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I really don't know how to interpret the graph. Can someone help me?

I thought of doing 0.6253+0.3751 to find the expected value of the lottery but where is the sure bet?

  • $\begingroup$ @bakerstreet Can you check this out plz? $\endgroup$ Commented Dec 12, 2022 at 11:41
  • $\begingroup$ The bold line is an indifference curve. To answer the questions try to find the equation for the indifference curves. What do you notice about the slope? How would it change based on risk aversion? $\endgroup$ Commented Dec 15, 2022 at 3:08
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    $\begingroup$ @LudwigNagasena you said indifference curves as in plural, where is the other one? The problem is I have no idea how I can find the equation. Any tip? $\endgroup$ Commented Dec 15, 2022 at 12:39
  • $\begingroup$ You can deduce other curves from the equation. Indifference curve is just a curve that solves $U(p1, p2) = c$ for each constant $c$ where $U(p1, p2)$ is expected utility. I suggest to try to work it out yourself, but if you still struggle, there is an explainer (just the first two paragraphs would suffice): econport.org/econport/request?page=man_ru_advanced_icurves $\endgroup$ Commented Dec 15, 2022 at 21:05
  • $\begingroup$ What concerns these indifference curves, understand that all other indifference curves are parallel to the one illustrated in the graph. The indifference curves to the North-west bring you more utility than those in opposite direction. $\endgroup$
    – Athaeneus
    Commented Dec 16, 2022 at 8:52

1 Answer 1


the first thing to start with is to realize how the graph works. Any point inside the graph gives you three probabilities: $Prob_1$, $Prob_2$ and $Prob_3$. Remember that while you only see the first and the last, the $Prob_2$ is given as the probability remaining: $Prob_2 = 1 - Prob_1 - Prob_3$ and in the graph you can compute it as the distance from the hypotheneuse (see attached graph below):

Triangle diagram

What you can do in your situation then is to count the expected income for extreme cases. You know that in first extreme case you get with $Prob_3 = 1/2$ the income $I = 3$ and with $Prob_2 = 1/2$ the income $I = 2$.

This way you can count both extreme cases:

$$ E[I_{right}] = \frac{5}{8} * 3 + \frac{3}{8} * 1 = 2,25 $$

$$ E[I_{left}] = \frac{1}{2} * 3 + \frac{1}{2} * 2 = 2,5 $$

What you can see is that the right case gives you less expected income than the left one but they are both prefered the same way. And here's the thing: The right one is more dramatic (either you get the best prize or the worst), whereas the left one is more modest (lower probability of getting the best prize but on the other hand you do not ger the worst one either). Since the right one has smaller expected income AND it is more dramatic AND the consumer is indifferent between this and the other one, it means that he is RISK LOVER. In other words, he is willing to pay something for the adrenaline coming from more risky bet. One short remark: You can compare the slope between expected-iso income line and utility function to come to the same conclusion.

What concerns the second question, it tells you the same story, John prefers the right case to obtaining $2.50$ for certain (which is the expected income from the left case btw). Based on the first information given, I tried to fit the coefficient $k$ of his utility function which I assumed to be $U(I) = I^k$, while the coefficient I obtained was around $k = 3.2$. Here it is important to note that this might be biased solution because I do not know exactly his utility function $U(I)$ but for the information given it might be all right...

Then what you do is to compare:

$$ A = (2)^{3.2} \approx 9.2 $$

$$ B = 0.35 * (2.5)^{3.2} + 0.65*1 \approx 7.2 $$

Which results in a lottery A to be prefered.

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    $\begingroup$ I can't thank you enough. But what do you think about this answer tho? freeimage.host/i/Hob092f $\endgroup$ Commented Dec 16, 2022 at 18:02
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    $\begingroup$ Actually, it is not that different from the answer I have provided (the answer to the first question is esentially what I have done but without the intuition behind). However, what concerns the answer to the second question, I must admit I really like the trick with the normalization since it helps you to avoid fitting some function you just assume to be appropriate... In short, I find their approach in the second part to be objectively better than mine (in this case). $\endgroup$
    – Athaeneus
    Commented Dec 16, 2022 at 18:16
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    $\begingroup$ OK. I really appreciate it! Have a good day! $\endgroup$ Commented Dec 16, 2022 at 18:28
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    $\begingroup$ @Disintegrators I have just realized that previously I have swapped the names of lotteries. Not that it changed the idea and procedure but could make some mess. Therefore, I have corrected it so now my solution completely corresponds to your question (unified marking). $\endgroup$
    – Athaeneus
    Commented Dec 16, 2022 at 21:15

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