0
$\begingroup$

I'm unconvinced that my refusal to accept the fourth of the von Neumann-Morgenstern axioms is irrational. Wikipedia claims that there is a Dutch book argument against me, but I do not see how that can be possible if I don't assign definite utilities to bets at all, which is what the Dutch book arguments for the axioms of probability assume. Does anyone know of such an argument?

$\endgroup$
4
  • $\begingroup$ Wierdly enough this is probably more philosophy than math. Look at this entry from the Stanford Encylopedia of Philosophy: plato.stanford.edu/entries/dutch-book and references within. The reason for this being studied in philosophy is that the argument is about more about modeling utility and rational behavior than mathematical theorems. $\endgroup$
    – twnly
    Mar 6, 2019 at 7:43
  • $\begingroup$ I am pretty sure that claim is false in general. The definition of Dutch book does not make sense in this context. Its probably true in more specific circumstances, but not in this case, I believe. $\endgroup$ Mar 7, 2019 at 4:15
  • $\begingroup$ Actually, you could define dutch books in terms of cyclic preferences, which makes sense in this context. However, it is then false: you only need axioms 1 and 2 to avoid cyclic preferences. $\endgroup$ Mar 7, 2019 at 4:18
  • $\begingroup$ @PyRulez Cyclic preferences imply Dutch Books, but a Dutch Book does not imply cyclic preferences. A Dutch book means a series of bets that a person will agree to, even though these bets will leave the person making these bets worse off 100% of the time. $\endgroup$ May 1, 2022 at 17:17

1 Answer 1

0
$\begingroup$

The VNM theorem is a proof that if you have preferences, one of the following holds --

  1. You can rewrite your preferences as a utility function. This doesn't imply you think of them as a utility function, or that a utility function is a natural representation of them. Just that you could, in principle, rewrite them like this.
  2. Your preferences can be Dutch booked. In other words, there exists some series of bets you'd agree to that will leave you worse off with probability of 100%. (Or, equivalently, there exists some series of bets that will always leave you better off, but you'll refuse to take them.)

I'll try and give a sketch of a Dutch book for refusing to accept independence of irrelevant alternatives:

Start with alternatives ${A, B}$, where you prefer A to B. Say there exists some C such that you prefer $.5 B + .5 C$ to $.5 A + .5 C$ (where $.5 x$ means $x$ has a 50% chance of happening). I start by offering a choice of lotteries ${A', B'}$, where $A'$ gives you $A$ if a coin toss is heads, and similarly for $B'$. You take $B'$. Now I offer bets $'B$, $'A$, which are the same as before, except I flip the case where I pay you. Once again you take $'B$ to $'A$, because $.5 B > .5 A$. Now realize: There is a 100% chance you will receive $B$. This is worse for you than a 100% chance of $A$. You've intentionally refused the pair of bets $A'$, $'A$ that would have left you better off with probability 1, leaving you worse off.

The key to the proof is that by introducing and removing $C$, I can reliably get you to switch between preferring $A$ and $B$. This lets me turn you into a money pump, by introducing option $C$ and then letting you trade between $A$ and $B$. I gave another explanation of this here.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .