The St. Petersburg Paradox

I have been to Vegas at least ten times, and have never gambled a dollar. I do not actually begrudge anyone that goes into a casino and treats it for what it is, entertainment, but that has never appealed to me.

What has always interested me, however, is gambling-related paradoxes, and here we actually have two of them: the St. Petersburg Paradox and its corollary, the Martingale Game.

The St. Petersburg Paradox describes a game that, despite its infinite expected value (EV), nobody would reasonably pay much to play. The classical game is simple: the pot starts at $2, and on a heads the pot doubles and on a tails the pot pays out.

The EV is a simple divergent series: 1/2 × 2 + 1/4 × 4 + … + 1/2^N × 2^N. This is chosen because it simplifies to 1 + 1 + 1 + … + 1, though my version starts from a $1 payout, so the series is instead 1/2 + 1/2 + 1/2 + … + 1/2. While half as quick, the effect is the same: its expected value is infinite, and therefore ANY real number would be a reasonable amount to pay to play this game. However, in a world where no one will ever see 1,000 flips of heads in a row during their lifetime, there are practical limits to how much anyone should be willing to spend before choosing to flip.

From my playing the ‘half as good’ version, $5 from $50 has a reasonable chance of hitting a sufficient fat-tail payout that stabilizes your bankroll and allows you to play indefinitely, but $10 from $100 will almost certainly bust.

The Martingale Game, or betting system, is one that is rumored to have bankrupted casinos before there were table limits, but that is probably apocryphal.

Imagine we are in a casino that has infinite money, and we have infinite money. Well, in this world I have a strategy for you that will always pay out N dollars, whatever you want to win. The system is simple. Start by betting N dollars in a game that pays out double your bet and, we will round up and say, has 50/50 odds. When you win, you net N dollars; when you lose, double your bet.

Payout equals the total bet + N. If you start betting 1 dollar and lose 6 times before winning, that is 1 + 2 + 4 + 8 + 16 + 32 = 63, with the final payout being 64. 64 − 63 = 1 (or N). The problem with this system comes in, in that your EV becomes 0 the moment you bound the number of times you can double on your losses, because the net winnings will, on average, exactly cover a scenario of ruin when you can no longer double.