I was asked this question at an interview and was stumped by it. The problem is as follows:
Alice is playing an online game. She wins the first round and loses the second. The probability she then wins subsequent rounds is proportional to the number of wins. Calculate the probability that after 100 games, the number of wins and losses are equal.
I am unsure of how to proceed. At the interview, I was originally thinking of using a simple sum of
$$\sum_i \alpha w_i \delta_i,$$
where $\alpha$ is normalisation constant, $w_i$ is the number of wins at the $i$th game and $\delta_i$ is either 0 or 1. However, I doubt this is a correct approach, and $w_i$ is dependent on the $\delta$’s for $j<i$.
I was discussing with a mathematician friend who said that this is a possible variation of Polya’s urn, but I can’t see the connection. Could anyone help me with this visualisation and provide hints on how to move forward? Thank you!
Let $W_i$ and $L_i$ denote the number of wins and losses after the $i$-th rounds. Then, assuming that the probability of winning the $(i+1)$-th round is $W_i/(W_i+L_i)$, \begin{align} &\mathsf{P}(W_{100}=L_{100}\mid W_2=L_2) \\ &\qquad=\binom{98}{49}\frac{W_2(W_2+1)\cdots (W_2+48)\times L_2(L_2+1)\cdots (L_2+48)}{(W_2+L_2)(W_2+L_2+1)\cdots(W_2+L_2+97)} \\ &\qquad=\binom{98}{49}\frac{(1\cdot 2\cdots 48\cdot 49)^2}{2\cdot 3\cdots 98\cdot 99}=\frac{1}{99}. \end{align}