Are these results consistent with expected utility theory?

Question

Results show people prefer gamble A over B. The same group of people prefer gamble C over D.

A: $2400 for sure

B: 0.33% chance of $2500, 0.66% chance of 2400, and 1% chance of 0

C: 33% change of $2500 and 67% chance of 0

D: 34% chance of $2400

For gamble A over B I wrote:
U(2400) > (.0033)U(2500) + (.0066)U(2400) + (.01)U(0)
U(2400) > (.0033)U(2500) + (.0066)U(2400)
(.9934)U(2400) > (.0033)U(2500)

For gamble C over D I wrote:
(.33)U(2500) + (.67)U(0) > (.34)U(2400)
(.33)U(2500) > (.34)U(2400)

First, is this correct? Second, what does this mean?

Gamble B being 33% and 66% makes sense, but the problem clearly states .33% and .66%, which is why I'm having so much trouble. Also, I know gamble D is incomplete, but can't you leave (.67)U(0) out anyways?

Okay, turns out my teacher made a typo. You guys are right. Thank you all for you help.

Answer 1

Presumably gamble B is meant to read $33\%$ and $66\%$, else the probabilities do not add to $1$.

Assuming this: Then the statement that the group prefers $A$ to $B$ means that $$U(24)>.33\times U(25)+.66\times U(24)\implies .34\times U(24)>.33 \times U(25)$$

where we have dropped the irrelevant $0's$.

If we assume that $D$ is meant to suggest that $66\%$ of the time you get $0$ then, as you point out, this comes to $$.33\times U(25)> .34\times U(24)$$

Easy to see that these are in conflict, a violation of Expected Utility.

Answer 2

You don't mean .33% and .66% in B, you mean 33% and 66%. So all of your .0033 and .0066 numbers should read .33 and .66. That gives for the first comparison $$ 0.34 U(2400) > 0.33U(2500) + 0.01U(0) $$

For your second, as pointed out, you forgot something in the statement of (D); probably it should have been 0.34U(2400)+ 0.66U(0). $$ 0.34U(2400)+ 0.66U(0) < U(2500) $$

Then we can combine the two observations to say $$ 0.33U(2500) + 0.01U(0) < 0.34 U(2400) < U(2500) - 0.66U(0) \\ $$ Thus $$ 0.33U(2500) + 0.01U(0) < U(2500) - 0.66U(0) \\ 0.67U(2500) > 0.67 U(0) \\ U(2500) > U(0) $$ and $$ U(2500)-U(0) = [ U(2500) - 0.66U(0)] - 0.34U(0) > 0.34 U(2400) - 0.34U(0) \\ \implies [U(2500)-U(0)] > 0.34 [U(2400)-U(0)] $$ and also $$ 0.34 U(2400) - 0.34U(0) > 0.33U(2500) +0.01U(0) - 0.34U(0) \\ \implies .34 [U(2400) - U(0)] > 0.33[U(2500)-U(0)] $$

The meaning is that the utility of \$2400, relative to that of \$0, is somewhere between $\frac{33}{34}$ and $1$ times the utility of \$2500, relative to that of $0.

We could take the utility of \$0 to be $0$, as you did above, and simplify our statement to say that the utility of \$2400 is between $\frac{33}{34}U(2500)$ and $U(2599)$.

What we see, according to this experiment, is that there is at least some interval between zero and \$2400, where the marginal utility of extra money is greater than the marginal utility averaged in the interval between \$2400 and \$2500.