How Poker Explains Who Saw The Interview Where

Apparently I still can’t stop talking about The Interview and how you can figure out who saw it where (theater vs. video). I thought for sure I was done after this post, but then I came up with a few more takes on the problem. I really wasn’t kidding when I said that a huge part of thinking like a math person is being able to look at problems from multiple points of view! The first two approaches below are inspired by a discussion of a very similar problem here (h/t Mark Lakata).

The set up again: The New York Times told us that The Interview made \$15M over Christmas weekend, from a total of 2M paid customers, each of whom either rented it for \$6 or bought a ticket for \$15. We want to know how many rented and how many bought tickets.

1.  We know that if all 2M people rented, the total revenue would have been \$12M. To reach the actual revenue of \$15M, we need to make an additional \$3M by converting some renters into ticket buyers. Changing a single renter into a ticket buyer nets an additional \$9 (from \$6 to \$15). The number of conversions we need to make is then the total incremental revenue, \$3M, divided by the incremental revenue per conversion, \$9. So we make \$3M / \$9 = 1/3 M conversions and end up with 1/3 M ticket buyers. (This is essentially another form of shifting the sliding scale from my last post.)

2.  For a mathematically equivalent, but more vivid version of this argument, imagine a poker game with 2M players (!) and a \$6 ante. When everyone antes up, we have \$12M in the pot. Cards are taken. The first bet is (an additional) \$9. Some players see the bet, put in \$9, and stay in, the rest fold. If we find that over this round, \$3M more was added to the pot, for a total of \$15M, we can deduce that the number of players who put in \$9 more and stayed in (they correspond to the ticket buyers, who paid \$15 in total) has to be \$3M / \$9 = 1/3 M. This formulation suggests a longer poker game, with multiple rounds of betting. Which in turn would correspond to a larger algebraic system, with more equations and more unknowns. Rentals, full price tickets, matinees, anyone?

3.  If you like to think in terms of averages, you can observe that if in total 2M people paid \$15M, then the average person paid \$7.50. If some people paid \$6 and others paid \$15, and if the average of \$7.50 is 1/6 of the way between \$6 and \$15, then \$6 must contribute 5/6 of the mass of the average, while \$15 contributes 1/6. So 5/6 of the people paid \$6 (rented) and 1/6 paid \$15 (bought tickets). This is very close to my original weighted average argument, but with a change of variables, so we’re averaging over people (between \$6 and \$15) rather than over possible outcomes (between \$12M and \$30M), which seems a little more intuitive.

All three of these arguments are clean (I think), illuminating (they give you insight, one way or another, into why the answer is what it is), and elementary (you don’t need to know any math beforehand to come up with them). On the other hand, they aren’t exactly obvious. Moreover, as we’ve seen, they’re connected to some powerful mathematical techniques (solving systems of equations, changing variables, and so on). The connections work both ways. Having the general math techniques at your disposal makes it more likely that you’ll come up with your own nice take on the problem. On the other hand, if you’re lucky enough to come up with such an approach from first principles, it’ll help you understand why the general techniques work in the first place. The insight can go either way; math does not have to flow in any particular direction. To paraphrase Andrew Wiles, who proved (with help) Fermat’s last theorem: when you do math, you make your way, blindfolded, through a door into a dark room, you feel your way around some, you might bump into some dead ends, you might stumble once or twice, but if you’re thoughtful and persistent, you end up with a pretty good map of the territory.