This is what the USG says:

Quote

three shots. On the other hand, when you use the laws of

probability, you actually come out ahead using an auto-shot,

since each shot is calculated independently.

For example, let's say chance of hitting in auto shot is 25%, and

snap shot is 30% (average). The chance of getting at least one

hit out of three is 1-(chance of no hits). So 25% hit prob AND

no hit in three tries is 0.75 * 0.75 * 0.75 = 0.42

Chance of getting at least one hit is then 1 - 0.42 = 0.56, or

56% which is much better than 30% (but no higher than the aimed

percentage). When your accuracy improves, you increase your

chances of hitting the target multiple times, which ensures a

kill, and you use less TUs than ONE aimmed shot.

Now this assumes each shot being calculated independently but without considering full TU usage shooting, plus the difference in his weapon stats doesn't correspond to the stated values (in most cases snap has more 20% accuracy that auto).

Assuming this and using his math for a laser rifle (but with the UFOPedia stats, not game values), it would go:

0.54 * 0.54 * 0.54 = 0.15 or a 85% chance of hitting with at least 1 shot out of the 3

It beats the odds of snap with laser rifle (65%) right? But what if you consider the unit spending all of its TUs?

0.54 * 0.54 * 0.54 * 0.54 * 0.54 * 0.54 = 0.024 or a 99% chance of hitting with at least 1 shot out of the 6

but with 4 snaps from the laser rifle

0.35 * 0.35 * 0.35 * 0.35 = 0.01 or 99%, but on this case there is a bigger probability of hitting with more than 1 shot (my recent game observations show it).

Is there a way to prove this mathematically? I think the USG is right if you consider that the unit fires either snap or auto but its logic (advice) doesn't apply to when the unit fires several shots each turn.

PS - The firing of 2 auto bursts also allow for an extra snap shot for the laser rifle, which would make the odds more even related to firing 4 snaps.