Hilbert’s Bank: Why the St. Petersburg Paradox Presents a Problem For Expected Utility
Infinity is messing things up
This is a follow-up to Achilles, the Tortoise and the Hotel: Exploring the Infinite. The first part of this post was already posted there, and reappears here — slightly edited — as a recap.
Next to Hilbert’s Hotel, on Paradoxica, lies Hilbert’s Bank. Obviously, this bank has a countably infinite amount of money. Curious, you walk into the bank, and the manager proposes a game.
“I’ll toss a fair coin,” she says. “If it’s tails, you win $2. If it’s heads, I’ll toss the coin again. Is it tails? Then you win $4. If it’s heads, I’ll toss again, and so on.”
“So your stake starts at $2 and doubles when heads appears, and the game stops if (and when) tails appears.”
“How much money are you willing to pay to play this game?”
You respond: “$10, no more. It seems like fun to play the game, but I won’t make much money.”
“Ah, but that’s where you’re wrong,” say the manager. “Your expected payoff is infinite! Since the coin is fair, there’s a ½ probability it comes up tails on the first run. So you have a ½ probability of making $2. If the coin comes up heads on the first run, there’s a ½ probability it comes up tails on the second run, in which case you’ll make $4. This outcome has a probability of ½ x ½ = ¼. You’ll make $8 with a probability of ⅛, etc.”
“Therefore, your expected payoff is $2 x ½ + $4 x ¼ + $8 x ⅛ + … = $1 + $1 + $1 + … = $∞. You should therefore be willing to pay all the money you have to play this game.”
You probably wouldn’t pay all your money to play this game: after all, what is the probability you’ll even make $50? (It’s 1/32.) So paying even $50 seems (and probably is) ridiculous. But then you have to believe there’s an error in the bank manager’s reasoning. What is that error?
A standard response is to say monetary payoffs have diminishing marginal utility: that is, each extra dollar you win has less value than the previous one. This makes sense: winning, say, $100 is awesome, but if you already won $10,000, an extra $100 isn’t that awesome anymore.
Utility measures how much something is actually worth to you personally: watching a cool movie probably has higher utility than cleaning a dirty toilet. Where temperature is measured in e.g. Kelvin, Celsius or Fahrenheid, utility is measured in utils. Winning $100 might be worth 100 utils to you. If we except the principle of diminishing marginal utility here, than winning $10,000 isn’t worth 100 times as much as winning $100 (which would be 10,000 utils).
So how should we value money? Well, we could use the natural logarithm (ln):
Now the expected monetary payoff is still infinite, but the expected utility isn’t: 0.69 x ½ + 1.39 x ¼ + 2.08 x ⅛ + … isn’t infinite. That appears to resolve the paradox of not wanting to pay anything for an expected infinite payoff.
But this only solves the paradox “at the surface”: the actual paradox remains. What if the monetary payoff doesn’t double on each round of the game, but instead equals $e², $e⁴, $e⁸, … (where e is Euler’s Number)? Then our V() gives the following utilities:
Now the utilities are the same as the original monetary payoffs, which means our expected utility is infinite. The paradox is back: our expected utility is infinite! Or we just directly measure utility in dollars, which is more intuitive: that is, we just act like the original version of the St. Petersburg game already specifies utility with the $2, $4, $8, etc. outcomes. We’ll do that from now on.
This infinite utility thing is a problem: as we’ve seen, it means we should be willing to pay all our money to play this game. But also, consider the Petrograd game: it’s the same as the St. Petersburg game, except that for every outcome of the game, it gives you exactly $1 more than the St. Petersburg game would have given you. So instead of $2, $4, $8, etc., the payoffs are $3, $5, $9, etc:
Which game would you prefer to play?
I’d play the Petrograd game: it simply gives $1 more for every possible outcome. But the expected utility of both games comes out to infinity, which suggests it doesn’t matter which game you play. In case it isn’t obvious why this is a problem, consider that instead of $1, the Petrograd game pays $1,000,000 more than the St. Petersburg game on every outcome: the payoffs are now $1,000,002, $1,000,004, $1,000,008, etc. Once again, the expected utility principle is indifferent between this game and the St. Petersburg game — but any rational person would choose the Petrograd game and with it, the guarantee to become a millionaire.
To be absolutely clear: I am not saying the expected utility of the Petrograd game is higher than that of the St. Petersburg game. I am saying these expected utilities are exactly equal — they are both infinity — and that that presents a problem.
There is another tool we can use here that does conclude the Petrograd game is better: dominance reasoning. Like we discussed, for every possible outcome to this game, the Petrograd game has a higher payoff than the St. Petersburg game. It will be tails at some point during any of these games, and whenever that happens, the Petrograd game simply pays more. This means the Petrograd game dominates the St. Petersburg game, and directly suggests it’s better to play the Petrograd game.
So dominance reasoning has an intuitive advantage over expected utility in these problems. But even dominance reasoning doesn’t solve everything. Consider yet another problem: the Leningrad game. Like the Petrograd game, the Leningrad game pays $1 more for every outcome than the St. Petersburg paradox, except for one highly unlikely outcome: for example, the outcome where tails happens on the 100th run. In that highly unlikely outcome, the Leningrad game pays $1 less than the St. Petersburg game.
The Leningrad game still has infinite expected utility — and it doesn’t dominate the St. Petersburg game: after all, it doesn’t pay more for every outcome. Yet any rational person would still prefer the Leningrad game over the St. Petersburg game! The Leningrad game almost certainly pays out more — I might even say it is guaranteed to pay out more. Yeah, that 100th run still has some probability, you might say. But if the word guaranteed is to have any meaning at all, I should be allowed to use it for an event whose probability of not happening is smaller than 1 in 1 thousand billion billion billion.
As an aside (and for fun) let’s dive into how small that probability — 1 1,267,650,600,228,229,401,496,703,205,376 — really is. It is, of course, the probability the coin comes up tails at the 100th round in the Leningrad game. But if you played the Leningrad game every second from the moment of the Big Bang till now, you still — in all probability — wouldn’t have seen it end at the 100th round. Heck, make that every nanosecond. That’s how small the probability is. So yes, if you play the Leningrad game, I can guarantee you you won’t lose that dollar.
Anyway, back to the Leningrad game. The point was that dominance reasoning — while powerful in the Petrograd game — doesn’t help here. But relative expected utility does! Let’s start by calculating the relative expected utility of the Petrograd game over the St. Petersburg game. We do this by substracting the utility of each outcome of the St. Petersburg game from the utility of that outcome in the Leningrad game. We multiply each result by the probability of the outcome and sum everything together. For every outcome, the utility of the Petrograd game is $1 higher than the St. Petersburg game. That means the relative expected utility of the Petrograd game over the St. Petersburg game is just 1/2 x $1 + 1/4 x $1 + 1/8 x $1 + … = $1, which is higher than $0. So the relative expected utility principle favors the Petrograd game over the St. Petersburg game — just like dominance reasoning.
So how about the Leningrad game vs. the St. Petersburg game? The calculation is almost the same as before, with one small difference: the relative expected utility here is 1/2 x $1 + 1/4 x $1 + 1/8 x $1 + … + 1/2¹⁰⁰ x $-1 + … That’s not exactly $1, but close, and it’s certainly greater than $0. Even though standard expected utility and dominance reasoning are indifferent here, the relative expected utility principle does favor the Leningrad game over the St. Petersburg game!
However… Consider another two games. I’m getting a bit lost in the city names for these games, which different sources seem to use differently, so we’ll call the following games Game A and Game B. They are both like the St. Petersburg game, but in Game A, if you’re at round 2, you get to play another instance of the St. Petersburg game as a bonus. In Game B, this is the case in round 3.
We should prefer Game A over Game B, as it gives a higher chance of the bonus St. Petersburg game. And yet… It’s clear by now the expected utility is equal (and infinite) for both games. Dominance reasoning doesn’t suggest anything either: neither Game A nor Game B gives more payoff in every round. And unfortunately, relative expected utility also doesn’t help: for every outcome except those in rounds 2 and 3, subtracting the utility of Game B from Game A gives $0. In round 2, it gives $∞, and in round 3, it gives $-∞. Multiplying them by their corresponding probabilities still gives $∞, and summing the results together is a problem, because it requires doing $∞ + $-∞ which is, eh, undefined. Relative expected utility simply doesn’t have an answer here, which is too bad for such an elegant principle.
So how do we resolve the St. Petersburg paradox?
I’m not sure. One idea we haven’t discussed is using the median instead of the mean. That is, expected utility calculates the average value — the mean value — of a bet. The median of a series of values is that value for which it is true that half the values of the series are below it and half are above it. In the St. Petersburg problem, it would be a value for which it is true that the probabilities of the outcomes below that value sum up to 1/2 — as well as the probabilities of the outcomes above that value, of course. So the median could be set at $3: there’s one 1/2 probability outcome below it (at $2) and an infinite amount of outcomes above it ($4, $8, etc.) whose probabilities sum up to 1/2. This median suggests paying no more than $3 to enter this game, which doesn’t seem unreasonable to me. At least it doesn’t suggest paying all your money!
And remember the Petrograd game? Here’s a reminder of the payoffs:
While the mean expected value of the Petrograd game is equal to that of the St. Petersburg game, the median isn’t: it’s $4, and thus $1 higher. That’s another advantage of using the median!
How about the Leningrad game?
The median expected value is still higher than that of the St. Petersburg game. In fact, it seems to me it’s just $4, like the Petrograd game.
What about Game A vs. Game B? Game A has a bonus St. Petersburg game at round 2, which now gives a value of $3. Game B has this bonus $3 in round 3. This gives outcomes of $2, $7, $8, $16, etc. for Game A and $2, $4, $11, $16, etc. for Game B. A reasonable median for Game A would be the average of $2 and $7, so $4.50. For Game B it would be $3, meaning the median gives the advantage to Game A here. Great! However, the problem resurfaces if we shift the bonus St. Petersburg game to e.g. rounds 10 and 11: then the outcomes for both Game A and Game B start with the old $2, $4, $8 again, putting the median for both at $3. That’s unfortunate.
Still, it seems the median gives a lot of benefits in this area. But it’s still not perfect, and whether the median is generally a good substitute for the mean is questionable and something I’d need to research. For now, thanks for reading!
References
Bernoulli, D. (1896). Specimen theoriae novae de mensura sortis. Gregg.
Colyvan, M. (2008). Relative expectation theory. The Journal of philosophy, 105(1), 37–44.
Peterson, M. (2013). A Generalization of the Pasadena Puzzle. dialéctica, 67(4), 597–603.
