Sisyphus' Probability Tables(updated April 8, 2008)

[twilkerson]

You miss 100% of the shots that you don't take.
-Wayne Gretzky

[Ullar/utgar own]

very good idea. these are extremely useful.

[jaamartin]

Originally Posted by Sisyphus

The specialized tables in this thread come from this formula with only slight modifications.

Originally Posted by StarofEarendil

Sisyphus, I like the formulas, but have found I cannot get the problems to work out right. So I was wondering if you could give a demonstration of sorts to show some of the younger nerds on the site how to work these formulas.

Star, I shortened the post to highlight Sisyphus's statement about modifying the formula a little. The thing is, Sisyphus never speaks of the changes made. Looking at the formula's just about anything can be changed a tad, and without getting the original formula to match up with his generic table, I don't trust myself to just make changes.

Here is an example of what I get.

For A=1 :

D       Result
0       0.5
1       0.333
2       0.222
3       0.148
4       0.099
5       0.066
6       0.044
7       0.029
8       0.0195
9       0.013
10      0.0087

But when I try the next column, I get differences.

For A=2

D       Result
0       1
1       0.8333
2       0.6667
3       0.5185
4       0.3951
5       0.2963
6       0.2195
7       0.1610
8       0.1171
9       0.0845
10      0.0607

This does not correlate properly with the posted values at the beginning.

Sisyphus, what errors am I making? I have only looked at P(w) at the moment since that is the one I will most likely use for a while until it become ingrained in my brain...

Jason

[dok]

Well, without looking over your shoulder neither I nor Sisyphus can check your work. But I have looked at the formulas, and they are accurate.

Just to give you one example... looking at the A=2, D=1 entry, you should get

.5*(2/3) + .25*(2*2/3 + 1/3)

Which does reduce to .75, not .833.

[jaamartin]

dok, I am not sure what I am messing up though. Here is what I have entered into MathCAD:



Again, I am looking at the function P(w). I will run this out by hand tonight when I get to my folks house and see if it isn't just MathCADs uncanny ability at yielding incorrect results.

[Sisyphus]

Originally Posted by jaamartin

dok, I am not sure what I am messing up though. Here is what I have entered into MathCAD:



Again, I am looking at the function P(w). I will run this out by hand tonight when I get to my folks house and see if it isn't just MathCADs uncanny ability at yielding incorrect results.

Well, at least you understand your biggest mistake is using MathCAD.

I think your problem is that you need to pass "little a" as a parameter to your BigA function. IIRC, MathCAD evaluates everything from top to bottom which means that the "little a" in your BigA function is not the same "little a" as in the Pw function.

[jaamartin]

I know that software will corrupt the brain when it comes to math, but I was in a pinch at work and supposedly doing other things.

Thanks for pointing that out! Everything is matching up now.

[Obsidian]

I saw that there was no table for counterstrike against a normal attack when defending with HEROIC DEFENSE AURA, so here are some numbers that I found using my heroscape probability calculator.

The odds that an attacker will receive at least one wound when attacking a figure with COUNTERSTRIKE and HEROIC DEFENSE AURA.

	Att	0	1	2	3	4	5	6	7	8	9	10
Def												
1		0.500	0.250	0.125	0.063	0.031	0.016	0.008	0.004	0.002	0.001	0.000
2		0.750	0.500	0.313	0.188	0.109	0.063	0.035	0.020	0.011	0.006	0.003
3		0.875	0.688	0.500	0.344	0.227	0.145	0.090	0.055	0.033	0.019	0.011
4		0.938	0.813	0.656	0.500	0.363	0.254	0.172	0.113	0.073	0.046	0.029
5		0.969	0.891	0.773	0.637	0.500	0.377	0.274	0.194	0.133	0.090	0.059
6		0.984	0.938	0.855	0.746	0.623	0.500	0.387	0.291	0.212	0.151	0.105
7		0.992	0.965	0.910	0.828	0.726	0.613	0.500	0.395	0.304	0.227	0.166
8		0.996	0.980	0.945	0.887	0.806	0.709	0.605	0.500	0.402	0.315	0.240
9		0.998	0.989	0.967	0.927	0.867	0.788	0.696	0.598	0.500	0.407	0.324
10		0.999	0.994	0.981	0.954	0.910	0.849	0.773	0.685	0.593	0.500	0.412

[galvornman]

Greetings all,

After having my ass handed to me by Jexik on a few occasions, I asked myself - how many attacks would it take for a figure with M attack dice vs. a defender with N defense dice to inflict one wound, on the average?

Rather than deal with probability calculations - I ran simulations. I used R (code available to anyone who wants it), and each element in the table represents an average over 1000 simulated battles. Essentially the algorithm is the following:

Roll attack dice & count skulls.
Roll defense dice and count shields.
Is the number of skulls greater than the number of shields?
If it is, the number of attacks necessary to inflict one wound is 1, and we move on to the next simulation.
If it isn't, we try again, and keep counting the number of attacks until I get skulls greater than shields. This number will necessarily be something greater than 1, and I record it.

I do this process 1000 times, for every combination of attack and defense dice (up to 10 attack and 10 defense), then taking the average of all 1000 results.

Below I attach a table with my results. Note that these results have been rounded. Example interpretation: For an attacker with 3 dice (A3) against a defender with 3 defense dice (D3) I can expect to attack twice (2) in order to inflict one wound.

(apologies for the poor formatting!)


D1 D2 D3 D4 D5 D6 D7 D8 D9 D10
A1 3 5 7 10 15 23 33 51 81 116
A2 2 2 3 4 5 8 11 15 20 28
A3 1 2 2 3 3 4 5 7 9 12
A4 1 1 2 2 2 3 3 4 5 7
A5 1 1 1 1 2 2 2 3 3 4
A6 1 1 1 1 1 2 2 2 2 3
A7 1 1 1 1 1 1 2 2 2 2
A8 1 1 1 1 1 1 1 1 2 2
A9 1 1 1 1 1 1 1 1 1 2
A10 1 1 1 1 1 1 1 1 1 1

I was a bit surprised at first by the first row, specifically the A1 vs. D1. The average is, of course, not the whole story - simulations show that about 34% of the time one attack is enough to inflict one wound against a defender with 1 die (this is not surprising). In 22% of the simulations in this case it took 2 attacks, and in 13% of the simulations it took 3. The average, being sensitive to outliers, is pushed up by the presence of a few large cases - 12% of the observations were greater than 5 in this particular case.

For me, this table demonstrates what we've known all along - rolling a few attack dice many times trumps rolling a lot of dice only a few times.

Why is Q9 so good? He rolls a few dice many times, resulting in more net wounds.

Why are 4-man squads so dominant? If they have at least 3 attack dice and are ranged - again, lots of opportunities to roll dice and inflict wounds.

Why are rats good? If attacked by a 2-4 dice attacker, it takes an average of two attacks to take each one down - and in this time they get to move (assuming normal attacks) and cause trouble.

Why do the Deathwalkers die? Even 3 attack dice squads need only roll against DW9K an average of 9 times before he falls.

Nilfheim/Zelrig good? 3 attacks.

Why is bonding so awesome? More attacks.

Apologies if this work offers no new insight into this great game - it only confirms what we already know by experience. Excluding counter drafting and gimmicks (Thorian speed vs. normal ranged attacks), winning armies are those that allow you to roll a few dice as often as possible. Squadscape isn't going anywhere anytime soon.

[padlock]

Originally Posted by galvornman

For me, this table demonstrates what we've known all along - rolling a few attack dice many times trumps rolling a lot of dice only a few times.

That of course depends on who you are attacking. I did the math, and it turns out that if you had a large pool of dice and were able to split them up as you saw fit when attacking a group of single life figures, then the optimal number was always equal to defense of the figures you were attacking.

For example, if you were given a hundred dice and told to attack a horde of stingers, you would be best off doing 33 attacks of 3 (followed by 1 attack of 1). If however, you were attacking a group of minions, you'd be better of doing 16 attacks of 6 (followed by 1 attack of 4).

When fighting multi-life heroes, the numbers change and it becomes much more beneficial to have fewer large attacks then many small ones.

If higher defense squads and more multi-life heroes ever become really popular, then I suspect that relatively low attack squads (like the 4th mass) will become less dominant.

[bmaczero]

Thanks for sharing the formulas, Sisyphus! I've got a quick question, though. What's the meaning of the "A" over "a" term (the first one inside the sum). I've never seen that before

EDIT: Nevermind, I stumbled across it on Wikipedia .

[CheddarLimbo]

Could someone help me calculate the odds of an outcome?

What is the probability someone will (a) get initiative (b) roll a 14 or higher to summon Retchets with Iskra, and the (c) roll all skulls on a 3 die attack to Lethal Sting a figure?