When I was in college, I recall learning of a very cool theorem (actually, it may not have been a theorem) that involved a situation that went something like the following:
Assume that there are four bags, each containing red and blue balls. There are certain circumstances in which the probability of pulling a red ball out of bag 1 are higher than pulling one out of bag 2 and the probability of pulling a red ball out of bag 3 are higher than pulling one out of bag 4, but the probability of pulling a red ball out of bags 1 and 3 combined are lower than pulling one out of 2 and 4 combined. Does this scenario ring a bell with any mathematicians out there? I’ve been trying to remember this for some time now. Thanks…
Let R[sub]i[/sub] be the event of pulling a red ball out of bag i. What you’re saying is that P(R[sub]1[/sub]) > P(R[sub]2[/sub]), and P(R[sub]3[/sub]) > P(R[sub]4[/sub]), but P(R[sub]1[/sub] U R[sub]3[/sub]) < P(R[sub]2[/sub] U R[sub]4[/sub]).
Expanding by inclusion-exclusion, we need that P(R[sub]1[/sub]) + P(R[sub]3[/sub]) - P(R[sub]1[/sub]R[sub]3[/sub]) < P(R[sub]2[/sub]) + P(R[sub]4[/sub]) - P(R[sub]2[/sub]R[sub]4[/sub]). I think it’s reasonable to assume independence here, so let’s do that, and denote P(R[sub]i[/sub]) as p[sub]i[/sub].
That gives us p[sub]1[/sub] + p[sub]3[/sub] - p[sub]1[/sub]p[sub]3[/sub] < p[sub]2[/sub] + p[sub]4[/sub] - p[sub]2[/sub]p[sub]4[/sub]. Rearranging, we have p[sub]1[/sub] - p[sub]2[/sub] + p[sub]3[/sub] - p[sub]4[/sub] < p[sub]1[/sub]p[sub]3[/sub] - p[sub]2[/sub]p[sub]4[/sub].
All the steps are reversible, so if you find a set of numbers that satisfy this, it’ll work.
Hmm…let’s set p[sub]2[/sub] = p[sub]4[/sub] = a, and p[sub]1[/sub] = p[sub]3[/sub] = a + b. That gives us 2b < 2a(a + b), or b < a[sup]2[/sup] + ab. We’ll solve for b; that gives us b < a[sup]2[/sup]/(1 - a). For a in [0, 1), a[sup]2[/sup]/(1 - a) is also in [0, 1].
It looks like a = 1/2 and b = 1/4 should work.
Shoot, I screwed up. That equation should be 2b < (a + b)[sup]2[/sup] - a[sup]2[/sup], or 2b < 2ab + b[sup]2[/sup]. We’ll solve for a this time, to get a > 1 - b/2. We also need that a + b < 1, or a < 1 - b.
Oops, hit submit too soon. 1 - b/2 > 1 - b, so there are no values of a and b that meet these criteria. Doesn’t mean there’s no solution, but that there are at least three distinct probabilities of drawing a red ball.
I think you’re making this a bit too complicated, ultrafilter.
Suppose the bags are filled as follows:
Bag 1: 1 blue ball, 1 red ball
Bag 2: 10 blue balls, 9 red balls
Bag 3: 10 blue balls, 7 red balls
Bag 4: 1 blue ball, 0 red balls
Bag 1 has a higher proportion of red to blue balls (1/2) than bag 2 (9/19), while bag 3 has a higher proportion (7/17) than bag 4 (0/1).
But bags 1 and 3 combined have 11 blue balls and 8 red balls, while bags 2 and 4 combined have 11 blue balls and 9 red balls, which is a higher proportion of red to blue.
This is sometimes called a “baseball averages” problem, because you can rephrase it as follows: how can player A have a higher batting average than player B in both the first half of a season and the second half of a season, but have a lower batting average for the season as a whole? Bags one and three represent player A’s at-bats (red is a hit, blue isn’t) while bags two and four represent player B’s at-bats.