Two four-sided dice, expected value of the larger

Question

You roll two fair four-sided dice, each showing 1, 2, 3, or 4. You look at the larger of the two faces (if they tie, that shared value is the larger). On average, what is the value of the larger die?

Accepted Answer

There are 4 	imes 4 = 16 equally likely rolls. Find how often the larger die equals each value k. How often is the larger die exactly k? The larger die is k or less when both dice are k or less, which happens in k^2 ways. So the larger is exactly k in k^2 - (k-1)^2 = 2k - 1 ways, the same rule as for six-sided dice. For k = 1, 2, 3, 4 that is 1, 3, 5, 7 ways, which add to 16. Average. Multiply each k by its count 2k-1 and add: 1(1) + 2(3) + 3(5) + 4(7) = 1 + 6 + 15 + 28 = 50. Divide by 16: EV = \frac{50}{16} = \frac{25}{8} = 3.125. The larger of two four-sided dice averages 3.125, above a si…

Two four-sided dice, expected value of the larger

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