Handling the infinities in Pascal’s Wager
January 23, 2014 — 12:01

Author: Alexander Pruss  Category: Afterlife Atheism & Agnosticism General  Tags: , , , , ,   Comments: 24

In its classical formulation, Pascal’s Wager contends that we have something like the following payoff matrix:

 God exists No God Believe +∞ −a Don’t believe -b c

where a,b,c are finite. Alan Hajek, however, observes that it is incorrect to say that if you don’t choose to believe, then the payoff is finite. For even if you don’t now choose to believe, there is a non-zero chance that you will later come to believe, so the expected payoff whether you choose to believe or not is +∞.

Hajek’s criticism has the following unhappy upshot. Suppose that there is a lottery ticket that costs a dollar and has a 9/10 chance of getting you an infinite payoff. That’s a really good deal intuitively: you should rush out and buy the ticket. But the analogue to Hajek’s criticism will say that since there is a non-zero chance that you will obtain the ticket without buying it—maybe a friend will give it to you as a gift—the expected payoff is +∞ whether you buy or don’t buy. So there is no point to buying. So Hajek’s criticism leads to something counterintuitive here, though that won’t surprise Hajek. The point of this post is to develop a rigorous principled response to Hajek’s criticism entailing the intuition that you should go for the higher probability of an infinite outcome over a lower probability of it.

A gamble is a random variable on a probability space. We will consider gambles that take their values in R*=R∪{−∞,+∞}, where R is the real numbers. Say that gambles X and Y are disjoint provided that at no point in the probability space are they both non-zero. We will consider an ordering ≤ on gambles, where XY means that Y is at least as good a deal as X. Write X<Y if XY but not YX. Then we can say Y is a strictly better deal than X. Say that gambles X and Y are probabilistically equivalent provided that for any (Borel measurable) set of values A, P(XA)=P(YA). Here are some very reasonable axioms:

1. ≤ is a partial preorder, i.e., transitive and reflexive.
2. If X and Y are real valued and have finite expected values, then XY if and only if E(X)≤E(Y).
3. If X and Y are defined on the same probability space and X(ω)≤Y(ω) for every point ω, then XY.
4. If X and Y are disjoint, and so are W and Z, and if XW and YZ, then X+YW+Z. If further X<W, then X+Y<W+Z.
5. If X and Y are probabilistically equivalent, then XY and YX.

For any random variable X, let X* be the random variable that has the same value as X where X is finite and has value zero where X is infinite (positively or negatively).

The point of the above axioms is to avoid having to take expected values where there are infinite payoffs in view.

Theorem. Assume Axioms 1-5. Suppose that X and Y are gambles with the following properties:

1. P(X=+∞)<P(Y=+∞)
2. P(X=−∞)≥P(Y=−∞)
3. X* and Y* have finite expected values

Then: X<Y.

It follows that in the lottery case, as long as the probability of getting a winning ticket without buying is smaller than the probability of getting a winning ticket when buying, you should buy. Likewise, if choosing to believe has a greater probability of the infinite payoff than not choosing to believe, and has no greater probability of a negative infinite payoff, and all the finite outcomes are bounded, you should choose to believe.

Proof of Theorem: Say that an event E is continuous provided that for any 0≤xP(E), there is an event FE with P(F)=x. By Axiom 5, without loss of generality {XA} and {YA} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0≤xP(XA), just choose a∈[0,1] such that aP(XA)=x and let F={XA&Ua}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=−∞} and B={Y=−∞}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=−∞ if xP(A) and X1(x)=0 otherwise, and Y1(x)=−∞ if xP(B) and Y1(x)=0 otherwise. Since P(A)≥P(B) by (7), it follows that X1(x)≤Y1(x) everywhere and so X1Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXBY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+∞)<P(Y2=+∞), X2* and Y2* have finite expected values and X2 and Y2 never have the value −∞. We must show that X2Y2. Let C={X2}=+∞. By subdivisibility, let D be a subset of {Y2}=+∞ with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+∞)=P(Y2=+∞)−P(X2=+∞)>0.

Choose a finite N sufficiently large that NP(Y3=+∞)>E(X3)−E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+∞)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3Y4. By Axiom 1 it follows that Y3>X3. but DY2CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

• Jeremy Gwiazda

“….you should go for the higher probability of an infinite outcome over a lower probability of it.”

I haven’t worked through your details, but the above strikes me as false. Consider. Gamble B is the St Petersburg Game payouts are \$2 prob ½, \$4 prob ¼, \$8 prob 1/8…. Gamble C is the St Petersburg Game payouts are \$8 prob ½, \$16 prob ¼, \$32 prob 1/8…. After running the games/gambles, there is a 100% chance of keeping the payout in B, and you have a 50% chance of keeping the payout in C. This is thrown together quickly, but doesn’t your principle say go for B (100% being > 50%)? That is, the higher probability of an infinite outcome. But in repeated plays C outperforms B.

For the failure of reasoning involving expected value, infinity, and repeated cases, here’s a short paper on the topic:

But the point being that some sort of counterexample has to exist to the principle above (or at least so it seems to me on a quick pass).

January 24, 2014 — 13:57
• Alexander Pruss

Very nice example! But not a counterexample to what I said.

First, note though that I said: “lower probability of *it*.” Here the “it” is that particular infinite outcome. In other words, in the comparison I had in mind, one needs to keep fixed the infinite outcome, or at least keep fixed its value. Consider two outcomes:
D: Bliss of degree 1000 every day for eternity.
E: Bliss of degree 1 every day for eternity.
It seems pretty rational to go for a 99.99999999999% chance of D over a certainty of E.
So it’s important to keep the infinities fixed.

Second, in your case there is no infinite payoff, just an infinite expected payoff in each round.

January 24, 2014 — 15:11
• Mark Rogers

Very interesting post, thank you Dr. Pruss. My concern is that you and Alan Hajek are not squaring off face to face. You are making the payoff matrix believe or don’t believe. Hajak is making the decision matrix try to believe or do not try to believe. Disclaimer here, I have not been able to access Hajak’s paper nor do I follow yet all of your work. So reasonably enough the heart of Hajek’s claim is, I think, that there is a chance if you try to believe there is a certain nonzero chance you will not come to belief as well as a nonzero chance that you will come to belief even if you never try to believe. Reasonable as this proposition may sound it is not a part of Pascal’s Wager. Pascal clearly says in my 1910 Trotter translation  “…that you will at last recognize that you have wagered for something certain and infinite, for which you have given nothing.” in other words you will certainly come to believe. This of course is predicated on the individual seeking belief in the correct way. Pascal says “Learn of those who have been bound like you, and who now stake all their possessions. These are people who know the way which you would follow, and who are cured of an ill of which you would be cured. Follow the way by which they began; by acting as if they believed, taking the holy water, having masses said, etc.” It could be though that Hajek has shown something important. That is, if you go after belief in the wrong way your chances of attaining eternal bliss are no greater than if you had never sought belief.

January 25, 2014 — 18:00
• The theorem still makes Pascal’s Wager work with the assumption that one might not come to faith even if one tries to believe, as long as one is more likely to come to faith when one tries than when one doesn’t try.

January 25, 2014 — 21:40
• Jeremy Gwiazda

Gotcha. Looking forward to working through the details in more detail over the next few days. Question on a first pass. You write:

“Choose a finite N sufficiently large that NP(Y3=+∞)>E(X3)−E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions).”

Does this also mean that we can’t be working in a non-archimedean field?
(In part I”m trying to remember what Hajek says when the payouts are infinite numbers in a nonstandard model, or surreal number type things…. But it sounds like this is not the part of the argument you are addressing?)

January 26, 2014 — 10:27
• Jeremy Gwiazda

“We will consider gambles that take their values in R*=R∪{−∞,+∞}, where R is the real numbers.”

Think I may have just answered that one by (re)reading. Interesting argument.

January 26, 2014 — 10:31
• Yeah, R is archimedean. The only infinite values that the gambles can take are +∞ and -∞, and these are plain “flat-rate” infinities (unlike in nonstandard analysis where there are bigger and smaller infinities).

By the way, here’s a first take on the many-gods objection in the above framework. Suppose we have worldviews R1,…,Rn, with probabilities P_1,…,P_n of being true. Simplify by supposing that no religion offers an infinite negative payoff. Let H(i,j) be the probability of an infinite positive payoff (“heaven”) if view i is true and one tries to adopt view j. Then for most of the serious worldviews we will have H(i,i) larger than H(i,j) if j is not equal to i. Now the probability of heaven if one tries to adopt view j will be H(j) = sum_i P_i H(i,j). By the theorem, if there is a view j such that H(j) is maximal, then one should try to adopt view j.

This gets quite complicated. I’ve played with a little spreadsheet, where I assigned probability 0.5 to atheism, and then probabilities to other religions in proportion with the number of believers. The results become very sensitive to exactly what probabilities one assigns in the H(i,j) matrix. With some numbers I made up, I got Catholicism, Protestantism, Eastern Orthodoxy and Islam all very close together, each giving about a 1/3 chance of heaven if you adopt it. The numbers are too close to distinguish responsibly further. But perhaps some part of the decision between these can be made on evidential grounds.

January 27, 2014 — 10:17
• IanS

Instead of using an axiomatic approach, one could simply define the obvious decision rule as follows:
a) Break each gamble into a finite and an infinite part.
b) Compare the infinite parts using P(infinity) – P(minus infinity). If (and only if) this is a tie:
c) Break the tie using the expectations of the finite parts (if they exist).

This rule is consistent with your axioms. Variations are possible. In b) we could use any function that increases with P(infinity) and decreases with P(minus infinity). I suspect these are the only rules consistent with your axioms.

Such rules may be reasonable, but does reason compel us to use one of them? I doubt it. Even finite but highly skewed distributions of payouts raise questions. Suppose I had a once-only opportunity to buy for one dollar a single non-transferrable lottery ticket with one chance in a million of a two million dollar payout. Would I buy the ticket? Probably not, despite the positive expectation. What if the payout were a billion dollars? Or even eternal bliss? Still probably not. The dollar loss is certain. The gain, however great, is very unlikely. There is no possibility of repetition, let alone sufficient repetitions to make the laws of large numbers work. Of course, many people must think differently, or there would be no lotteries. It is not unreasonable to buy the ticket. But neither is it unreasonable to reject it. An atheist who gives a non-zero but very small probability to God’s existence might think the same of Pascal’s wager.

January 28, 2014 — 15:52
• One may also be able to justify expected utility comparisons without the Law of Large Numbers, with this argument.

February 3, 2014 — 14:45
• The rule of comparing infinite outcomes by probability of getting the outcome seems to me like it would beg the question again Hajek. Better to give independently plausible axioms from which that rule can be derived.

And, yes, it does seem that certain kinds of risk averseness can make it rational to reject bets with good expected utility. But one should be careful that this not simply be a case of ignoring small chances, which is a bad policy. Parfit has the example of where everybody in a large city is offered a small payment in exchange for undertaking a tiny risk–say, one in N where N is the population–of blowing up the city. If they all take the gamble, ignoring the tiny risk, boom!

January 28, 2014 — 18:37
• IanS

Yes, the merit of axiom-theorem-proof is that it sharpens the focus of disagreement. I have trouble with axiom 2, comparison of expected utility, at least as a requirement for reasonable choice. Work through the proof using Pascal’s wager, with “Don’t believe” as X, “Believe” as Y and the usual payouts. The key step is to replace the infinite payout with a large finite payout (N in the proof). N is chosen so that the modified “Believe” (Y3 in the proof) has greater expected utility than X. The smaller P(God exists), the larger is the required N. To make this work, you must be prepared to base your choice on the expectation of arbitrarily large payouts for outcomes with arbitrarily small probability. This I don’t accept, as with the lottery ticket in my example.
“But one should be careful that this is not simply a case of ignoring small chances, which is bad policy”. Yes, if many similar small chances accumulate, as in the example you quote. In Pascal’s wager, we each have one life, and are each saved or not.

January 29, 2014 — 15:39
• IanS:

I am with you on concerns about the applicability of decision theory outside the realm where something like a law of large numbers is relevant. But that’s not an objection to Pascal’s Wager as such, but decision theory at large.

I think that in practice, the issue you raise is somewhat alleviated by the fact that the probabilities aren’t that small. There are some facts, such as that the universe is governed by elegant and simple laws, for which there are very few decent explanations (theism and optimalism may be the only ones), all of them in some way extraordinary, and theism is one of them. Maybe one could say that these facts have no explanation, but we should prefer explanations to non-explanations. Conservatively speaking, I think thoughts like this get you to at least something like 1/3 for theism. And then it’s reasonable to divide the probability between theistic religions according to the numbers of adherents of given worldviews or something like that, and so you get something like 1/6 for Christianity. Which isn’t all that small.

January 29, 2014 — 15:56
• Define x+y only when both x and y are real numbers, or one of them is zero and the other is infinite. If I replace the requirement of disjointness in Axiom 3 with the requirement that the sums are well-defined (remember that x+y is only defined if both x and y are real numbers or if one of them is zero), then I think I can prove Axiom 2 from the other axioms. Haven’t written out the details, and am checking with a friend for a reference since I suspect this is known.

January 29, 2014 — 8:31
• Not quite prove Axiom 2, but only that if EX < EY, then X < Y. That’s slightly weaker but all I need.

January 29, 2014 — 17:13
• IanS

Are you sure? A first thought is that if F is any non-decreasing function, then ranking on E(F(X)) (if it exists) seems consistent with the other axioms. But E(X) < E(Y) need not imply E(F(X)) 0). But I may be totally off-beam.

January 30, 2014 — 4:10
• IanS

Something went wrong – the inequalities were read as HTML. Last line should read: E(X) < E(Y) need not imply E(F(X)) < E(F(Y)). For a simple example, rank on the probability that X is greater than zero. But I may be totally off-beam.

January 30, 2014 — 4:21
• Alexander Pruss

If F is non-linear, you’ll violate the strengthened Axiom 3. But I will need one more axiom: if Y is certainty of 1 and X is certainty of 0, then X<Y. Otherwise, my modified axioms can be satisfied by just letting X<Y and Y<X for all X and Y.

January 30, 2014 — 7:42
• Or maybe what I need is that if X is the certainty of a and Y is the certainty of b, and a<b, then X<Y.

January 30, 2014 — 8:43
• Jeremy Gwiazda

IanS wrote “To make this work, you must be prepared to base your choice on the expectation of arbitrarily large payouts for outcomes with arbitrarily small probability. This I don’t accept, as with the lottery ticket in my example.”

I’m getting a bit confused on a point related to this, one that seems like it must have a standard reply. Let’s say I go into a subway and say to a random person “You give me 10 dollars and I’ll utter some words that will grant you infinite reward after death.” Of course no person would offer up the 10 dollars. But are we forced to say that paying the 10 dollars has positive expected value, because it is not logically impossible that the words grant infinite reward? That seems absurd, but what is the way out?

January 30, 2014 — 10:10
• Jeremy:

That’s close to the “Pascal’s Mugger” story, where somebody comes up to you and says that he’s got superpowers and if you don’t give him \$10, he’ll cause great suffering to 10^100 people (say). That’s a tough one. No infinities there, which removes one distraction.

One answer to the mugger is that the greater the number he claims to be able to cause suffering to, the less the probability he can do it. So maybe the probability that he’s saying the truth is sufficiently small that the expected utility of giving him the ten bucks isn’t worth it. Another answer is that a weirdo who says stuff like this, even if he has the power, is too unpredictable. Maybe if you give him \$10, he’ll cause great suffering to 10^200 people?

The weirdo response might work in your case. Someone who would hold back eternal bliss because one didn’t give him \$10 when he hasn’t given you any evidence but the uncompelling word of a complete stranger is a weirdo, and maybe his actions are just apt to be too unpredictable.

Of course, one might try a similar move in the PW case–that would be a version of the many gods objection, I guess–but I think it wouldn’t work.

January 30, 2014 — 10:44
• Jeremy Gwiazda

Thanks Alex — good points. I may try to track down and reread Pascal’s Mugger and the reply.

And it does seem that other considerations are important in replying to many gods (as you write “I think that in practice, the issue you raise is somewhat alleviated by the fact that the probabilities aren’t that small…”).

January 30, 2014 — 11:34
• Mark Rogers

Dr. Pruss you say:

“The theorem still makes Pascal’s Wager work with the assumption that one might not come to faith even if one tries to believe, as long as one is more likely to come to faith when one tries than when one doesn’t try.”

I do not know about other religions but I think Christianity would say say that to make the assumption that one is more likely to come to faith when one tries to believe than one doesn’t try is error. It is considered pelagianism.

January 30, 2014 — 19:28
• Mark:

Pelagianism is the doctrine that one can come to salvation by one’s own efforts. The claim that one can increase the probability of salvation by one’s own efforts is a quite different claim. Surely our actions do make a difference with respect to the probability of salvation. Suicide bombing seems very likely likely to decrease the probability of salvation. On the other hand, it seems plausible that searching for the truth, trying to do what is right and so on are a preparation for grace. (This is true even if the way that trying to do what is right prepares one for grace is by showing what a wretch one is, how incapable of success one is.)

Moreover, I did not say that the trying to come to faith would be by one’s own efforts. God isn’t idle here. God gives some grace to all. The attempt to come to faith would itself, presumably, be a gift of prevenient grace.

February 1, 2014 — 11:12
• Mark Rogers

Dr. Pruss,