Option Pricing on a Dice Product
The value of an underlying asset $S$ is determined by the product of two independent rolls of a fair 100-sided die (faces $1$ through $100$). You want to price a European call option with strike $K = 9700$ that expires immediately.
What is the fair price of this call?
Hints
- The product $D_1 \cdot D_2$ can only exceed $9700$ when both dice roll in the high 90s. Start by figuring out which values of $D_1$ can possibly contribute.
- For each eligible $D_1$ value, find the minimum $D_2$ such that $D_1 \cdot D_2 > 9700$. This is $j > 9700 / i$.
- There are only 8 pairs $(i,j)$ with $ij > 9700$. Enumerate the payoffs for $i \in \{98, 99, 100\}$ and sum them up.
Worked Solution
How to Think About It: This is a discrete expected-value calculation. The option pays $\max(S - 9700, 0)$ where $S = D_1 \cdot D_2$ and each $D_i$ is uniform on $\{1, \ldots, 100\}$. The maximum possible product is $100 \times 100 = 10000$, so the option is deep out of the money and it only pays when both dice roll very high. The key is to enumerate the small number of $(D_1, D_2)$ pairs where $D_1 \cdot D_2 > 9700$.
Quick Estimate: We need $D_1 \cdot D_2 > 9700$ out of $10000$ equally likely outcomes. The product only exceeds $9700$ when both dice are in the high 90s. For $D_1 = 100$, we need $D_2 > 97$, giving 3 outcomes. For $D_1 = 99$, we need $D_2 > 98.0$, giving 2 outcomes. For $D_1 = 98$, we need $D_2 > 98.98$, giving 2 outcomes ($D_2 = 99$ and $D_2 = 100$, since $98 \times 99 = 9702 > 9700$). By symmetry, each of these also counts when the roles of $D_1$ and $D_2$ are swapped. The payoffs are small (at most $300$), and there are only about 8 favorable pairs. Rough guess: total payoff around $1000$, divided by $10000$, so the price is around $0.10$.
Approach: Enumerate all pairs $(i, j)$ with $ij > 9700$.
Formal Solution:
The fair price is: $$C = E[\max(S - 9700, 0)] = \frac{1}{10000} \sum_{i=1}^{100} \sum_{j=1}^{100} \max(ij - 9700, 0)$$
We need $ij > 9700$. For a given $i$, we need $j > 9700/i$.
- If $i \leq 97$: $9700/i \geq 100$, so no valid $j$. No contribution.
- $i = 98$: $9700/98 = 98.98\ldots$, so $j \in \{99, 100\}$.
- $j = 99$: payoff $= 98 \times 99 - 9700 = 9702 - 9700 = 2$
- $j = 100$: payoff $= 9800 - 9700 = 100$
- Subtotal: $102$
- $i = 99$: $9700/99 = 97.98\ldots$, so $j \in \{98, 99, 100\}$.
- $j = 98$: payoff $= 9702 - 9700 = 2$
- $j = 99$: payoff $= 9801 - 9700 = 101$
- $j = 100$: payoff $= 9900 - 9700 = 200$
- Subtotal: $303$
- $i = 100$: $9700/100 = 97$, so $j \in \{98, 99, 100\}$ (since we need $j > 97$, i.e., $j \geq 98$).
- $j = 98$: payoff $= 9800 - 9700 = 100$
- $j = 99$: payoff $= 9900 - 9700 = 200$
- $j = 100$: payoff $= 10000 - 9700 = 300$
- Subtotal: $600$
Total payoff sum: $102 + 303 + 600 = 1005$.
$$C = \frac{1005}{10000} = 0.1005$$
Answer: The fair price of the call is $\$0.1005$.
Intuition
This is a clean example of pricing a derivative by direct computation of $E[\max(S - K, 0)]$. The trick is recognizing that a deep out-of-the-money option on a bounded discrete distribution has very few states that contribute -- you do not need any fancy machinery, just careful enumeration.
In interviews, the point of this problem is to test whether you can organize a calculation efficiently. A common mistake is trying to compute the full distribution of $D_1 \cdot D_2$ over all 10000 outcomes. The right move is to start from the constraint $ij > K$ and work out which pairs matter. This mirrors real pricing work: for far OTM options, most of the probability mass contributes nothing, and the price is driven by a small number of extreme scenarios.