This is an exercise problem.
Suppose $X_1=200$ with probability $1/3$, $0$ with probability $2/3$.
$X_2=200$ with probability $p$, $0$ with probability $1-p$.
$X_3=200$ with probability $1-p$, $0$ with probability $p$.
$X_4=200$ with probability $2/3$, $0$ with probability $1/3$.
I am asked to show that the preference $X_1\succ X_2$, $X_4\succ X_3$ cannot be represented through expected utility for any choice of $p\in[0,1]$.
I tried to write it in the form $X_1\sim p_1\sim1/3\delta_{200}+2/3\delta_0$, etc. and was trying to find a contradiction with the independence axiom by $X_1\succ X_2$ implying $X_3\succ X_4$ using the independence axiom. But I don't know how to deal with the general p term. I'm greatly thankful if anyone could help.
Let $U$ denote the utility function.
Then $X_1\succ X_2$ implies $$\frac{1}{3}U(200) + \frac{2}{3}U(0) > pU(200) + (1-p)U(0), \\\ \Big(\frac{1}{3}-p\Big)U(200) > \Big(\frac{1}{3}-p\Big)U(0).$$
But $X_4\succ X_3$ implies $$\frac{2}{3}U(200) + \frac{1}{3}U(0) > (1-p)U(200) + pU(0), \\\ \Big(\frac{1}{3}-p\Big)U(0) > \Big(\frac{1}{3}-p\Big)U(200).$$
Hence, we have a contradiction.