Putnam probability problem Let $X_i$ be chosen uniformly and independently from $(0,1)$ for $i=0,1,2,\dots$
and let $Y_i=X_0X_1^{-1}X_2 \dots X_i^{(-1)^i}$
Find the probability that there exists N such that $Y_N<1/2 $ and $ \forall i<N: Y_i<1$
Any tips on how to start would be welcome, my approach was to try and compute the probability of the complementary event, which is $P(\bar A)= \prod\limits_{N\in\mathbb{Z}_+} P_N$ where $P_N$ is the probability that $Y_N\geq 1/2 $ or $ \exists i<N: Y_i\geq1$ for the integer $N$.
We can see that $P_0=1/2$ , 
$P_1= P(X_1\geq2X_0)$ and 
$P_2= P(X_1\geq X_0 \ or\ 2X_0X_2\geq X_1) $
...
But how to go from here ?
 A: This is a way to do it but not the approved solution:

*

*To fail, the partial product must exceed $1$ before it is below $\frac12$, so we could see this as a game between two players, trying to break the lower boundary or the upper boundary, and they take alternate turns

*If you test it empirically, the probability (starting at the empty product of $1$) is just over $0.59$.

*You do not have to start at $1$ when testing: other values from $\frac12$ to $1$ might be interesting:

*

*Clearly starting at $\frac12$, the probability is $1$

*Starting at $\frac34$, the probability is about $0.76$

*starting at $\sqrt{\frac12}\approx 0.7071$, the probability is just under $0.8$, which is about halfway between the values starting from $1$ and $\frac12$, so perhaps something logarithmic is involved



*Let $q(x)$ be the probability of the player aiming low winning when starting from $x$ where $\frac12 \le x \le 1$. Then using the antisymmetry of the game you can say the first player either wins immediately or can win if the second player then fails to win, giving

$$ q(x)=\frac{1}{2x} + \int\limits_{y =1/2}^x \frac1x \left(1-q\left(\frac1{2y}\right)\right)\, dy  \\= 1 - \frac1x \int\limits_{y =1/2}^x q\left(\frac1{2y}\right)\, dy$$

*

*I do not see an easy way of solving that directly, but making some intuitive guesses based on the empirical observations, and then checking that it satisfies the integral equation, the solution is $$q(x)=\frac{1- \log_e(x)}{1-\log_e(1/2)}$$


*We want the probability of winning, starting at $1$, i.e. $$q(1)=\frac{1-0}{1-\log_e(1/2)} = \frac{1}{1+\log_e(2)} \approx 0.5906161$$


*As a further check, $q(3/4)\approx 0.7605258$ and $q(\sqrt{1/2}) \approx 0.7953081$
