Probability with a mix of speed and time.

[DoMeASolid]

I'm playing a game and this math question goes beyond what I can do.
I have tried to understand the math behind it. Please do me a solid and help me out.

What is the chance and how fast would it take so that I will have exactly 10 critical attacks over the duration of the weapons magazine..?

The weapon in my game has:
31.2% chance of a critical hit.
Firing out 5 bullets a shot.
All bullets can critically hit, but have separate chances of 31.2%. (5 bullets but only one/two may be a critical)
Shooting 12 shots a second.
With a total of 43 shots possible in the magazine.
About 3.5-3.6 seconds overall.
215 possible bullets fired over that time.

 

[DoMeASolid]

Basically I want to know if this is viable to use in game. I need the weapon to critical attack 10 times before it does what I want it to.
I need to know how fast, and at what chance I have that it will critically strike 10 times. I'm sure it is almost guaranteed chance. I mostly would like to know the probability of getting all 10 critical shots within the shortest amount of time.

 

[Romsek]

I don't understand what you mean when you say all 5 bullets in a shot can have a critical hit with probability 31.2% but then say that only 1/2 out of 5 may be a critical hit. Which is it?

 

[Romsek]

I don't understand what you mean when you say all 5 bullets in a shot can have a critical hit with probability 31.2% but then say that only 1/2 out of 5 may be a critical hit. Which is it?

Assuming all 5 bullets are capable of a critical hit we have the number of critical hits in some number of shots is binomially distributed
with parameters [MATH]p=0.312[/MATH], and [MATH]5N[/MATH], with [MATH]N[/MATH] being the number of shots fired. Remember there are 5 bullets per shot.

We cannot compute a number of shots that guarantees 10 critical hits. It's always possible you'll get fewer, random is random, but we can find a number of shots such that the probability you get 10 critical hits is greater than or equal to some value, typically 90, 95, or 99%.

P[\text{at least 10 critical hits out of N shots}] = 1 - P[\text{9 or fewer critical hits in N shots}] = \\

[MATH]1 - \sum \limits_{k=0}^9~\dbinom{5N}{k}(0.312)^k (1-0.312)^{5N-k} = \text{a big mess of a formula in N}[/MATH]
But we can throw technology at it and come up with some numbers.

[MATH] \begin{array}{cc} \text{number of shots} &\text{probability of 10+ crit hits}\\ 8 &0.85\\ 9 &0.93\\ 10 &0.97\\ 11 &0.99 \end{array}[/MATH]