FAQ

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What is sigma (σ) and what does it mean?

Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.

σ ("sigma") is a single number that characterizes precision. In statistics σ represents standard deviation, which is a measure of dispersion, and which is a parameter for the normal distribution.

The most convenient statistical model for shooting precision uses a bivariate normal distribution to characterize the point of impact of shots on a target. In this model the same σ that characterizes the dispersion along each axis is also the parameter for the Rayleigh distribution, which describes how far we expect shots to fall from the center of impact on a target.

Shooting precision is described using angular units, so typical values of σ are things like 0.1mil or 0.5MOA.

Rayleigh distribution of shots given σ

With respect to shooting precision the meaning of σ has an analog to the "68-95-99.7 rule" for standard deviation: The 39-86-99 rule. I.e., we expect 39% of shots to fall within 1σ of center, 86% within 2σ, and 99% within 3σ. Other common values are listed in the following table:

Name Multiple of σ Shots Covered
1 39%
CEP 1.18 50%
MR 1.25 54%
2 86%
3 99%

So, for example, if σ=0.5MOA then 99% of shots should stay within a circle of radius 3σ=1.5MOA.

σ also tells us what to expect from other precision measures. For example, on average a five-shot group has an extreme spread of 3σ. So if σ=0.5SMOA and we are shooting at a 100-yard target we would expect the extreme spread of an average 5-shot group to measure 1.5".


How meaningful is a 3-shot precision guarantee?

The first answer is that any performance standard that is expressed in terms of Extreme Spread is probably meaningless. After all, even the most precise gun will occasionally shoot a wide 3-shot group, and a shot-out barrel will with some probability still put three rounds through the same hole. The inherent precision of a gun is revealed over a large number of shots, and it doesn't matter how those shots are sampled: neither the gun nor the target has a memory. If we shoot 100 rounds through a rifle, we could pick any three at random as a 3-shot group. Therefore, a precision guarantee has to hold over a large number of shots and is more aptly expressed in terms of probabilities. E.g., "99% of shots with our rifle will fall within a ½MOA radius." As we will see in Closed Form Precision such a 99% claim is the same as saying 3σ = ½MOA, or that the rifle's precision σ = 1/6MOA. Which is the same as saying Circular Error Probable = 0.2MOA.

There are two qualifications to the preceding:

  1. We do know that the sample size of a 3-shot group is smaller than the sample size of larger groups, simply because the sample center is almost certainly some distance from the true center. This effect is covered in the section on Correction Factors. For example, the average 3-shot group has a mean radius \(1/\sqrt{c_B(3)} \approx 80\%\) the size of the mean radius measured from twenty or more shots.
  2. It is not exactly true that guns have no memory: The effects of heating and fouling can create second-order changes in both the precision and point of impact. So it might be necessary to qualify a precision guarantee with limits on barrel temperature and/or prescriptions on how many shots be fired before and after bore cleaning. And in the limit barrels erode and have a finite lifespan, so especially with higher-power calibers practical limits do exist on how long precision can stay constant.


What is the best number of shots per group?

Relative Efficiency of Extreme Spread estimation by group size.

Five or six.

If you intend to use "group size" (e.g., Extreme Spread) to estimate precision then you'll spend 13% more bullets shooting 3-shot groups to get the same statistical confidence.

Four-shot groups are only 3% less efficient than five-shot groups, so practically just as good.


How many shots do I need to sight in?

99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.

It is impossible to perfectly align a sight with a gun's center of impact because if there is any dispersion in the gun's point of impact the center can only be estimated. The problem of sighting in a gun is the problem of estimating the location of the center of impact.

We can express the expected distance of our sample center from the true center ("zero") in terms of σ. The more sighter shots we take the better our estimate of the true zero's location and hence the lower our sighting error. This is shown in the following table for sighter groups of various sizes:

Sighter
Group Size
Average Distance
from True Zero
Error at 100 yards
for σ = 0.5MOA
Shots Lost to
Sighting Error
on 50% Target
Shots Lost to
Sighting Error
on 96% Target
3 0.7 σ 0.4" 8% 4%
5 0.6 σ 0.3" 6% 3%
10 0.4 σ 0.2" 3% 1%
20 0.3 σ 0.15" 2% <1%

The number of sighting shots you should take depends on your sensitivity to sighting error (not to mention your patience and supply of ammunition).

The last column in the table might be typical of a hunter, who can hit an area the size of his quarry's vital zone at least 96% of the time. If he only takes 3 sighter shots the resulting sight error would only cause him to miss that vital zone an additional 4% of the time.

The 50% Target column might be more typical of a competitive shooter. For example, if he can hit an area the size of the bullseye with 50% of his shots, but then he spends only 3 shots sighting in for a match, then he should expect to miss the center ring an additional 8% of the time due to sighting error. If he spends 10 shots sighting in then his sighting error falls to 3%, so he would expect one additional bullseye every 20 shots (8% - 3% = 5%).


How do I tell whether A is more accurate than B?

This question appears frequently in various forms. For example:

  1. Is gun A more precise than gun B?
  2. Does ammo load A shoot better than load B in this gun?
  3. Does this accessory affect the precision of my gun?