Fraction comparison trick revisited

A few years ago I wrote about a curious result: a/bis usually greater thanc/dwhena+dis greater thanb+c. And usually" means with probability 11/12.

As stated in the post, you can make this statement rigorous by choosing integers up to sizeN at random and taking the limit asN goes to infinity.

We can also choosea,b,c, andd uniformly from [0, 1]. This is similar to choosing four integers up to N and dividing all the integers byN, which doesn't change the probabilities.

Nowa/b >c/d if and only ifad >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 thanb +c, it's likely thata/b is greater than c/d. It seems plausible that ifa +d is much greater than b +c, it's more likely that a/b is greater than c/d. Or in terms of cross products, the more the sum of one pair is bigger than the sum of the other, the more likely the product of the former is to be greater than the latter. That is

f(r) = Prob(ad >bc | (a +d)/(b +c) >= r)

should be an increasing function ofr. We know the probability above equals 11/12 whenr = 1. We expect that the probability should approach 1 as r goes to infinity.

This is indeed the case, as shown in the plot below.

In particular, f(2) 0.99 andf(6) 0.999. So ifa +d is two (or six) times bigger thanb +c, there is a 0.99 (or 0.999) probability thatad is larger thanbc.

You can calculatef(r) analytically or estimate it by simulation. I had Claude do both and they produce comparable results. There is a closed-form expression forf(r) but it's really complicated.

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