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InyokaMadoda:

A quick question since you apparently have the credentials to answer it and you gave the initial formula I used:

Let n = the number of shots / attacks fired
Let w = the number of wounds you would like
Let p = the probability of scoring a wound (as calculated in your formula)

The formula is then nCw * p^w * (1-p)^(n-w), where nCw is the expression for combinations (n!/(w! x (n-w)!))

What kind of numbers should be entered for n, w, and p? I would assume that n and w would be positive integers (whole numbers like 1, 2, 15, etc for those of you not up on your math lingo) and p would be a decimal in the 0 to 1 range. Is that correct? After “programming” the binomial distribution into excel I got some really odd results if I used decimals for n or w. I'm guessing that's due to them being used in the factorial (!) portions of the equation. Just want to try and make sure I did this correctly if I'm going to end up using it to make 40k decisions

JD21290 wrote:No offence, but this is a pure waste of time
no matter how well done a formula is done it will not predict even a few shots.

While it won't do that great of a job at predicting a few shots, it will do a great job at predicting a lot of shots.

working on probability you can roll a D6 6 times, all formula's say there is an equil prob. of rolling each number, so you should roll 1,2,3,4,5,6.
have a quick go, roll 6 dice and see what happens

Actually, the chances of rolling that number combination on 6d6 is rather low. It's extreamly low if you need to roll them in that order. However, if you roll 600d6 and then count up how many times you rolled each number (and you are using well balanced dice) you'll find that you get very close to 100 of each. Go ahead and try it, it really does work (again assuming you have well made dice).

So I think part of the problem here is that you may not fully understand what the numbers mean. The things we are discussing here are probability, which is the chances of a certain event happening. Again going back to the dice, if you roll a d6, you have a 1/6 chance of rolling any particular number. So what are your chances of rolling a certian combination of numbers (a roll to hit of a certain value or hight, a roll to wound of a certain value or hight, and a roll to fail an armor save of a certain value or lower)? That is the sort of thing we are talking about here. So with this I can figure out how likely a unit is to damage another unit. It doesn't tell me that the first time I try, I'll fail and the second time I try I'll succede or anything like that, it just gives a % chance of it occuring which can then be used as a guidline for play. If GW had used this sort of thing more, we wouldn't have had some really broken things in 40k like holo fields and 4th edition assault cannons.

JD21290 wrote:if you work out that you need 20 shots with a lascannon to destroy a landraider (just picked a number) the chances of killing it in around 20 shots is vague, and doing it in dead on 20, that would rarely happen.

This is very true, but again, its in how you use the numbers. If I know that it takes 20 las cannon shots on average to kill a land raider, but only 10 melta gun shots then I can use that information to more effectively direct my fire power. So if I'm looking to pop said land raider, I'll try to get my melta guns in range and spend more of my time shooting my las cannons at other targets. Conversly, if I really need to destroy that land raider, I will devote every las cannon I have to the task since I understand that its going to take a lot of shots to get the job done.

The other place this information is useful is when trying to figure out the best weapon for a particular target. So here is a question for you, which is better for taking down holo field falcons, a las cannon, an auto cannon, a scatter laser, or a melta gun? Do you know the answer? Without running numbers, I certainly don't. All the weapons function so differently that its difficult to figure it out without some more investigation of some sort. I suppose you could roll out 100 turns of shooting with each and see which one works better but without running numbers, its really hard to tell.

look at battle reports from any WD mag going back as many years as you wish, you will notice in each that there are some stupid rolls, either things like shooting a fex with 5 lascannons and doing nothing, then shooting it with a bolter and causing an unsaved wound.

Again, while this is true, would you rather have 5 bolters to kill a fex with or would you rather have 5 las cannons to try and do it with? You'd rather have the las cannons because they have a significantly higher probability of hurting the fex. So sure, the bolter can (and sooner or later will) hurt the fex, but over the course of many shots, the las cannons are going to be the ones really doing the damage...unless you manage to get a whole lot of bolters in there.

if you can predict a rough idea what the dice will be then gambling wouldnt exist

Funny enough, probability is why gambeling exists. Do you think casinos would still be in buisness if they lost money? No they wouldn't be. The probability of all the games you see in casinos favors the house over the players and over the course of many many games the house earns money on them all. Ever heard of a casino losing too much money on its tables? I certainly haven't. That's how they afford to put up all those huge buildings on the Vegas strip. Its all fuled by probability and peoples lack of understanding about it (or desire to believe that they can beat it). And while some people go home with lots of money in their pockets from winnings, a significantly larger portion go home empty handed. That's probability, not luck.