The science of Hillary Clinton's coin-toss victories

If you’re like most people, you’d say it wouldn’t be unusual if Clinton and Sanders each won 3, and it probably wouldn’t be that unusual if either Clinton or Sanders won 4, while the other won 2. But 5:1 or 6:0 seems too unlikely to occur at random, doesn’t it?

Unfortunately, this is one of those cases where our mathematical intuition and what actual probabilities are don’t line up at all. If you have a fair (50/50) coin at play in each instance, it’s true you’re more likely to have three “wins” for each candidate than any other specific outcome. But it’s still not all that likely: there’s only a 31.25% chance that Clinton and Sanders would have walked away with three delegates apiece. Furthermore, the odds that Clinton would win four and Sanders would win two is only a little worse: 23.44%. But if you combine that with the odds that Sanders would’ve won four with Clinton winning two, you get that a 4:2 outcome has a 46.88% chance of happening. Meaning the “unlikely” outcomes of 5:1 or 6:0? They actually have a 21.88% chance of occurring, which is about the same as your odds of winning any prize at all (most likely, $4) if you buy six random Powerball tickets.