Counting Dice Rolls by Their Minimum for a Payoff
Summary
The document shows how to count ordered outcomes for three six-sided dice according to their minimum value, then use those counts to reason about the expected payoff of a call-like contract with a strike of three. The counting method starts from rolls whose dice are all at least a given face value, then subtracts rolls whose minimum is higher. For instance, the count with minimum four comes from the rolls containing only four, five, or six, less those already counted with minimum five or six.
It also presents a combinatorial breakdown for each possible minimum, checking that the counts sum to the full set of rolls. The discussion corrects the initial outcome counts and illustrates a general way to partition outcomes by minimum. The contract is described as a probability exercise rather than a standard financial option; the document does not complete the expected-value calculation or address discounting, market pricing, or broader option valuation assumptions.
Key ideas
- Count rolls with minimum m by subtracting outcomes whose minimum exceeds m from rolls where every die is at least m.
- For three dice, the minimum-four count is found from rolls using faces four through six after removing higher-minimum cases.
- A combinatorial breakdown can classify outcomes by the dice faces present and their positions.
- The counts across all possible minima should sum to the full number of ordered rolls.
- The example frames the payoff as a probability exercise rather than a standard traded option valuation.
Tags
Full text
# Option on a dice game with three dices and min. value # Option on a dice game with three dices and min. value We have a call option on 3 dices with strike 3. What's the fair value of the call when it pays the min value of the 3 dices? E.g if we throw and have 426, the min is 2 here and so call is OTM (S < K) 216 permutations (6^3) min(6) = 1/216 x (6-3) = 0.0139 EV (666) min(5) = 8/215 x (5-3) = 0.0744 EV (555 556 565 566 655 656 665 666) min(4) = ? I'm getting lost in the permutation formula to calculate the number of outcomes where the min. comes 4. Can someone help me figure out how to find out the correct # of outcomes with value 4? ## Answer by Valometrics.com (score 2, accepted) https://quant.stackexchange.com/a/50998 There is a simple way to find the number of rolls with the minimum of 4: the number of rolls with the minimum of 6 is 1. The number of rolls with the minimum of 5 is the number of rolls for which all outcomes are 5 or 6 minus the number of rolls with the minimum of 6: 2*2*2-1=7. The number of rolls with the minimum of 4 is the number of rolls for which all outcomes are 4,5,6 minus the number of rolls with the minimum of 5 or 6: 3*3*3-7-1=19 and so on. ## Answer by Attack68 (score 1) https://quant.stackexchange.com/a/50997 This isn't really an option as it is an exercise in probability. How many rolls have a minimum of 6? 1 = 3C3 (6s) How many rolls have a minimum of 5? 7 = 3C3 (5s) + 3C2 (5s6s) + 3C1 (5s6s) How many rolls have a minimum of 4? 19 = 3C3 (4s) + 3C2 (4s5s) + 3C1 (4s5s) + 3C2 (4s6s) + 3C1 (4s6s) + 3P3 (4s5s6s) How many rolls have a minimum of 3? ``` 3C3 (3s) 3C2 (3s4s) 3C2 (3s5s) 3C2 (3s6s) 3C1 (3s4s) 3C1 (3s5s) 3C1 (3s6s) 3P3 (3s4s5s) 3P3 (3s5s6s) 3P3 (3s4s6s) = 3P3 * 3C2 ``` = 1 + 9 + 9 + 18 = 37 How many rolls have minimum of 2? ``` 3C3 (2s) = 1 3C2 * 4C1 = 12 3C1 * 4C1 = 12 3P3 * 4C2 = 36 ``` = 61 How many rolls have minimum of 1? ``` 3C3 = 1 3C2 * 5C1 = 15 3C1 * 5C1 = 15 3P3 * 5C2 = 60 ``` = 91 Total is 216
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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.