probability of penalty (in soccer) being scored or being missed I have the following assumptions.
I am expecting 0.4 penalties in a match on average, and I am assuming that penalties follow a Poisson distribution. The probability of a penalty being converted is 82%. 
What is the probability of having a scored in the match?
I know using Poisson pdf that the probability of having no penalty is 0.67032
poisson(λ = 0.4, x = 0)
so the probability of having at least one is the remaining 0.32968
So I guess I need the conditional probability
P(penalty scored / penalty awarded) 
my guess is
p(awarded) * p (converted) for the first case and
p(awarded) * (1 - p(converted)) for the second case of having a missing penalty
What do you think, how should I calculate those conditional probabilities?
 A: Hint:
The distribution of scored penalties in a match can be shown to have Poisson distribution again.
This time with rate: $$0.4\times(1-0.82)$$
A: Hints:


*

*The probability of at least one penalty being converted is 1 minus the probability
that no penalties are converted
$$ P(converted) = 1 - P(\neg converted)$$

*No penalties are converted when all awarded penalties are failures.
$$ P(\neg converted) = \sum_{awarded} P(\neg converted | awarded) P(awarded)$$

*The awarded penalties are Poisson distributed.
$$ P(awarded) = Po(\lambda)$$
EDIT:
Further details


*

*$G$: at least one penalty is converted

*$\neg G$: no penalty is converted

*$C_i$: $i$ penalties are converted

*$\neg C_i$: $i$ penalties are failed

*$A_i$: $i$ penalties are awarded


We have that
$$P(G) = 1 - P(\neg G);$$
note that no penalty is converted if, for any awarded penalty in the match, all the penalties are failures. The events $A_i$ are independent, so
$$P(\neg G) = \sum_{i=0}^\infty P(\neg C_i|A_i)P(A_i)$$
here, $P(\neg C_i|A_i)$ is the probability of failing $i$ penalty kicks; if we assume each kick is independent of the others,
$$P(\neg C_i |A_i) = (1-0.82)^i$$
Using the formula gives a probability of approximately $0.2796$ that at least one penalty is converted. The following small julia code shows that this is indeed the case
julia> using Distributions

julia> simulate(N) = begin
       p= 0.0
       for i in 1:N
           awarded = rand(Distributions.Poisson(0.4))
           if any(rand(awarded) .< 0.82)
               p += 1/N
           end
       end
       return p
       end
simulate (generic function with 1 method)

julia> simulate(10000000)
0.27953870000700837

