# How to price a risk reversal for common dice gain with chance to re-roll

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?