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Shrike78 wrote:
so the formula for say sniper scouts vs MEQs would be

Normally 1/2*1/2*1/3

with rending X*(1/2*1/3(?)*1/3) + X/6(?)

Something like that. It'll be 1/2*((1/3*1/3)+(1/6))

So the non-rending is 1/12 (or 3/36) and rending is 5/56, or 66% better. Daaaaang.

Shrike78:

You are correct. It is not flaming to point out when someone gets something wrong, particularly if you explain how they are wrong. Thank you for crunching the numbers so I didn't have to!

Nurglich and Shrike:

I love where your discussion is going. Nurglich, I never thought of it in those terms, or, I should say, extrapolated the figures far enough to see the difference in expected returns vs maximum potential returns. That's a great way to see the "upper limit" and work backwards, too.

Shrike, I like fractions for mathhammer, too. It's cleaner, but as my final step, I just divide them in to numerals for clarity to my American mind. Most if not all of my longhand work is done in fractions.

To whoever asked about converting back and forth, just do the math ( "/" means divide!). Each side of the dice that can succeed is 1/6 chance of success, or roughly 16.667%.

Nurglitch wrote:Speaking of evaluating the value of various combat/shooting configurations, there's something I'm calling "weighted expected value".

It's basically the expected value of a set of attacks (the product of the total number of attacks multiplied by the likelihood of any particular attack having some effect), multiplied by expected utility if those attacks hit at a 100% rate, or the maximum potential value.

Weighted Expected Value:
EV = (Number of Attacks x Likelihood of Effect)
PV = (Number of Attacks x Likelihood of Effect or 100%)

WEV = EV x PV

I am ashamed of my mental faculties. I understood the rest of this post, but I am completely lost as to this part's meaning or even application. Being something of a mathhammer geek (hey, i'm not too great at tactics, so I might as well know how much of what kills what right? ) I thought I could handle the majority of statistical figures in relation to dice roles...

This has proven me wrong

Mathhammer is an advantage, why shun it when it is so readily availibale. To know how a unit should preforme is really useful, also since warhammer is a game of averages it really becomes useful. (wow my spelling sucks)

Stepping back to the rending issue...wouldn't it be:

1/2(1/3*1/3 + 1/6)

To explain, it is your 50% chance to hit plus the probability to get either a normal wound with a failed armor save or a rending shot with no armor save allowed. Which is about 13.89% (5/36, not 5/56)

Then, if you want to find the chance to get a single wound for X amount of shots:

1-(31/36)^X

Axyl wrote:Stepping back to the rending issue...wouldn't it be:

1/2(1/3*1/3 + 1/6)


To explain, it is your 50% chance to hit plus the probability to get either a normal wound with a failed armor save or a rending shot with no armor save allowed. Which is about 13.89% (5/36, not 5/56)

holy crap man! that equation is absolutely right! High five! you win a cookie.

Then, if you want to find the chance to get a single wound for X amount of shots:

1-(31/36)^X

This model is incorrect i believe. If for no other reason than because according to your equation you will never have a 100% statistical likelyhood of wounding a model. While practically there are never 100% guarantees, statistically speaking there are.

Yes, Axyl, that is the correct rending formula.

And, Shrike, you can never, ever have a 100% probability. Yes, it gets very very likely the more dice you throw, but you can never be sure. 100% - (the chance not to wound)^(number of rolls) is correct. Yes, the (1-p)^N gets very small, but it will never actually reach zero.

Shrike78:

The notion is to compare the combat potential of weapons and models. This potential is not merely by the expected value of their attacks, or averages, as despoiler52 puts it, but by the maximum potential values of attacks in relation to the reliability of those attacks.

In theory, there's the average, and then the margins of error, or deviations from the average. Given the small sample of dice rolled in a game of Warhammer 40k then predictions based solely on expecting the average are prone to error when the results of the dice deviate from the expected average. You'll miss something like 2/3 of the bell curve.

But that's okay, because there's hard and fast limits on what might happen. Potentially, thanks to the Warhammer system, the floor is always 0, and the ceiling is limited by the number of attacks or shots.

So we weight the expected value only by the maximum potential ceiling, because the floor is the same in all calculations and can be safely ignored. This gives you a value, the greater the better.

There's a number of objections to doing this:

(1) If the expected value happens most of the time, why worry about an unlikely potential?

A: The answer to this is that you should worry about an unlikely potential because Warhammer is about capitalizing on good luck as much as it is about relying on average luck.

(2) This over-values effectiveness over reliability!

A: No, this values potential only so much as increases in potential outweigh losses in reliability, rather than over-valuing reliability as the expected value does by ignoring the unlikely in favour of the likely when both are a part of the game.

(3) This doesn't tell you how the weapon should be expected to perform!

A: No, that's what the expected value does. The weighted expected value lets you compare the value of a weapon to its comparative price when determining a model/unit/army configuration. After all, this is how spam works: you throw out enough shots or attacks and something will stick, and sometimes it all sticks. If you're looking for more familiar numbers, divide by 2, but be careful not to confuse the ensuing number with the expected value as it will over-state the reliability of the weapon.

(4) You count the number of attacks/shots twice!

A: Sure, but we also count the likelihood of both the average and the maximum potential occurring. It's just that the maximum potential always occurs when it occurs, by definition.

The ratio of expected value to weighted expected value is probably indicative of something, though what I haven't figured out yet. I find the weighted expected value reflects the notion that "Quantity has a quality all its own".