Whoosh..

[mugger2006]

Homework: if a shot has x% chance to hit, how many shots do you have to fire to ensure y% to kill?

Hi there,
So as we saw before if you got a 20% chance to hit and you fire 3 shots you get a 0,8^3=0,512 chance not to hit.

So if you got a 20% to hit and you want to have a 90% hit rate you have to consider the following: 0,8^x=0,1. Since you want a 10% chance of not hitting which means you got a 90% chance to hit at the end.

You need to calculate the logarithm:
x=log[0,8](0,1) [=base], (=number) which you calculate with the natural logarithm (ln) like: ln(0,1)/ln(0,8)~10,32.
So you need about 10 shots. But as it's all about probability your first shot may hit.
So this only tells us that 9 out of 10 times you fire 10 shots at a target with a hit chance of 20% you'll score at least one hit.

To keep it short the more you'll fire the better your chance to score a hit.

As for spike:
when he fired 6 PIAT shots (with a hit chance of 20%) at my Panzer IV H he had a probability of 0,8^6 (~26,2%) of not hitting after the 6th shot. So if he retrys it 10 times with same conditions he will hit my tank at least roughly about 7 times (of course he can score more hit then this...)

But as often said that's war....:cool: