Expectation of picking a number from a uniform distribution

Question

Ms. A selects a number X randomly from the uniform distribution on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers $Y_1$,$Y_2$,.... from the uniform distribution on $[0,1]$ until he gets a number larger than $X/2$, then stops. The expected number of draws that Mr. B makes equals (a) 2ln2 (b) ln 2 (c) 2/e (d) 6/e

I tried reaching a solution through integrating from X/2 to 1 to find the probability that a random variable Y would achieve a value greater than X/2. I then defined an RV Z which is equal to the maximum of all Ys picked, and imposed the condition that Z > X/2. But I got a hunch that my method was wrong and stopped midway, I cant come to think of any method either. A solution that I saw uses conditional expectation and the law of iterated expectations, but I'm finding it difficult to grasp in one stretch. This is the link to the solution that I saw,

https://sites.google.com/a/econschool.in/econschool/stuff-of-interest/anotherpost/dse-2014-q22

How do I solve this question, can somebody please help me out with a hint ?

Answer 1

Hint:

If $N$ denotes the number of draws needed by Mr. B then $N$ follows geometric distribution with "random parameter" $1-X/2$.

That gives: $$\mathbb EN=\mathbb E\left[\mathbb E[N\mid X]\right]=\mathbb E \left[\frac{1}{1-X/2}\right]$$

Answer 2

For a fixed $x\in [0,1]$, let $w$ be the expected number of rounds until $y > {\large{\frac{x}{2}}}$.

Then we have the equation $$w=\left(1-{\small{\frac{x}{2}}}\right)(1) + \left({\small{\frac{x}{2}}}\right)(1+w)$$ which yields $$w=\frac{2}{2-x}$$ hence, given that $x$ is uniformly distributed on $[0,1]$, the expected number of rounds is $$\int_{0}^{1}\frac{2}{2-x}\;dx = 2\ln(2)$$