[Left of West]

Expected value is actually just a property of a random variable. I'm truncating the distribution of the random variable, and the expectation of that new is distribution is exactly what i'm saying it is.

My random variable is "the outcome of a squad firing at a carnifex", your random variable is "the outcome of 1 gun firing at 1 carnifex", so with an infinite supply of carnifexes (in a single squad) for you, your expectation is fine.

I see what you're saying, but you're still off.

In actuality, the number I'm is the outcome of a squad firing at a single carnifex. Not just one gun at one 'fex, but any number of guns at a single 'fex. While it's true that wounds beyond the fourth are "wasted," this isn't relavent in the context of a single squad shooting at a single fex.

Where it becomes relavent is, as you point out later, when multiple squads are shooting at multiple 'fexes. You give the following scenario:

"Say we had a 10 man fire dragon squad firing at a carnifex. We calculate the state of the carnifex (ie 4 wounds, 3 wounds, 2 wounds, 1 wound, or 0 wounds) after all 10 fire dragons fired, and repeated this 1000 times. If you were to take 4 minus my expectation you would get the average number of wounds remaining on all the carnifexes (or a very very close number). "

This is certainly true, but the method you use here is correct when the event in question is a single squad shooting at a single 'fex--and you want expected value for a set of such events.

It is not correct for an event which is a single gun shooting at a single fex and what you want is expected value for a set of those events.

So, I do see what you're saying. If you're calculating a certain thing, namely expected results for a series of non-binary events (i.e. several squads firing on several carnifices) you'll need to do things differently. You have to have some way to prevent "rollover" wounds from affecting the outcome.


When you're only interested in one squad and one 'fex, though (as is virtually always the case) you don't have to worry about roll-over wounds. In fact, eliminating roll-over wounds isn't even accurate, since, as I said, those wounds do occur--even if they have no actual effect on the game.

The method you propose is indeed correct for a certain type of evaluation, but it is roundabout and indeed inaccurate when dealing with the situation that most people are interested in: namely a single set of weapons all firing at a single target.

To put it another way, you're finding expected value in terms of the 'fex's state. When I do 'mathhammer' that's not what I'm finding. I'm finding expected value in terms of the number of wounds done. While you have demonstrated that the two are different (imagine that) and that the former requires a different method than the latter (go figure) it remains a fact that, to find the expected number of wounds inflicted, my method is the correct one. Oddly, the expected number of wounds inflicted will not, because of those differences in method, dovetail perfectly with expected number of wounds remaining.

That isn't a reason to ditch the one in favour of the other, though, and it remains the case that nothing you've said justifies your initial suggestion that finding expected value in terms of wounds can't be done properly in the method I support. Clearly, it can.

 

[Moogash]

"In actuality, the number I'm is the outcome of a squad firing at a single carnifex. Not just one gun at one 'fex, but any number of guns at a single 'fex. While it's true that wounds beyond the fourth are "wasted," this isn't relavent in the context of a single squad shooting at a single fex."

Yes. Yes it is relevant...

"When you're only interested in one squad and one 'fex, though (as is virtually always the case) you don't have to worry about roll-over wounds. In fact, eliminating roll-over wounds isn't even accurate, since, as I said, those wounds do occur--even if they have no actual effect on the game."

Yes. Yes you do have to worry about roll-over wounds

Again, I will say:

Suppose we had a 10 man fire dragon squad firing at a carnifex. We calculate the state of the carnifex (ie 4 wounds, 3 wounds, 2 wounds, 1 wound, or 0 wounds) after all 10 fire dragons fired, and repeated this 1000 times. If you were to take 4 minus my expectation you would get the average number of wounds remaining on all the carnifexes (or a very very close number).

Whereas, if we took 10 firedragons each independent, each firing at his own carnifex, and repeated this 1000 times, the average number of wounds done to each group of 10 carnifexes would be equal (or rather very very close) to your expectation.

Your scenario works if we are talking about heavy weapons from different squads, and they have other targets worth shooting at (such as other Fexes) once the first is dead... Mine is for one large target or a squad shooting at one target.

"To put it another way, you're finding expected value in terms of the 'fex's state. When I do 'mathhammer' that's not what I'm finding. I'm finding expected value in terms of the number of wounds done. While you have demonstrated that the two are different (imagine that) and that the former requires a different method than the latter (go figure) it remains a fact that, to find the expected number of wounds inflicted, my method is the correct one. Oddly, the expected number of wounds inflicted will not, because of those differences in method, dovetail perfectly with expected number of wounds remaining."

What I care about is the interpretation through the law of large numbers. Mine has a reasonable interpretation through that, your distribution frankly does not if it is only 1 fex and 1 squad as is largely the case (according to you). The whole point of expectation is that it appears in the law of large numbers and the central limit theorem, other than that there is really nothing special about it.

 

[uber]

it is a point well taken that the expected value can easily be misused.

i used to teach a course on the binomial theorem, and i question i loved to give students on an exam was this:

1 out of 8 cars on the lot is black. if you pick 8 cars randomly, what is the probability of exactly 1 being black?

you can't imagine how many students told me the answer was 1. that actually is impossible.

i just diddled with some numbers, and i noticed a pattern. if you find the expected value to be 1, such as in the test question, it actually translates to around a 36% probability of occurring. what this means is that even though you can expect to pick 1 black car, the actual probability of that happening is just over 36%.

this casts the discussion over destroying falcons in an entirely different light. what you want to know is the smallest number of missiles to fire to give yourself at least a 50% chance of blowing up the target, yeah?

you don't need all your wiki hardware.

all you need to know is that getting at least one successful outcome (blowing up the target) is just the complement of having all failed shots. as soon as the probability of failing all shots drops below 50 percent then you have your answer.

for any binomial distribution the p(all failures) = q^n where q is the probability of a failed trial and n is the total number of trials.

in your example q=25/27

so we get the equation .5= (25/27)^n

taking log of both sides gives us log .5 = n log (25/27) remembering our power rule of logs, of course.

solving for n gives us 8.98. this means that the probability of getting all failures in 8.98 shots is precisley 50%. so if you round up to 9 then you get that the probability of getting all failures in 9 shots is less than 50 percent.

if you are uncomfortable with using logs, then really all you have to do is figure out what the probability of a failed shot is that just start raising it to higher and higher powers until the value you get is acceptably low to you. whatever the exponent you stopped at is the number of shots you will need to shoot to get the probabilty of at least one success. if you picked a high enough exponent this probability should be over 50%. after all you don't want to shoot a shot that you are odds on to miss. right?

i still think expected values are useful, but keep in mind that they can be overkill. in the previous example the expected value came out to be 13.5. this means that you would have to shoot on average 13.5 times before you destroy the falcon. the actual probability of getting at least one destroyed falcon in 13.5 shots is about 65%.

the moral of the story is that if you just play on the expected value then you will be fine. worrying about smallest number of shots to ensure a 50% probability of success is a fun little exercise, but outside of us mathematicians i don't think you need to worry about it.

 

[uber]

and i didn't use any truncation in any of my calculation. since attacks are done by the set, and not the model. i fail to see any need for any truncation.

 

[uber]

i think this discussion has been useful in showing that using the expected value can be misleading, but i really think that most of this discussion has been unnecessary.

the crazy gun example posed earlier leads to peculiar results if you think about what happens when you fire it once. but expected value is really designed to be the mean of a data set.

think about this, say you have a squad of 100 soldiers who all have the option to either take the crazy gun that 90% of the time does pretty much nothing and 10% of the time causes 1000 wounds. using your truncated calculation would suggest that it would be foolish to arm them such. but the truth is that the expected value is vindicated here as on average every time the soldiers fire you will cause 10,000 wounds. that is pretty damned awesome. a gun with a lower expected value, but with a higher truncated value would look better to you but would actually be worse in this example.

the point is that expected values have to be understood in their proper context.

if you want to gauge the wisdom of using a certain gun, i don't really see the benefit of consulting the expected value. what i do is look at the number of wounds per shot. the higher the value the better the choice.

ex: plasma gun vs melta gun. the melta gun is stronger and has lower AP. the plasma gun gets two shots and can overheat. since overheating only can happen on a miss it has nothing to do with how many wounds it can deliver.

both guns hit the same, but melta guns have a 16% better chance of wounding than plasma. taking .66x.84=.55 which is the number of wounds per shot of melta against a T4 target. .66x.66=.43 is the number of wounds per shot of plasma, but since plasma is rapid fire we double that to get .86. thus for plasma is better at killing T4 targets than melta.

once you know what your goal is then it really isn't all that hard to figure out what the best tool for the job is.

i have yet to encounter a moment in a game where i needed to know more than that.

and in case you are wondering i have never lost a game.

 

[uber]

"Therefore 3. Evaluating a binary event as suggested in 2b results in an expected value equal to the probability of getting a success."

actually this isn't right. the expected value is E(X), and it is not equal to the probability of getting a success, p. the expected value is defined as you said. what it means though is the average number of successes you can expect to get for a given number of trials. i don't think you meant what you actually wrote, because everything else you said is correct.


as to the whole issue that started this debate (truncation), it only has meaning if you resolve each shot separately, not if you fire each shot separately. until saves are rolled, the carnifex still has all its wounds available. you are only disallowed from further shots if it is dead. thus the reall issue is do you roll saves as the dice come out, or do you wait until all attacks are rolled. the book is quite clear on the order. the rules are designed to reflect reality, and since one guy doesn't wait until his buddies bullets hit...neither will the models. they all fire at the same time, and whatever happens after the bullets leave the barrels is up to fate.

also, i looked at that article, and it wasn't very helpful as only the abstract was available. there is simply no way to determine if it is modelling what we are talking about.

 

[uber]

Suppose we had a 10 man fire dragon squad firing at a carnifex. We calculate the state of the carnifex (ie 4 wounds, 3 wounds, 2 wounds, 1 wound, or 0 wounds) after all 10 fire dragons fired, and repeated this 1000 times. If you were to take 4 minus my expectation you would get the average number of wounds remaining on all the carnifexes (or a very very close number).
doing this experiment would give you a data set with 1000 elements all ranging between 0 and 4. simply adding them all up and dividing by 1000 would give you the average number of wounds on the carnifexes. there is no reason to take 4 minus the expected value.

Whereas, if we took 10 firedragons each independent, each firing at his own carnifex, and repeated this 1000 times, the average number of wounds done to each group of 10 carnifexes would be equal (or rather very very close) to your expectation.

this answer would be exactly the same as the former part. wounds are independent events. the occurrence of a wound is independent of how the rolls prior to it have turned out. the probability of the 10th shot scoring a wound is exactly the same as that of the first shot. the whole reason that the binomial formula works at all this case is because of this. you seem to keep falling back on the fact that a carnifex can only suffer 4 wounds. so what? it's possible that the carnifex may have to make 25 saves off a given attack. if you want to compute the probability of an unsaved wound, then you have the problem of what to do after you have scored the 4th one, but that isn't what the other guy is talking about. make sure everybody is having the same discussion. are you now asking how many shots should you fire to have a 50% chance of killing a carnifex? that is very different from the falcon question