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Old November 17, 2012, 01:06 AM   #156
barnbwt
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Join Date: May 17, 2012
Posts: 704
If you run the odds numbers accouting for the fact that every shot you get to take, each of the bad guys gets one as well, you'll see that surviving a multi-attacker encounter based solely on capacity (i.e. number of turns or innings) rapidly becomes statistically impossible. Even if they lack guns, as you spend time expending ammo, and whether or not you even hit, the baddies have the opportunity to incapacitate you; if they are anywhere close to as effective as you are with a gun, you cannot win.

I ran some rough probability numbers and got these for a successful 2 on 1 encounter where everyone has a 60% hit rate, and 2 shots to stop:

1st turn (you shoot and hit one of them, they both get to shoot back since 2 hits are assumed to incapacitate)
~82% probablilty you will receive a shot (whether you hit or not)
2nd turn (you incapacitate Guy 1, Guy 2 gets to a chance to shoot back)
~49% probability you receive a second shot and are incapacitated
3rd turn (you hit Guy 2 and he shoots back)
~74% probability you receive 2 or more shots by this point in the fight
4th turn (you hit Guy 2 and finally end the fight or you miss and he gets a shot)
~97% probability you receive 2 or more shots by this point in the fight

Granted, there are assumptions in this analysis, and granted, any attack is a statistical outlier. The point is, capacity is not the important variable in determining victory, but speed and placement. If you have to take fewer turns (i.e. one shot stops) and get more turns the the other guys, the odds quickly move in your favor. If you simply add the number of innings you could physically prolong the fight, odds of survival rapidly become insignificant.

Quote:
2 hits are necessary to subdue an enemy and the average hit percentage was 1/3 of the time. At this point, the trained officer had taken, on average 6 shots to subdue one criminal.
Odds don't work that way (1:3*2=6 bullets)
The odds of getting two shots of six are:
.3*.3 = .09 (0 miss)
.3*.3*.7 = .063*2 (1 misses, 2 ways)
.3*.3*.7*.7 = .0441*3 (2 misses, 3 ways)
.3*.3*.7*.7*.7 = .0309*4 (3 misses, 4 ways)
.3*.3*.7*.7*.7*.7 = .0216*5 (4 misses, 5 ways)
Grand Total: 58% chance of stopping one guy with 2 of six bullets
TCB

(If it looks like I made an error in these calcs please let me know and I will attempt to rectify the mistake; e-net numbers only have value if they are accurate. I can show the math I used to arrive at these numbers if anyone really cares that much about permutations)
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Last edited by barnbwt; November 17, 2012 at 01:26 AM.
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