IMPERIAL LECTURE SERIES - Calculating Probabilities Using Standard Deviation

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When calculating probabilities of causing mass carnage in 40k, there two main ways that people tend to approach the problem. The first answers the question "What's the probability of causing carnage?" and the other is "How much carnage am I likely to cause?"

The first method, that of calculating probability, basically boils down to finding the inverse of the chance to miss (1 - the chance to hit x the number of trials). The second, that of calculating standard deviation, basically boils down to multiplying your chance for a single result by the number of times you try to get that result. While the former method is the most common, it has a caveat: you can never get above 1 (100%), which means that this model can not be used to calculate the ammount of damage you can expect to do, as it doesn't properly account for the effect of multiple hits the higher the set gets. **

Thus, the more useful model winds up being one that takes into account the average number of kills, rather than the chance to get a kill. This, then, requires us to have a system of displaying chances based on the average, rather than the upper bound. Enter the "standard deviation". The end result of a standard deviation is an expression of the expected value (the average) with a calculated range of "close, but still plausable" results, as calculated by the standard deviation.

The steps basically go like this:

1.) Find the % probability of each possibly outcome as per model #1 (if necessary)

2.) Find the average.

3.) Dake the difference of the outcome and the average.

4.) Square the results from #3.

5.) Find the average of the squared results.

6.) Take the square root of #5.

7.) Add and subtract the results from #6 to the average from #1 (to calculate the bounds)

8.) To other math to taste.

Yes, this is tricky, and isn't likely to be done on the fly (thankfully, there are automated ways of finding the standard deviation out there). Here is a simple example:

Example #1: The heavy bolter

1.) This is a simple example, as, in this case, you are just as likely to hit as to miss (being a guardsman in this case). This means that your probability of hitting and missing are both .5 (50%).

2.) The average number of hits you get, then, is simply taking .5 times the number of shots (3) for an expected 1.5 kills.

3.) Here comes the tricky part. We could get 0, 1, 2, or 3 hits with a heavy bolter. Taking those numbers and subtracting the above average gives us -1.5, -0.5, 0.5, and 1.5, respectively.

4.) When we square these results, we get 2.25, 0.25, 0.25, and 2.25.

5.) Finding the average here requires adding all of the above results times their percent chance (.5) together. In this case, we get 2.5

6.) The square root of 2.5 is 1.58.

7.) This means that our "normal" results should be between -0.03 and 3.08 with 1.5 as the average.

This means that getting either 1 or 2 hits per round of shooting with the heavy bolter falls well within the realm of "normal" (only .5 off out of 1.58 possible). Both getting 0 or 3 hits are also within the realm of "normal", but are precipitously close to being anomylous results. So, complain or be merry when you get no hits, or three, but realise that it isn't abnormal to get either of those results.

Example #2: Hitting a hammerhead

1.) In this case, we are taking a priest (three attacks on the charge) who is trying to hit a skimmer. As he's a priest that's charging, he has the added complication of the ability to re-roll any misses. Doing a probability model, we get a .33 to miss completely, a .44 to hit once, a .20 to hit twice, and a .03 to hit three times.

2.) Finding the average of the above, we get about .93 hits per charge.

3.) The differences of hitting 0, 1, 2 and 3 times are -.93, .0049, 1.07, and 2.07, respectively.

4.) The squares of these are .8649, .0049, 1.145, 4.285

5.) The average (in this case (.8647 x .33) + (.0049 x .44), etc.) is 0.645.

6.) The square root of this is 0.832

7.) This means that "normal" encompasses .13 to 1.73 with .93 as the average.

This means that getting 1 hit is, by far, the most likely result (1 compared to .93 is well within the 0.832 limits). Note as well, that hitting zero times puts you .13 outside of the realm of normal results, as does hitting twice (.23 outside). Thus, you should really, really expect to hit once, and if you don't, you can consider yourself unlucky, or lucky, as the case may be (especially if you hit three times, which has a breach of deviation higher than the deviation itself!).

Final Thoughts:

Note that the above examples only deal with hitting. A similar process goes into wounding. In the case of using the standard deviation model to determine a hit, you simply have to multiply the result by the chance to wound (unlike in the probability model, which gets more confusing, and suffers from the same problem, as you can never guarantee a wound, no matter how many times you hit).

As well, though the process is farily complicated and involving, let me give you a shorthand for calculating approximate answers on the fly:

# of shots x % chance to hit x % chance to wound = the amount you should reasonably expect to kill.

For example, if I'm shooting a heavy bolter at space marines, I take the 3 shots times the half chance to hit (1.5) times the chance to wound (.666 which gives me about 1) times the chance that I make it past an armor save (.333) for about a third of a dead marine. Thus I know that in any one round of shooting, I'm not all that likely to kill a marine, and that if I take 3 heavy bolters against said marines, I should kill about one.

While this doesn't give you a great sense of if you were lucky or not if you didn't nail the average on the head (which is what the standard deviation is there for), it does give you a decent shorthand way of calculating odds when needing to make the tough decisions.

As well, this model does not negate the probability model, but is merely an alternative that is more useful on these types of data sets. In the end, an argument can still be made that it is possible that a million guardsmen with a million tactical nuclear missiles could still fail to kill anyone, the useful tool of standard deviance can reply that that would, indeed, be a deviant result.

** Let's take a simple example. A heavy bolter fired by an imperial guardsmen should score 1.5 hits (50% x 3) as per model 2 (standard deviation). Likewise, the same heavy bolter gunner has an 87.5% of causing a hit as per model 1 (probability). Let's extrapolate this up to 1,000,000 heavy bolters. Model 1 would take the number of heavy bolters and multiply it by the chance for a heavy bolter to hit someone for a total of 875,000 hits. Model 2 would take the number of heavy bolters and multiply it by the average number of hits for a round of shooting with a heavy bolter for a total of 1,500,000 hits.

Note that model #2 provides an estimate nearly twice that of model #1. This gets exacerbated the higher you go up. Let's say that all of my guardsmen are now armed with tactical nuclear missiles ( a "HEAVY 1,000,000" weapon). A million guardsmen with a million tactical nuclear missiles, according to model #2 gets us 500,000,000,000 kills. According to model #1, we have a just barely shy of 100% chance to make a kill times 500,000 guardsmen hitting for a total of 500,000 kills. This discrepency is because model #1 can never get higher than 1 kill per trial, as you can't get higher than 100%.