Fraction comparison trick revisited

A few years ago I wrote about a curious result: a/b is usually greater than c/d when a + d is greater than b + c. And “usually” means with probability 11/12. As stated in the post, you can make this statement rigorous by choosing integers up to size N at random and taking the limit as N goes to infinity. We can also choose a, b, c, and d uniformly from [0, 1]. This is similar to choosing four integers up to N and dividing all the integers by N, which doesn’t change the probabilities. Now a/b > c/d if and only if ad > bc, so another way to state the theorem is that Prob(ad > bc | a + d > b + c) = 11/12. Or to state it yet another way, pick two pairs of numbers at random. The probability that the pair with the larger sum also has the larger product is 11/12. That seems more intuitive to me than the original statement. If a + d is greater than b + c, it’s likely that a/b is greater than c/d. It seems plausible that if a + d is much greater than b + c, it’s more likely that a/b is greater…

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