# Training Bicyclists on a Mountain Road

### Problem ### Solution 1

Let $q=1-p$ be the probability that a bicyclist won't tumble on his own. Then the probability that the bicyclist number $k$ (counting from the head of the line) does not fall equals $q^k.$

Let $I_{k}$ be the indicator function for the event that the bicyclist number $k$ tumbles down:

$I_k=\begin{cases} 1, & \text{if the }k^{th}\text{ bicyclist tumbles}\\ 0, & \text{otherwise} \end{cases}$

Then $P(I_k=0)=q^k,$ so that $P(I_k=1)=1-q^k$ and $\mathbb{E}(I_k)=1-q^k.$

We are interested in the expected number of the bicyclists that eventually tumble, $\mathbb{E}(n):$

\displaystyle\begin{align}\mathbb{E}(n)&=\sum_{k=1}^n\mathbb{E}(I_k)=\sum_{k=1}^n(1-q^k)\\ &=n-q\frac{1-q^n}{1-q}=n-\frac{1-p}{p}(1-(1-p)^n). \end{align}

### Solution 2

$n$ tumble with probability $p$ (after the first does).

$n-1$ tumble with probability $p(1-p)$ (in case the second one tumbles).

$n-2$ tumble with probability $p(1-p)^2,$ and so on.

$1$ tumbles with probability $p(1-p)^{n-1}.$ Summing up

$\displaystyle \sum_{i=0}^{n-1}p(n-i)(1-p)^i=\frac{-1+(1-p)^{n+1}+p+np}{p}.$

### Solution 3

The probability that cyclist $i$ is the first one to fall is equal to the probability that the previous $i-1$ cyclist did not fall, $(1-p)^{i-1}$ multiplied by the individual probability of falling $p$

$P_i = p(1-p)^{i-1}$

If cyclist $i$ falls then the remaining $n-i$ clyclist fall too for a total of $x_i = n+1-i.$ The expected value of $x$ is then equal to

$\displaystyle E(x) = \sum\limits_{i=1}^{n}P_i x_i$

Substitute and split the sum

\displaystyle \begin{align} E(x) &= \sum\limits_{i=1}^{n} p(1-p)^{i-1} (n+1-i)\\ &=\frac{p}{1-p} [(n+1)\sum\limits_{i=1}^{n} (1-p)^{i} -\sum\limits_{i=1}^{n} i(1-p)^{i}] \end{align}

Now we use the following results

$\displaystyle \sum\limits_{i=1}^{n} a^{i} = \frac{a(1-a^n)}{1-a}$

$\displaystyle \sum\limits_{i=1}^{n} ia^{i} = \frac{a(1-a^n)}{(1-a)^2}-\frac{n(1-a^{n+1})}{1-a}$

$\displaystyle E(x) =\frac{p}{1-p} [(n+1)\frac{(1-p)(1-(1-p)^n)}{1-(1-p)} -\frac{(1-p)(1-(1-p)^n)}{(1-(1-p))^2}+\frac{n(1-(1-p)^{n+1})}{1-(1-p)}]$

which yields

$\displaystyle E(x) =n- \frac{1-p}{p}(1-(1-p)^n)$

### Acknowledgment

This is a paraphrase of problem 65352 from a Russian problem collection. Solution 2 is by N. N. Taleb; Solution 3 is by Alejandro Rodríguez. 