Friday, May 14, 2010

Newcomb's Paradox and Pascal's Wager

Let Egalitarian Universalism (EU) be the doctrine that God exists and gives everyone infinite happiness, and that the quantity of this happiness is the same for everyone. The traditional formulation of Pascal's Wager obviously does not work in the case of the God of EU. What is surprising, however, is that one can make Pascal's Wager work even given the God of EU if one thinks that Bayesian decision theory, and hence one-boxing, is the right way to go in the case of Newcomb's Paradox with a not quite perfect predictor.

Here is how the trick works. Suppose that the only two epistemically available options are EU and atheism, and I need to decide whether or not to believe in God. Given Bayesian decision theory, I should choose whether to believe based on the conditional expected utilities. I need to calculate:

  1. U1=rP(EU|believe) + aP(atheism|believe)
  2. U2=rP(EU|~believe) + bP(atheism|~believe)
where r is the infinite positive reward that EU guarantees everybody, and a and b are the finite goods or bads of this life available if atheism is true. If U1 is greater than U2, then I should believe.

We'll need to use our favorite form of non-standard analysis for handling infinities. Observe that

  1. P(believe|EU)>P(believe|~EU),
since a God would be moderately to want people to believe in him, and hence it is somewhat more likely that there would be theistic belief if God existed than if atheism were true (and I assumed that atheism and EU are the only options). But then by Bayes' Theorem it follows from (3) that:
  1. P(EU|believe)>P(EU|~believe).
Let c=P(EU|believe)-P(EU|~believe). By (4), c is a positive number. Then:
  1. U1U2=rc + something finite.
Since r is infinite and positive, it follows that U1U2>0, and hence U1>U2, so I should believe in EU.

The argument works on non-egalitarian universalism, too, as long as we don't think God gives an infinitely greater reward to those who don't believe in him.

However, universalism is false and one-boxing is mistaken.


Mike Almeida said...

Alex, the argument is interesting, but once you assume non-standard infinities it becomes invalid. You need additional assumptions. Even if we assume that c is positive, it might be the case that [rP(EU|believe)-rP(EU|~believe)] < [bP(atheism|~believe)-aP(atheism|believe)].

If c is positive and finite, it might be less than the positive and finite payoff of not constraining yourself in an atheistic world. If that's true, then U1 - U2 is not positive, and it's not true that you should dispose yourself to believe. You need to assume that c is sufficiently large. But that I take it will be a contingent matter. So in some EU worlds it will be rational to believe, in others not. It will depend on things we would be unlikely to know.

Alexander R Pruss said...

If c is positive, and r is positive and infinite, then rc is positive and infinite. But rc = rP(EU|believe)-rP(EU|~believe).

Mike Almeida said...


I guess I see nothing in the case you describe that entails that rP(EU|believe)-rP(EU|~believe) is not finite. What ensures that it is not finite? I might be subtracting a countable infinite quantity from a countable infinite quantity + 1. The difference would equal 1, on non-standard analyses. So it might not pay all that much to dispose myself to believe.

Alexander R Pruss said...

In non-standard analysis, it's a theorem that all the rules of arithmetic work for infinities. Thus: rP(EU|believe)-rP(EU|~believe) = r(P(EU|believe)-rP(EU|~believe)). Since P(EU|believe)-rP(EU|~believe) is a positive number, and r is infinite, the right hand side is infinite.

Alexander R Pruss said...

There is, though, one minor gap in my argument. In (3), I need the left hand side to be bigger than the right hand side by more than just an infinitesimal. (I was assuming probabilities are standard numbers, in which case this follows from (3). But if probabilities are allowed to be non-standard numbers, then this needs to be stated as an explicit assumption.)