Soccer penalty shots

Imagine a scenario where we have a soccer player who is shooting penalty shots. He makes his first shot with probability of failure $1/3$, then his second with probability of failure $1/5$ and so on, and at the $n$-th throw the probability of failure is $1/(2n+1)$.

During practice, he shoots $19$ consecutive kicks. What is the probability he succeeds in an even number of kicks?

For each throw, the respective probability of success is $$1-\frac{1}{(2n+1)}= \frac{2n}{(2n+1)}$$

To find the total probability of success in an even number of throws, we must add the probability of $0$ successful shots, $2, 4, \cdots, 18$.

But the $2$ successful throws can be, for example, the $1$st and the $2$nd, or the $4$th and the $19$th. Then, obviously, we will multiply the $2$. So how do we calculate the individual probabilities?

Any help will be highly appreciated!

The probability generating function for the number of failures is

$$ f(x)=\prod_{n=1}^{19}\frac{2n+x}{2n+1}\;. $$

The probability of getting an even number of failures is

\begin{eqnarray*} \frac{f(1)+f(-1)}2=\frac12\left(\prod_{n=1}^{19}\frac{2n+1}{2n+1}+\prod_{n=1}^{19}\frac{2n-1}{2n+1}\right)=\frac12\left(1+\frac1{39}\right)=\frac{20}{39}\;. \end{eqnarray*}

Since the total number of kicks is odd, the probability of getting an even number of successes is

$$1-\frac{20}{39}=\frac{19}{39}\;.$$

Hint

Try to work out the recursive equality:

$$\left(2n+1\right)p_{n}=p_{n-1}+2n\left(1-p_{n-1}\right)$$

Where the probability that he succeeds in an even number of kicks by taking $n$ penalties be denoted by $p_{n}$.

This is based on the thought that an even number can be achieved in two ways:

• an even number in the first $n-1$ efforts followed by a failure
• an odd number in the first $n-1$ efforts followed by a success.

Here $p_0=1$.