A rude assembly

Seamus Bradley's blog

Dutch Booking the Logically Ignorant

Last time, we saw that logical omniscience is difficult to get rid of. We saw that once you “go syntactic” it becomes difficult to stop your set of beliefs being infinite. We used a quotient algebra like what the Lindenbaum algebra does for the standard account to keep things manageable. This time round, we’re going to abstract from these details. All we assume now is that there is some structure of the beliefs. That is, the beliefs have at least some logical structure and that these beliefs inform betting odds. What I want to demonstrate is that a logically ignorant agent can be Dutch booked.

So, I assume there are sentences $\phi$ and $\psi$ that express the same proposition, but that are believed to different degrees. This means that the odds you would offer on one are different from the odds on the other. Since in the standard Dutch book setting, the opponent is allowed to choose the “direction” of the bet, she can choose to bet against the one with the longer odds and on the one with the shorter odds. In either case exactly one bet wins. If she arranges her bets such that they would win the same, then her outlay will be less than her winnings: she will have Dutch booked you!

Consider bets of the following form:

  • Win $S (1-q)$ if $\phi$, lose $S q$ otherwise.
  • Win $T (1-p)$ if $\psi$, lose $T p$ otherwise.

The $S$ is the stake, and the $q$ is the betting quotient. Likewise for $T$ and $p$. The idea behind the Dutch book argument is that if your opponent can set the size of $S$ and decide whether it is positive or negative, then unless your $q$s have the structure of a probability measure, you are guaranteed to lose money: you will be “Dutch booked”. What I want to show here is that a logically ignorant punter can be Dutch booked. That is, if $q \neq p$ then an agent who knows that the two propositions are logically equivalent can guarantee herself a profit.

We have stipulated that $\phi$ and $\psi$ are logically equivalent. Importantly, this means that they are always true or false together. So there are two possibilities: both are true or both are false.

Now, the clever punter will set $T = -S$ . If $\phi$ (and hence $\psi$) is true, then the two bets together win:

If the proposition is false then the outcome for our agent is:

So in either case the punter’s wealth changes the same amount. Now here’s the thing: as in the standard Dutch book argument, we are allowing the opponent to decide the value of $S$. So depending on whether $q$ or $p$ is bigger, the opponent can set $S$ such that our agent loses money!

The only way to avoid this is to set $q = p$. But that is simply to say that the agent should be logically omniscient!

Next time, I’m going to look at what can be done about this state of affairs.