Eating cake probability problem

Question

Question: We want to divide a cake between Mike and Jack. First, Mike takes a portion of the cake. Then, Jack takes a piece of the remaining cake. Both of the pieces are random variables uniformly distributed over what is currently available. What is the probability that Jack gets more than half of the cake?

Here is my approach:

$X_M \sim Uni(0,1)$

$X_J \sim Uni(0,1)$

$P[ \ X_M(1- X_J) > 0.5 \ | \ X_J] = 1 - P[ \ X_M(1- X_J) \leq 0.5 \ | \ X_J]$

$ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - F_{X_m}(\frac{0.5}{1- X_J} | X_J)$

$ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

At this point I don't know how to proceed and I am not sure if I have modelled the problem correctly. Any help would be appreicated.

Answer 1

To make the algebra a little easier, let $X$ be the fraction that Mike doesn't eat (which is still $\mathcal U[0,1]$) and $Y$ the fraction that Jack eats. We want $\mathbb P(XY>1/2)$. Now do the integral: $$\mathbb P(XY>1/2)=\int_0^1\mathbb P(Y>1/2x)\cdot f_X(x)\mathop{}\!\mathrm{d}x=\int_{1/2}^1\left(1-\frac{1}{2x}\right)\mathop{}\!\mathrm{d}x=\frac{1-\log2}{2}.$$

Answer 2

If $X_M$ and $X_J$ are i.i.d. $U(0,1)$, then $P(X_M(1-X_J) > 0.5)$ seems to be what you need; I am not sure why you are conditioning on $X_J$.

Since the joint distribution of $(X_M, X_J)$ is uniform on the square $[0,1] \times [0,1]$, you can compute the probability by computing the area of the region $\{(x,y) \in \mathbb{R}^2 : x(1-y) > 0.5\}$ that intersects the unit square $[0,1] \times [0,1]$. The region looks like this and the area can be computed with an integral.

Answer 3

Here is another approach, using conditionals.

Notice how $X_M\sim \mathcal{U}(0,1)$ while $X_J|X_M \sim \mathcal{U}(0,1-X_M)$ You're looking to compute $P(X_J>1/2)$. We get $$P(X_J>1/2)=\int_0^{1/2}P(X_j>1/2|X_M=t)f_{X_M}(t)dt=\int_0^{1/2} \int_{1/2}^{1-t}\frac{1}{1-x}dxdt=\frac{1-\ln(2)}{2}$$

Answer 4

Let the cake be a circle.

Suppose the entire circle is split up into n sectors, each making an angle $d\theta$ with the centre of the circle. Then the probability that Mike eats all of $\theta$ and up until somewhere within the small sector of the cake marked on the diagram as angle $d\theta$ is equal to $$\frac{d\theta}{2\pi}.$$

We need to calculate the probability that Jack eats > half the cake, given that Mike falls into a sector $d\theta$, and then do this for every sector $d\theta$ all the way round the circle. This probability is

$$\approx \frac{\pi - \theta}{2\pi - \theta}\ \cdot \frac{d\theta}{\pi} \quad \text{ if }\ 0 < \theta < \pi,$$ $$\quad =0 \qquad \qquad \text{ if }\ \pi < \theta < 2\pi.$$

So $P(\ J$ eats $>$ half cake)

\begin{align} =\lim_{n \to \infty} \sum_{i=1}^{n}\frac{\pi - \theta_i}{2\pi - \theta_i} \cdot \frac{d\theta_i}{\pi} + \ 0 \\ \\ = \frac{1}{\pi}\int_0^\pi \frac{\pi - \theta}{2\pi - \theta}\ d\theta \\ \\ = \frac{1}{2\pi}\int_0^\pi 1 + \frac{\pi}{\theta - 2\pi}\ d\theta\\ \\ = \frac{1}{2\pi} \left(\ \pi + \pi \Bigl[ \ln \left|\theta - 2\pi\right| \Bigr]_0^\pi\ \right)\\ \\ = \frac{1 - \ln 2}{2} \end{align}