30 November 2012

Training

In the post below this I laid out how GURPS modeled the unskilled shooter pretty well.

What happens to that scenario if the shooter spends a point on Guns/TL8 (Pistol).

In GURPS that's 200 hours of training and instruction per point.  Or less than four hours a week.  It's all instruction and practice with a skill, not just range time.

Guns/TL8 (Pistol) is an Easy skill based on DX.

One point gives a level equal to DX.  So a skill of 10.

That means our shooter will have under a 50% chance of landing one round per turn.  In the convenience store scenario, that means one shot out of the five should hit the perp and there's a chance of getting four to land if he rolls well.

Untrained.  3d6 a 5 or less hits.  With Rcl 5, you need a roll of 0 for two hits; thus a second hit is impossible.
1 point.  A 9 or less hits and 4 or less get two hits.

The probabilities are harsh.

Chance to Roll Result or Less on 3d6...
3   = 0.4629%
4   = 1.8518%
5   = 4.6296%
6   = 9.2592%
7   = 16.2037%
8   = 25.9259%
9   = 37.5000%
10 = 50.0000%
11 = 62.5000%
12 = 74.0740%
13 = 83.7962%
14 = 90.7407%
15 = 95.3703%
16 = 98.1481%
17 = 99.5370%
18 = 100.0000%

Our unskilled shooter has a mere 4.6% chance of scoring a single hit per turn.  The skilled shooter has 37.5% to land one shot per turn and a slim but attainable 1.85% chance of getting a second hit per turn.

I've forgotten my statistics to tell what the percentage is of any hits at all over two turns.  I want to say they're additive, but I could easily be wrong.  Time to dig out the book! 

EDIT

WRONG!
I THINK...

Unskilled shooter has a 9.2592% chance of landing one round and a mere 0.2143% of landing them both.

The skilled shooter is a lot more complicated.  I think it's 75% to land one round in two turns with five shots, 14.0625% to get two hits, 0.6944% to get three shots and a whopping 0.0343% chance to get four.

Geff gives me this:

First, let's define some notation, so the math is easier to express.
P(Y) means the probability of exactly Y hits after it's all over.
P(X,Y) means the probability of exactly Y hits during turn X only.

P(0) = P(1,0)*P(2,0)
P(1) = P(1,0)*P(2,1) + P(1,1)*P(2,0)
P(2) = P(1,0)*P(2,2) + P(1,1)*P(2,1) + P(1,2)*P(2,0)
P(3) = P(1,1)*P(2,2) + P(1,2)*P(2,1)
P(4) = P(1,2)*P(2,2)

Unskilled shooter hits on 5-:
P(X,0) = 0.953703
P(X,1) = 0.046296
P(X,2) = 0.00
P(0) = 0.909549 or 90.95% of missing altogether in two turns.
P(1) = 0.088305 or 8.83% of landing one hit over two turns.
P(2) = 0.002143 or 0.20% of getting two hits over two turns.
P(3) = 0.00 or no chance of getting three hits over two turns.

Skilled shooter hits on a 9-:
P(X,0) = 0.625
P(X,1) = 0.3565
P(X,2) = 0.018518
P(0) = 0.390625 or 39.06% chance of a clean miss.
P(1) = 0.445603 or 44.56% chance of getting one hit.
P(2) = 0.150227 or 15.02% chance of scoring two hits.
P(3) = 0.013203 or 1.32% chance of getting three strikes.
P(4) = 0.000686 or 0.07% chance of connecting four times.


UPDATE:

CorvetteMinivanJohn refused to help with this evolution.

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