Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I'm practicing penalties. Initially my goalscoring probability is $p$. As I miss a chance, the probability increases by constant value $q$. So the probability that I score my first goal at $n$-th take is

\begin{align*} p_n &= [p+(n-1)q] \prod_{k=0}^{n-2}{(1-p-kq)} \\ &=nq^n\left[\frac{p}{q}+(n-1)\right]\prod_{k=0}^{n-2}{\left( \frac{1-p}{q}-k \right)} \end{align*}

And my first goal will arrive at

$$E(n)=\sum_{k=1}^{\lfloor(1-p)/q\rfloor} {kp_k}$$

But how can I sum this? Some other assumptions that may simplify the calculation without trivializing the problem could be made.

share|cite|improve this question
Should this expectation be an infinite sum? I would expect that after $(1-p)/q$ misses, the chance of scoring is 1. – JessicaK Apr 28 '14 at 11:49
@JessicaK my mistake... – arax Apr 28 '14 at 11:54
I need to leave and I was not able to complete the problem, but it may be easier to compute the expectation by conditioning, but I don't think it will be any easier than just calculating the sum numerically. If you let $E_{p}$ denote the expected value for the R.V. $X\sim Ber(p)$. Then $E_{p}(shots| goal) = 1$, and $E_{p}(shots|miss) = 1 + E_{p+k}(shots)$. Repeatedly conditioning will eventually get you to $E_{p+nk}(shots) =1$ since $p+nk$ will equal or exceed 1. Hopefully you can then solve it by going back through the chain. **I do not know if this technique will actually succeed. – JessicaK Apr 28 '14 at 12:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.