Four says, four faces with their own values. At first glance, it seems like a matter of choosing the strongest. But there's something strange about how they relate to each other.
The Jealous Says
Riddle statement
There are four special dice:
- A: 4, 4, 4, 4, 0, 0
- B: 3, 3, 3, 3, 3, 3
- C: 6, 6, 2, 2, 2, 2
- D: 5, 5, 5, 1, 1, 1
You choose a die first. The dealer then chooses one of the remaining three. You both roll your says once and whoever gets the highest number wins.
Is there a die that is better than all the others? What should the second player choose?
Show solution
Solution
Answer: there is no one die that is better than all. The second player can always choose a die that beats the one chosen by the first with probability $2/3$.
The key comparisons are:
| Comparison | Why it wins | Probability |
|---|---|---|
| A beats B | A wins if he rolls 4; B always rolls 3 | $4/6=2/3$ |
| B beats C | B wins if C rolls 2 | $4/6=2/3$ |
| C beats D | C wins if he rolls 6, or if he rolls 2 and D rolls 1 | $2/6+(4/6)(3/6)=2/3$ |
| D beats A | D wins if he rolls 5, or if he rolls 1 and A rolls 0 | $3/6+(3/6)(2/6)=2/3$ |
This is how a cycle is formed:
The second player always responds with the die that beats the one chosen by the first:
- against A, choose D;
- against B, choose A;
- against C, choose B;
- against D, choose C.
In all cases he wins with probability $2/3$.