Ms. A selects a number $X$ randomly from the uniform distribution on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers. 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 $\frac{X}{2}$, then stops. The expected number of draws that Mr. B
makes equals?
The answer to this is $2\ln(2)$. Could someone post the explanation? I got the result as $1.333...$ which is a bit off.
Edit1: I assumed that since it is a uniform distribution, the expected value of the number drawn by A will be o.5. The probability of Mr. B drawing a number bigger than 0.5/2 is then 0.75, which gives expectation of number of draws by Mr. B to be 1.333..

Given $X=x$, B draws a large enough number if they draw something between $x/2$ and $1$, which occurs with probability $1-x/2$. Using the expectation of the geometric distribution, you find that then the expected number of trials is $\frac{1}{1-x/2}$. 
Now we need to deal with the fact that $X$ is itself random. The density of $X$ is $1$ on $[0,1]$ and zero elsewhere. Using the total expectation formula, the expected number of trials is
$$\int_0^1 \frac{1}{1-x/2} dx.$$
I think you can probably work from here.
