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I'm writing a mathhammer application and i've come upon an interesting discrepancy

I should note that most of this post will be statistical technobabble, and i dont plan on explaining stuff like the Binomial distribution and expected value and such.

so my question is, i was calc'ing out stats on a fight between a Space marine Captain (no special gear) and a regular Space Marine. I calculated out an expected value of 0.33 unsaved wounds on the space marine (3 attacks * 0.67prob to hit * 0.5 chance to wound * 0.33 chance unsaved = 3 * 0.11 = 0.33 expected unsaved wounds).

however when i tried to use the binomial distribution to calculate the probability that the space marine is killed (cumulative probability of 1, 2 and 3 wounds respectively) i only got a 0.299 probability of a kill (0.263+ 0.032+ 0.001 probability).

What's with the discrepancy here? Have i overlooked an assumption with my expected value calculations?

If you are trying to calculate the chance he is going ot die, then just multiply everything together(possibility to hit x possibility to wound x possibility to fail armour saves(lets say 3 wounds = 1/3 x 3))


Automatically Appended Next Post:
If you are trying to calculate how many wounds he is gonna take in total, then ask your math teacher...I can't help anymore

a discrepency of only .04 is not a huge difference.

any actual miscalculation with that formula likely can't effect the dice all that much.

Rounding and significant figures are your friend and must be kept the same and consistant through both set of calculations.

Also check your addition.

I have checked my addition and i do all my calculations to about 8 decimal places, I just truncated them to make it easier to read.

And I'd argue that 4% is enough to be significant. it if were 1 or 2 % i could just write it off, but thats too big of a margin for me to pretend it's not there.

.33 wounds is the average amount of wounds, not the amount of times you'll get at least one wound. Since you only need 1 for a kill, but potentially have 0, 1, 2 or 3 wounds, 2 of the 4 possibilities are overkill. Because of that, a small amount of the average wounds occur in situations where really, you're getting 2 or 3 wounds. When you're just calculating the binomial distribution, you're going to find that the probability of a kill will be only .299 because the 2 and 3 wound situations increase the average number of wounds, but not the average number of kills (since occasionally with extra failed saves, they're overkill.)

This will always happen whenever the shooter can do more wounds than there actually are wounds on the target. The situations where the guy takes 3 (or 10 or whatever) wounds will factor into the average number of wounds, but don't do quite as much to adjust the likelyhood of actually killing the target.

If you look at your own numbers you should see what is missing to get .33. the .032 chance of a kill with 2 wounds and the .001 chance of a kill with 3 wounds, when added in AGAIN, gets you to .33 (actually .34 since you'd ad the .001 twice). But you don't get to add them in again, since you're just looking at probability of kills, not wounds. You have a .032 chance to get 2 wounds. That shows up in the average as .032 extra wounds. But it won't show up here because you're looking for kills.

Ahhh alright thanks. That makes sense.

This is actually pretty important, because there are are a decent amount of situations where the attacker can overkill the target. In all of these situations, calculating the average number of wounds will cause you to overestimate your likely hood of actually killing all of the target.

good thing i found this out then. that probably accounts for why i've seen small units of space marines survive ridiculous onslaughts (4 space marines surviving 26 kroot charging :-P)

Hah, yeah, that would be one example, plus well unlikely things happen, thems the breaks.

I find this to be most important when people are calculating average results against vehicles from multiple shots. It doesn't matter if you destroy it 3 times or shake it 3 times, so when you're calculating the likely hood that a IG heavy weapons team will kill, shake, stun, whatever a tank, you need to eliminate all of those events where they kill it 2 or 3 times to really find out your odds of success. Or more accurately, just count all the times you fail to do anything, instead of just doing a simple average.

Hey lambadomy, could this phenomenon also factor into the endless huge debate about the mathammer of assault cannons versus lascannons against AR14 vehicles? Could it be that the overkill rolls on the penetration for the rending assault cannon are pulling the average damage rolls artificially high? Or that the multiple shot thing is pulling the average artificially high for the 4-shot assault cannon?

Or have the assault cannon proponents already taken this into account?

Because that's the biggest case for me where my experience on the tabletop is nothing like what the mathammer I've seen suggests it should be.

It's relatively easy for me to figure out the odds of a single lascannon shot doing nothing against AR14. But figuring out the odds of the assault cannon doing nothing are way beyond my rudimentary math skills.

Yeah, that would definitely influence it. Those times you get 2, 3, or 4 destroyed results are being counted in the average and are skewing the results. Of course, that doesn't mean that assault cannons aren't better still, just that just calculating the average gets you the wrong answer unless you can't overkill the target.

however when i tried to use the binomial distribution to calculate the probability that the space marine is killed (cumulative probability of 1, 2 and 3 wounds respectively) i only got a 0.299 probability of a kill (0.263+ 0.032+ 0.001 probability).

How did you calculate this?

wuestenfux wrote:

however when i tried to use the binomial distribution to calculate the probability that the space marine is killed (cumulative probability of 1, 2 and 3 wounds respectively) i only got a 0.299 probability of a kill (0.263+ 0.032+ 0.001 probability).

How did you calculate this?

looks like i has a little rounding error or two. either way it came out to just under 30%.

or were you referring to how to calculate binomial probability?

<----Math teacher

It looks like all your math is correct. The reason for the discrepancy is simply if you have event A and event B the chances of A or B occurring is P(A)+P(B)-P(A)*P(B). In the classic Venn diagram model you have to subtract out the overlap.

For those that don't know it the formula to use to calculate causing a specific number of wounds is:

S: Chance of success
F : Chance of failure
X: Total number of attacks
Y: Number of wounds you want to cause

(X!/(Y!)*(X-Y)!)*(S^Y)*(F^(X-Y))

If you want to calculate doing at least 1 wound just calculate the chances of doing 1,2, and 3 wounds.

**Edit**

I did the calculations with no rounding and I got 0.297668...

Or you can just make a fancy program in your calculator and have it all done in a couple seconds.

I actually made a fancy program in a webpage :-P also, factorials in the binomial coefficient? Ick. All about the summations, bro! for loops ftw!

Flavius Infernus wrote:Hey lambadomy, could this phenomenon also factor into the endless huge debate about the mathammer of assault cannons versus lascannons against AR14 vehicles? Could it be that the overkill rolls on the penetration for the rending assault cannon are pulling the average damage rolls artificially high? Or that the multiple shot thing is pulling the average artificially high for the 4-shot assault cannon?

Or have the assault cannon proponents already taken this into account?

Because that's the biggest case for me where my experience on the tabletop is nothing like what the mathammer I've seen suggests it should be.

It's relatively easy for me to figure out the odds of a single lascannon shot doing nothing against AR14. But figuring out the odds of the assault cannon doing nothing are way beyond my rudimentary math skills.

The probability of overkill with an assault cannon is so low that I can imagen that it would not effect the result that much.

ThatEdGuy wrote:<----Math teacher

It looks like all your math is correct. The reason for the discrepancy is simply if you have event A and event B the chances of A or B occurring is P(A)+P(B)-P(A)*P(B). In the classic Venn diagram model you have to subtract out the overlap.

For those that don't know it the formula to use to calculate causing a specific number of wounds is:

S: Chance of success
F : Chance of failure
X: Total number of attacks
Y: Number of wounds you want to cause

(X!/(Y!)*(X-Y)!)*(S^Y)*(F^(X-Y))

If you want to calculate doing at least 1 wound just calculate the chances of doing 1,2, and 3 wounds.

**Edit**

I did the calculations with no rounding and I got 0.297668...

Or you can just make a fancy program in your calculator and have it all done in a couple seconds.

From one Maths Teacher to another, remember that a far easier way of calculating the probability of at least one wound is to do 1 - the probability of doing 0 wounds. This will save you a lot of time!

InyokaMadoda wrote:

ThatEdGuy wrote:<----Math teacher

It looks like all your math is correct. The reason for the discrepancy is simply if you have event A and event B the chances of A or B occurring is P(A)+P(B)-P(A)*P(B). In the classic Venn diagram model you have to subtract out the overlap.

For those that don't know it the formula to use to calculate causing a specific number of wounds is:

S: Chance of success
F : Chance of failure
X: Total number of attacks
Y: Number of wounds you want to cause

(X!/(Y!)*(X-Y)!)*(S^Y)*(F^(X-Y))

If you want to calculate doing at least 1 wound just calculate the chances of doing 1,2, and 3 wounds.

**Edit**

I did the calculations with no rounding and I got 0.297668...

Or you can just make a fancy program in your calculator and have it all done in a couple seconds.

From one Maths Teacher to another, remember that a far easier way of calculating the probability of at least one wound is to do 1 - the probability of doing 0 wounds. This will save you a lot of time!

I did say in my initial post that that's not the way i wanted to do it, seeing as i won't always be calculateing p(any wounds), and it'll often be p(>x wounds). I'm considering adding an if statement to calculate 1-p(0 wounds) for when i need p(any wounds). I'm a little obsessed with minimising time-complexity :-P

Do you have a programmable calculator? I wrote a program for my TI-84 to do this fairly quickly.

No, but i carry my laptop with me pretty much everywhere.