I was just thinking about an extension to the common dice throwing interview expected value question:

Question: Imagine a game where you throw a die and get a payoff equal to the number shown by the dice; you are able to re-roll the die once and then your payoff will be the number shown on the second dice. What is the fair-value of:

(a) a call option with strike = 4

(b) a put option with strike = 2

(c) a risk reversal (pointing to the call) with strikes at 2 and 4 respectively (call at 4 and put at 2)

Attempt: In general, I think the expected value for this game can be calculated by doing: $$ E[game] = \frac{1}{2} (3.5) + \frac{1}{2} (5) = \frac{7}{4} + {10}{4} = 4.25 $$ because we will re-roll the dice half of the time (i.e. when we get 1, 2, or 3 on the first roll (this leads to EV of 3.5 for second roll as all outcomes possible); otherwise, we will keep the first roll if we get 4, 5, or 6 (thus an average of 5)

For (a), the call option with a strike of 4. I thought of getting the probability distribution and then thinking about the pay-offs relative to the strike. So for 1, 2, or 3, the probability of ending up with those numbers is when we get them on the re-roll: $ = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12} $. For the numbers 4, 5, or 6, we get those either on the first roll or the re-roll $ = \frac{1}{6} + \frac{1}{2}\cdot\frac{1}{6} = \frac{1}{4} $. As a quick check, the probabilities add to 1 which is a good sign. Now the fair price of the call, I think, is: $$ \text{Fair value of call} = \sum (\text{payoff} - \text{strike})\cdot P(\text{payoff)} = \frac{1}{4} (4 - 4) + \frac{1}{4} (5 - 4) + \frac{1}{4} (6 - 4) = 0.75 $$ Does this seem right?

For (b), following the same method, I got: $$ \text{Fair value of put} = \sum (\text{payoff} - \text{strike})\cdot P(\text{payoff)} = \frac{1}{12} (2 - 1) + \frac{1}{12} (2 - 2) = \frac{1}{12} $$

For (c), then I think I would just do: $ \text{price of call at 4} - \text{price of put at 2} = 0.75 - \frac{1}{12} = \frac{2}{3} \approx 0.67 $

Do these answers seem right? Also, does the fair value of these options have any dependency on being able to (delta) hedge the options?

Thanks in advance

  • $\begingroup$ in a world where a gain of X is exactly as good as a loss of X is bad (i.e. there is no utility function, there is no value ascribed to taking risk, no charges for var, etc.) then the delta hedge is an independent trade and should have no impact on the value. In the real world however, then the ability to hedge will absolutely impact where people are going to be willing to buy and sell derivatives. $\endgroup$
    – will
    Sep 10, 2021 at 12:35


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