Probability your d6 beats their d4

Question

You roll one fair six-sided die; your opponent rolls one fair four-sided die. You win if your number is strictly higher. What is the probability that you win?

Accepted Answer

Let X be your d6 roll (1 to 6) and Y your opponent's d4 roll (1 to 4). There are 6 	imes 4 = 24 equally likely ordered pairs. Count those with X Y by fixing your roll: - X = 1: no Y in \{1,2,3,4\} is below 1, so 0. - X = 2: only Y = 1, so 1. - X = 3: Y \in \{1, 2\}, so 2. - X = 4: Y \in \{1, 2, 3\}, so 3. - X = 5: Y \in \{1, 2, 3, 4\}, so 4. - X = 6: Y \in \{1, 2, 3, 4\}, so 4. Total favorable: 0 + 1 + 2 + 3 + 4 + 4 = 14. Therefore P(X Y) = \frac{14}{24} = \frac{7}{12} \approx 0.5833. Because the bigger die is yours, you are the favorite, which the 58.3\% confirms. Note the symmetry shortcut…

Probability your d6 beats their d4

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