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These are mistakes which almost everyone makes, and, while often not a big deal in the spur of the moment, when trying to mathhammer something out they can lead to wrong conclusions.
The first problem is that people miscalculate expectation. In a binomial distribution (which is what one normally uses for die rolls) the expectation is n*p, where n is the number of rolls, and p is the probability of success. This is all well and good in theory, but we quickly run into a problem... The problem is that sometimes you factor impossible events into your distribution. If I have 10 marines shoot at a carnifex, in our model they have the possibility of doing 10 wounds, which is impossible, all wounds after the 4th would be truncated, thus you can't actually use a straight up binomial model and you can't really use n*p (though n*p will usually be close).
This leads us into the second problem, which is that people use expectation even though they really shouldn't and are interpreting it incorrectly. It pains me to count the number of times i've heard, "well, i'm expected to do at least 4 wounds to X, so half the time I'll do 4 or more wounds to X." UNTRUE! You are talking about the median not the mean (aka expectation) of the distribution! The median number of wounds is the number (not necessarily unique) such that half the time you will do better, and half the time you will do worse. This solves our first problem because in the truncation example, unless the median of a distribution is in the truncated, it will not be changed by truncation.
The mean minimizes the square sum of the error terms, while the median minimizes the absolute sum, so Expectation is much less robust to outliers (ie if a gun has a 1/1000 chance of doing 5000000 wounds to the target, it's going to totally screw up the mean but do almost nothing to the median)
Now, even with the median people can still misinterpret in a different way! I also hear people say "If I shoot with 10 guys at the Falcon and my expected number of vehicle destroyed results is exactly 1, then the expected number of guys I need to kill the Falcon in one round of shooting is 10." WRONG AGAIN! In order to determine what the expected number of guys needed to kill the Falcon is you need to look at a geometric distribution. You figure out the probability to kill a carnifex with 1 guy, then plug that in for p, and find its expectation (which you probably dont want anyways...) or find the median (which is more useful).
Anywho, all this said, the median is a huge pain in the ass to calculate and usually must be done by simulation, and outliers are usually not that fair out or that heavily weighted (like 100 bolter shots doing 100 wounds to a carnifex) so on the fly expectation is ok, but when people say "I mathematically proved that X is better than Y" they better have their statistics right... (or i'll melee them with my bolt pistol)
Last edited by Moogash; August 2nd, 2008 at 00:55.
Unfortunately you've made a mistake as well.
The only people who will understand what you just said, are the people who wouldn't make the mistakes in the first place!
I'll sum it up here:
Expectation ie. n*p DOES NOT MEAN WHAT YOU THINK IT MEANS!!!
If you are really interested in getting a good 50%-of-the-time estimate, figure out the median of the distribution.
And, if you are interested in "how many units on average do I need to destroy the falcon" look into a geometric distribution.
I have an advanced high school understanding of probability and suchforth (I know, that's not a degree, but it's better than the average).....and while I understand the concepts of your.....wall-of-text.....translating it into something useful is just beyond me.
Maybe you should try to make it more "layman" friendly.
I think you probably need to define all the probability jargon so that everyone knows what they actually refer to.
Mysterious Member of the ANZAC Clan
Since people are having trouble with this let me give an example of using the median and geometric distributions:
Ok, say we want to apply the above to shooting a ML with a marine at a falcon.
For a single ML we have:
2/3*1/3*1/3 = 2/27 = p = probability a single shot will destroy a falcon
Now we want to figure out the number of shots we need to take in order to have at least a 50% chance of destroying the falcon, so we look at a geometric distribution on wikipedia, it tells us that the median is the smallest integer greater than
so we plug that in and get 9, thus in order to have a 50% chance of destroying a non-holofield falcon, we need to shoot 9 shots.
If we used expectation instead of the median, we would have gotten the usual answer of 27/2 or 13.5, which may have a useful interpretation (especially if you are considering total number of shots to kill 2 or 3 falcons), but is not the "50% interpretation" that most use it as.
I agree with you that binomial distribution would be the best way to calculate how much firepower you actually need. I'd say you're going to have issues applying that in a game by game basis however due to the relatively small number of shots. You're not going have a finite variance unless you start adding up data from a long series of games or one very very large one. In either case, your shifting your parameters away from an individual game which may lead to some bias or skewness in the results.
I think this could work well with bolters, or other weapons that are going to have several hundred shots a game. Heavy weapons, however, maybe at best be shot 10 or 12 times a game, and not always at a falcon/carnifex/whatever. It's going to be difficult to show a match between the theoretical probability you calculate and actual probability experienced on and individual game by game basis. At that point, I think the math hammer most people use is just as good, considering the numbers you'd produce would probably within a standard deviation .
Binomial distributions always have finite variance. You don't need to sample anything or estimate any parameters, you can figure out the probabilities of each result by just plugging the probability of a single success into the binomial distribution formula.
Thus, it should work equally well with heavy weapons as it does with bolters.
Though, like I said, a geometric distribution would work much better for tanks (where all you care about is how many shots until something happens, not the expected damage of a single shot).
Also, the numbers that you produce (expectation etc...) are not random, so "standard deviations" don't apply. (though the 2 numbers may still be close)
Truncation of impossibly large numbers is neither necessary nor desirable. In just about any case where you're actually bothering to do math-hammer, expected results against a theoretically infinite target are more useful to a decision making process than expected results against an actual target--not to mention far less work.
If you really need info regarding specific numbers of wounds, and work is not an issue, the method suggested by Onlainari produces considerably more useful information than anything you've suggested here.
Once again, the conservative, sandwich-heavy portfolio pays off for the hungry investor!
the point here is that expectation is often not useful, in cases such as shooting at falcons. Also, if one wants to combine probabilites say kill both a falcon and a wraithlord, using a geometric distribution will be way more useful.
"In just about any case where you're actually bothering to do math-hammer, expected results against a theoretically infinite target are more useful to a decision making process than expected results against an actual target"
making a statement without support isn't an argument. Care to explain why?
Also, the point is that expected results often aren't useful. Expectation is useful insomuch as it is used in the law of large numbers, so for tons of bolters wounding hormagaunts, sure. Not so much for killing a falcon with MLs...
My point re: number of wounds was just that expectation isn't really a great thing to look at for a small number of shots, because it really doesn't say much about probabilities which is what one is really looking at. So sure, look at probabilities or a binomial distribution (same thing...), but the strong law of large numbers doesn't actually say anything about the rate of convergence (and in terms of warhammer orders of magnitude it's not fast...) so expectation is really not all that useful.
furthermore, the median is actually fairly trivial to calculate for a geometric distribution.
Last edited by Moogash; August 6th, 2008 at 06:09.
"the point here is that expectation is often not useful, in cases such as shooting at falcons. Also, if one wants to combine probabilites say kill both a falcon and a wraithlord, using a geometric distribution will be way more useful. "
Sure, I never said it was. What I did say is that for this kind of question, you can either use:
1-(1-p)^n (for situations where only one success is necessary, like the Falcon)
P = p^r * (1-p)^(n-r) * nCr (for situations where more than one success is necessary, like a Carnifex)
P = probability of getting exactly r successes
p = probability of success on a single trial
n = number of trials
and get more detailed information than you would following your suggestion.
For situations in which you need to know how likely you are to get one or a handful of successes (which is what I was talking about when I mentioned specific numbers of wounds) these two formulas are better.
For situations where you just need a rough idea of how many wounds to expect, using the simple expectation calculation (n*p)e right way to go.
It is not correct to include truncation as you suggest because, in fact, your results are never truncated. When shooting at a Carnifex, wounds beyond the fourth don't do you any good--that's certainly true. However, you can get more than four wounds. You can shoot ten bolters at it, get ten hits, ten wounds, and it can fail all ten of its saves--viola: ten wounds.
Truncation is necessary in situations where you would resolve each attack before moving on to the next one--in which you would actually stop rolling after getting the fourth wound, even if you had attacks left. That isn't the case in 40k. You roll them all at once--and in a situation like that it is actually correct to use n*r and actually incorrect to factor in a maximum number of wounds. There is no maximum which it would be appropriate figure in beyond the limit imposed by the amount of shots--and that, obviously, has already been accounted for.
So, actually, your first point is just flat-out wrong and your second point had already been addressed--and better--by Onlainari before you made this thread.
That enough of an argument for you?
Once again, the conservative, sandwich-heavy portfolio pays off for the hungry investor!