2019
Problem - 4484
Yannick is playing a game with $100$ rounds, starting with $1$ coin. During each round, there is a $n\%$ chance that he gains an extra coin, where $n$ is the number of coins he has at the beginning of the
round. What is the expected number of coins he will have at the end of the game?
Answer
$1.01^{100}$
Assuming Yannick has $N \le 100$ coins at round $k$. Then the expectation of coin count for the next round is $$\frac{N}{100}\cdot (N+1) + \frac{100-N}{100}\cdot N=1.01N$$
This means that the expected number of coins is always $1.01$ times of his current holding regardless of the actual value of his current holding. Therefore the answer is $$1\times 1.01^{100}=\boxed{1.01^{100}}$$