Two mismatched dice

EasyProbability~4m

You roll a fair $40$ -sided die and a fair $60$ -sided die (faces $1$ through $40$ and $1$ through $60$ respectively). What is the probability that the $60$ -sided die shows a strictly greater value than the $40$ -sided die? Round to the nearest hundredth.

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Condition on the $40$ -sided die showing $a$ ; then the $60$ -sided die beats it with probability $(60-a)/60$ .
Average over $a = 1,\dots,40$ . The relevant sum is $59 + 58 + \dots + 20$ .

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0.66 (± 0.01)

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Two mismatched dice

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