=S(F)^10=0.4^10=1.05*10^3 which translates into a pretty good 99.99% chance to kill that poor target. [size=12pt]3 Some examples[/size] Lets calculate some actual probabilities. How great is the chance that a single Fire Warrior with a Pulse Rifle manages to inflict a wound on a Marine EQuivalent target within 12”? First of all we know that: P(Scoring hit with BS3) = 1/2 P(S5 inflicting wound on T4) = 4/6 P(Making 3+ Sv) = 4/6 In this case the easiest way to calculate the joint probability is to multiply the different cases we would consider success, since we need to succeed on all three roll to inflict a wound. P(Scoring hit with BS3) * P(S5 inflicting wound on T4) * P(NOT making 3+ Sv) = P(Scoring hit with BS3) * P(S5 inflicting wound on T4) * (1P(Making 3+ Sv)) = 1/2*4/6*(14/6)=3/6*4/6*2/6=1/9 P(Wounding MEQ)=1/9. But our guns are Rapid Fire ones. So we actually make two shots with each gun. Again we do not add them up so that P(Wounding MEQ w. Rapid Fire) would equal P(Wounding MEQ) + P(Wounding MEQ). Rather we calculate the complementary event to the risk of failing both shots: P(Failing to Wound) = (1P(Wounding MEQ)) * (1P(Wounding MEQ)) P(Inflicting at least one W on MEQ with RF) = 1 – ((1P(Wounding MEQ)) * (1P(Wounding MEQ))) = 1( (1(1/9) ) * (1  (1/9) ) ) = 1 ( (8/9) * (8/9) ) = 17/81 So the probability for a single Fire Warrior to bring down a marine is about 21%. Take care to understand that we have 21% chance to inflict at least one wound. This means that the same Fire Warrior doesn’t have the same chance to inflict exactly one wound on a two strong MEQ squad, which brings us into the next part where we handle the chance to bring down entire units. [size=12pt]4. The formula[/size] ??? (skip if you’re REALLY not interested) Since the probability for a single Fire Warrior to bring down a marine is about 21% A good guess is that we should bring about five FWs to the party, right? To calculate how well five FWs would fare against a single MEQ we could more or less repeat what we did under “3 Some examples”, but that is rather tedious and boring. There are a few ways to simplify this process. Be warned though I will not take my time explaining why and how these really works, rest assured they do. Our little baby today is a formula that looks like this: It is called the binomial distribution. And its notation is b(x; n; p). In the way we are going to use it b(x; n; p) is the probability to inflict exactly ‘x’ wounds with ‘n’ shots, each having ‘p’ chance to wound the target. If you are unfamiliar with the notation in the formula but still want to try it out I’ll try to simplify some of the parts to “easy math”. The above is a notation of the numbers of combinations of x in n. If you like math you know this already, if you don’t you don’t want to know. It could also be written as Where n! is a more comprehensive way of writing 1 * 2 * 3 * … * n2 * n1 * n, thus 4! = 1*2*3*4 = 24. Knowing this you can write out pretty much any given situation with no more then basic high school math (depending on where you’re from and all that). That way you could write the whole thing as: b(x; n; p) = n! * px * (1p)nx x!(nx)! [size=12pt]5. Using the formula[/size] Lets do the calculation in section 3 again. Remember that our mission is to calculate chance of killing that poor MEQ with two Pulse Rifle shots. What we need is our ‘x’, ‘n’ and ‘p’. WE know already that we will shoot two shots (n=2) and that the chance of inflicting a wound is about 11% (p=1/9) and we will kill him if we cause one or two wounds. This would translate into b(1;2;1/9) + b (2;2;1/9). But this is just excessive and we could think backwards again and calculate the chance of the MEQ to survive, which would be if we didn’t inflict any wounds (x=0). b(0;2;1/9) is the probability for us to inflict exactly zero wounds with two Pulse Rifle shots on a MEQ target or in other words, the chance of the MEQ to survive. Let’s plug some numbers into that formula. P(MEQ suvives) = b(0; 2; 1/9) = 2!/(0!(20)!)*(1/9)^0 * (1(1/9))^(20) = 64/81 So P(MEQ survives) = 64/81 of course makes P(Inflicting at least one W on MEQ with RF) = 17/81 and that’s exactly the same result as we calculated in section 3. The main use of the binomial distribution is that it’s very easy to change it for different setups. Ever asked yourself how big the chance is that you kill off an entire squad of five MEQs with a full squad of Fire Warriors? Well it’s a bit more involved when it comes to the calculations but we realise if that we fail to kill the squad if we less than 5 wounds of if x = 0 through 4. We fire 24 Pulse Rifle shots (12 * Rapid Fire) and the probability to cause a wound is still 1/9. This could be written as P(fail to wipe) = b(0; 24; 1/9) + b(1; 24; 1/9) + b(2; 24; 1/9) + b(3; 24; 1/9) + b(4; 24; 1/9) P(wipe out) = 1 – P(fail to wipe) ? 12% So where are the calculations then, I her you say. Well I’m not stupid enough to print them out here. If you are familiar with sums you can write the so called cumulative binomial distribution as: Anyway, now you know that it can be made by hand and in principle how you would do it. Then we should learn to make our computer do it for us. [size=12pt]6. Using Excel to do it for you[/size] Almost any serious math/statistics program or calculator will be able to do this for you. I guess most of you have access to Excel. Excel has indeed that ability to make these calculations for you. The programs contains a function that is b(x; n; p) but is called BINOMDIST(exact number of wounds to cause; number of shots; chance to wound; cumulative?). The first three arguments to these functions hold no surprises, while the last one is some serious candy for us, but leave it as FALSE for a moment. Let’s redo the calculations in section 5 again. How great is the chance for a single MEQ to survive the Rapid Fire of a single Fire Warrior? This will translate into: P(MEQ suvives) = b(0; 2; 1/9) In Excel this would be written as: =BINOMDIST(0; 2; 1/9; FALSE) and it does yield the desired answear 64/81. But we want to know the chance to kill it and therefore we put in: =1BINOMDIST(0; 2; 1/9; FALSE) This works wonders on single wound single model enemies. What if you want to kill a whole squad? Back again to the problem posed in section 5. How big the chance is that you kill off an entire squad of five MEQs with a full squad of Fire Warriors? We saw then that it would be: P(fail to wipe) = b(0; 24; 1/9) + b(1; 24; 1/9) + b(2; 24; 1/9) + b(3; 24; 1/9) + b(4; 24; 1/9) P(wipe out) = 1 – P(fail to wipe) ? 12% It’s still very inflexible and has to be tailored to every situation. That’s when the mystery argument in the BINOMDIST function steps into the picture. If you set the mystery argument to TRUE it will work as something called the cumulative binomial distribution. It looks like this: Without getting into what it really is it’s an easy way to do the above unaccounted for equation. If you set the cumulative argument to TRUE the BINOMDIST function will return the probability to inflict 0‘x’ wounds, or inflict at the most ‘x’ wounds. As always we calculate the chance of not managing to kill them all. In excel P(F to wipe 5 MEWs with 12 FW) = b(0; 24; 1/9) + b(1; 24; 1/9) + b(2; 24; 1/9) + b(3; 24; 1/9) + b(4; 24; 1/9) will be written as: =BINOMDIST(0; 24; 1/9; TRUE) The chance to kill them will be the complementary event. 1 BINOMDIST(0; 24; 1/9; TRUE). If you know the number of wounds in the squad ‘x’ and the number of shots ‘n’ and the probability to cause a wound with a shot ‘p’ you can calculate the chance to wipe the squad in Excel with the following formula: =1 BINOMDIST(x1; n; p; TRUE) Back to our example, we have 5 wounds 24 guns and a 1/9 probability of causing a wound: =1  BINOMDIST(4; 24; 1/9; TRUE) This formula will in Excel return the answer 0,12026013, or about 12%. The probability to kill of the whole team is about 12% and that my friends if why we don’t FoF MEQs. Acknowledgements: Thank you for the input and proofreading Yawgmoth1111 [ mace ] Abanim Well I’m done, please give me feedback to se if it’s legible in any kind of way or if we can change it into something that is. Happy Theoryhammering. /fork 

28 Dec 2005, 21:26  #2 (permalink) 
Shas'Ui
Join Date: Jul 2005
Location: Florida, USA, Earth, Sola segumuntum, third arm of spiral milky way galaxy, the universe.
Posts: 647

Re: forks guide to Theoryhammer
you deserve some seroius karma.
i should'nt have slpet in math... :
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28 Dec 2005, 21:40  #3 (permalink) 
Shas'Saal
Join Date: Oct 2005
Posts: 147

Re: forks guide to Theoryhammer
I like what you are trying to do, but I must say, that it is a bit thick to get to.
Also, instead of percentages, I usually try to deal with expected number of wounds, it works out more cleanly and it is just more easily applied to the game dynamic. 
28 Dec 2005, 21:52  #4 (permalink)  
Kroot Shaper
Join Date: Dec 2005
Posts: 82

Re: forks guide to Theoryhammer
Quote:
Perhaps its heavy, but then again  just for the interested.* Quote:


29 Dec 2005, 00:16  #5 (permalink) 
Shas'La
Join Date: May 2005
Posts: 350

Re: forks guide to Theoryhammer
I'm glad someone is tackling the probability myths I've seen hanging around there, yet I also feel that a better example might be needed to clarify one thing. I hope you don't mind the semihijack, I love what you're doing here
In this example: "1/2*4/6*(14/6)=3/6*4/6*2/6=1/9 ...Again we do not add them up so that P(Wounding MEQ w. Rapid Fire)* would equal P(Wounding MEQ) + P(Wounding MEQ). Rather we calculate the complementary event to the risk of failing both shots: " "1 ( (8/9) * (8/9) ) = 17/81 So the probability for a single Fire Warrior to bring down a marine is about 21%." 1/9 = 18/81 ~ 17/81 for the sake of estimation and common useage. This result might be misleading. Here's an example where the effects are a touch more pronounced: A Shas'O Crisis Suit doubletaps a Tactical Marine with a single Plasma Rifle. (5/6) * (5/6) = 25/36 ~ 69% chance of success on one of the shots Method1 (Just Doubling It): 138% chance of success. As Fork said. We KNOW that this event can fail, so this result is garbage. Method2 (Complimentary Event): 1 ( (11/36) * (11/36) ) ~ 91% chance of success. This is a more intuitive and natural result. A lot of people find themselves disapointed with the performances of their units, without realizing that they're overestimating the effect their firing should have. They expect the kill to be guarenteed, whilst cursing dice gremlins for bending 'natural law'* *: 
29 Dec 2005, 00:49  #6 (permalink) 
Shas'O

Re: forks guide to Theoryhammer
.. My head hurts. And I usually get/write this stuff. Though it makes perfect sense to me..
If I've done my own math properly, Doesn't it take 9 Pulse rifle shots to result in a wound on a marine? My interpretation of Percentages exceeding 100% is that once it does, it means it's overkill. E.g. taking 18 Fire Warriors to kill 1 Marine = 200% chance of killing him = Overkill? But the reality is  the chance to kill a target still remains the same (given the ratio of shooters and targets/wounds remains the same). If you double the amount of firepower you're putting out, you're doubling the number of targets you can kill. And so the chance of 18 Fire warriors killing 2 MEQ is equal to that of 9 Fire Warriors Killing one MEQ. But then when you come back to firing upon one target, the chance to kill him exceeds 100%.. The chance to Kill the Target  e.g. inflict ONE wound is 120%. The chance to inflict TWO wounds is still 60%. [hr] Note: I think the actual chance of killing him is 84% or something, not 120%. Edit: OH! That's exactly what you did above! Now I get it... So really, there's no such thing as 120%, as it's a miscalculation of just doubling something and applying it to a situation where the ratio of shooters to targets/wounds is different. 4/10 refers to the chance of not wounding, in said scenario 6/10 similarly refers to the chance of wounding Chance that he doesn't receive a wound = 4/10 * 4/10 = 16/100 = 16% Chance that he receives one wound (from the first set of shooting) = 6/10 * 4/10 = 24% Chance that he receives one wound (from the second set of shooting) = 6/10 * 4/10 = 24% Chance that he receives two wounds = 6/10 * 6/10 = 36/100 = 36% Totals: [table][tr][td]Unwounded[/td][td]16%[/td][/tr] [tr][td]Receives ONE Wound[/td][td]48%[/td][/tr] [tr][td]Receives TWO Wounds[/td][td]36%[/td][/tr] [tr][td] Total [/td][td]100%[/td][/tr][/table]The last two results would obviously kill him if he has one wound, so the actual chance of killing him is 84%, Likewise, Chance of survival is 16%. [hr] P.s.  Rapid firing is just doubling the amount of shots you put out. So don't you just add *2 to the equation? And wouldn't it come out to be 2/9 (22/99) chance of killing a marine (e.g. double 1/9) Edit: Ok, reread the post, and now I get what you're saying there, though now my head hurts again (this time from understanding the math itself, not from being confused at what you were saying :S).
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29 Dec 2005, 01:41  #7 (permalink) 
Shas'El

Re: forks guide to Theoryhammer
Very good... but my own heuristics work well enough for me
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Watch me rebuild my army!  Now with pics of my new scheme! Open invitation to all: Send me a pm, and I'll be have a look at your army list. [url=http://forums.tauonline.org/index.php?topic=11198.0]Crisis Suit Theory, Army Composition, and why Fireknives suck  You've got to read this classic by T0nka 
29 Dec 2005, 07:19  #8 (permalink)  
Kroot Shaper
Join Date: Dec 2005
Posts: 82

Re: forks guide to Theoryhammer
Quote:
Quote:
Quote:
Mace: I’ll try to answer some questions and go back to reedit for clarity, but for now I to jump into your discussion. Although I realise you get most of this stuff by now I’ll comment it anyway for others. Quote:
Quote:
Quote:
Quote:


29 Dec 2005, 08:27  #9 (permalink)  
Shas'Ui
Join Date: Nov 2005
Posts: 903

Re: forks guide to Theoryhammer
Congratulations, fork, on an excellent explanation. The first section in particular is a clear explanation of a common statistical fallacy. Unfortunately, any work so elaborate is bound to have some minor flaw, and I have a small correction to make to your use of binomial expansion.
You've stated the formula correctly. The one we're using is: [tt]B(n; x; p) = n! * p[sup]x[/sup] * (1p)[sup]nx[/sup] x!(nx)![/tt] ... where n is the number of shots, x is the number of wounds, and p is the probability of a single shot causing a wound. However, the way you've used it suggests that you're slightly misled as to what the output of the formula actually means. The figure we get out of it, B, is the probability of exactly x events occurring  not the probability of more than x events. (An event in this case is an unsaved wound.) A simple example should illustrate this. Look at the case you were examining earlier, where a single fire warrior is rapidfiring at a lone marine. You considered the case: [tt]n = 2 p = [sup]1[/sup]/[sub]9[/sub] x = 0[/tt] ...that is, where there are two shots, each with a one in nine chance of wounding, and zero wounds are inflicted in total. You (correctly) worked out that this figure is [sup]64[/sup]/[sub]81[/sub]. Now, there are two other possibilities  that one wound is inflicted, or that two wounds are inflicted. (If two wounds are inflicted, the marine is removed and the second wound is wasted; and we can thoroughly rule out the possibility of three or more wounds from only two shots.) [tt]B(n=2; x=1; p=[sup]1[/sup]/[sub]9[/sub]) = [sup]16[/sup]/[sub]81[/sub] B(n=2; x=2; p=[sup]1[/sup]/[sub]9[/sub]) = [sup]1[/sup]/[sub]81[/sub][/tt] (Feel free to check my working, but I'm quite confident of these results.) Since these three possibilities (of inflicting zero, one or two wounds) are complementary their probabilities should add up to 1, and they do: [sup]64[/sup]/[sub]81[/sub] + [sup]16[/sup]/[sub]81[/sub] + [sup]1[/sup]/[sub]81[/sub] = 1. Now, to apply it to your final example. We have a full squad of fire warriors firing at five marines, and we want to work out the probability that we kill all of them. In this case it is easier to work out the chance that we fail to do so, then subtract that probability from 1, as you did earlier in the example with the single marine. If we inflict zero, one, two, three or four wounds, we will have failed to eliminate the entire squad of marines. The probability of inflicting x wounds in this case is: [tt]B(24; x; [sup]1[/sup]/[sub]9[/sub])[/tt] ...so we want to work out the sum of the probabilities where x = 0, 1, 2, 3 or 4. In full: [tt]B(24; 0; [sup]1[/sup]/[sub]9[/sub]) + B(24; 1; [sup]1[/sup]/[sub]9[/sub]) + B(24; 2; [sup]1[/sup]/[sub]9[/sub]) + B(24; 3; [sup]1[/sup]/[sub]9[/sub]) + B(24; 4; [sup]1[/sup]/[sub]9[/sub]) = 0.0592 + 0.1776 + 0.2553 + 0.2340 + 0.1536 = 0.8797[/tt] That's interesting. That's the same figure as you obtained earlier on when you wrote: Quote:


29 Dec 2005, 09:32  #10 (permalink) 
Shas'O

Re: forks guide to Theoryhammer
Argh! Indeed, my head still hurts (writing a 40k Calculator Spreadsheet :P)  though my post was a combination of before/after thoughts when I was still reading through and not understanding what was said..
(For some reason all the math stated seems to be correct.. so..) I think there is a need to distinguish between what we're dealing with here. The chance to kill a target (e.g. a marine) can be worked out, as an exact/absolute percentage, taking into account the number of firers. Obviously the more firers, the higher the chance the target will die. Though this chance will always lie between 0% and 100%  though never reaching 100%  as there will always be the chance that the shooter will roll one's or that the the target will pass all saves  however remote that chance might be. Note that the chance to kill a target is to cause more wounds than it has. In the case of a one wound model, there is a decent chance that you can cause more than one wound, however that just contributes to the chance of killing the target. Example  9 Fire Warriors shooting on one marine  has a 65% chance of Killing the Marine. (Chance of not inflicting a wound whatsoever is 8/9 [to the power of 9] = ~35%) On the other hand, the number of wounds inflicted by X number of firers is just an average ratio/most probable successful outcome/median result (but not an exact kill percentage!). a) 1 Fire Warrior, Shooting One at 1 Marine  Is most likely to inflict 1/9 Wounds  (e.g. on average, one out of 9 times, he will inflict a wound) b) 9 Fire Warriors, Shooting Once each at a squad of marines (at least a few marines)  It is most probable that they will inflict one wound on the squad. (On average they will inflict one wound) c) 1 Fire Warrior, Shooting Twice 1 Marine  will inflict 2/9 wounds on average. The actual working out, based completely off statistics/averages etc: a) 1 Shot. 0.5 hits. 0.333 Wounds Caused, and 0.111 Wounds Unsaved. b) 9 Shots. 4.5 Hits. 3.000 Wounds. 1 Wound Unsaved. c) 2 Shots. 1.0 Hit. 0.666 Wounds. 0.222 Wounds Unsaved. a) b) and c) are in the exact same ratio, and b) shows that with 9 Fire Warriors, it is most probable that they will cause 1 wound. It obviously isn't the only outcome, but it is the most common/probable (e.g. see the example in my previous post/the first post with the target having a 48% chance of receiving one wound). (I'm starting to get confused again) What I'm basically trying to say is  that there is a difference between the Chance to Kill  which is an Exact Percentage  and the Average Ratio of Firers to inflict a wound. The latter basically describes the number of firers needed to achieve the most common result from the former. Though the confusing bit is that it's all very similar to one another.. Lets look at the Chance to Kill , 1 Shot Vs 1 Marine. This is all fairly straightforward, and just so happens to be the same as the average no of wounds that will be caused, which is 1/9, or 11% Chance to Kill. The reason it is so straightforward/the same? You can only inflict one or zero wounds, and the target only has one wound. The Next situation, 2 Shots Vs 1 Marine. Chance to Kill: 21% (Chance that he'll survive both shots is 79%, or 8/9 * 8/9)  0 Wounds: 79.01%  1 Wound: 19.75%  2 Wounds: 1.23% Average number of wounds caused from two shots: 0.222 (2 Shots. 1 Hit. 0.666 Wounds. 0.222 Wounds Unsaved.) The Chance to Kill is the total percenage, of all situations where the marine will be killed (receiving one or more wounds). The Average no. of wounds caused is a statistical average, purely based off the game mechanics and dice probabilities. What I'm getting at is that we are dealing with two different things, the Chance to Kill is different from that of the Average Wounds caused by X Shooters, and that the two should not be confused.. When you're talking about the chance to kill, doubling the firepower won't double your chance to kill a target, rather, it will increase the chance to kill (um.. inverse exponentially?). When you're talking about average wounds caused, doubling the firepower will double the amount of wounds caused, as the ratio will remain the same. An important difference between the two is that the first takes into account the number of targets, whereas the second simplisticly looks at the statlines and works out the average number of wounds caused. Correct me if I'm wrong, but it is a common misconception that you can just 'wrap wounds' from the average, when the average number of wounds caused exceeds the number of wounds of models taking the wounds, in order to create the chance to kill. As can be shown above, that isn't entirely correct or accurate, in getting the actual chance to kill. Note: Sorry I introduced the latter bit here, I got them both confused earlier with my wording too.. now to get my head around parts 4 and 5.. I don't think we're even up to that in school, or if it's even in our course :S (I sure don't remember doing it in earlier years)
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