Playing Games with Eternity

Surely what you need to do here is to do a proper analysis using some form of Non-Standard Analysis. Let I be the infinite goodness of heaven. Let D be the extra badness from spending a day in hell. Then, the payoff for playing today is I/2, the payoff for playing tomorrow is 2I/3-D, the payoff for playing the day after tomorrow is 3I/4-D, etc. Is it better to play today or tomorrow?

Payoff(tomorrow)-Payoff(today) = 2I/3 - D - I/2 = I/3 - D.

Since I is infinite and D is finite, this is positive, so it's better to play tomorrow.

Of course this leads to the paradox that if the probabilities keep on going up, you should wait.

Now, in the Pascal case, we need to be similarly careful. If I procrastinate believing until tomorrow, I get an extra day of earthly pleasure of utility E. But there is a non-zero probability p that I'll die before tomorrow and not get to heaven at all. Let's say that the chance God exists and rewards all and only the believers is 1/2. Let the infinite reward of heaven be I, and ignore hell.

Then:
Payoff(believe today) = I/2
Payoff(believe tomorrow) = ((1-p)/2)I + E.

So:
Payoff(believe tomorrow) - Payoff(believe today) = -pI/2 + E.

And this is negative, because I is infinite and E is finite. So it's more prudent to believe today.

One gets the results in your post only if I is an infinity such that I + E > I but (1-p)I = I. But that would be an incoherent account of infinity.

I suspect you can't get a coherent rational decision calculus that handles infinities if you use a single infinite quantity. You need a method, such surreal numbers or non-standard analysis, which lets you have many infinite numbers that can be coherently arithmetically handled just like finite numbers. There will, then, be multiple infinite numbers (a number is positively infinite iff it is bigger than every standard natural number; a number is a positive infinitesimal iff its reciprocal is positively infinite). If I is infinite, so is I/3, and I - I/3 = 2I/3, just as with finite numbers.

"But that's not right. You neglect entirely the infinitely bad payoff if you land in hell eternally."

Sure, we can consider all these things, and others, but we still get the same answer. Let D be the badness of a day in hell. Let H be the value of a day in heaven. Let J be the infinite badness of hell counting from tomorrow on. Let I be the infinite goodness of heaven counting from tomorrow on.

So, then in Gracely's story (assuming your summary of it):

Payoff(today) = (I+H)/2 - (J+D)/2
Payoff(tomorrow) = 2I/3 - J/3 - D

Payoff(tomorrow)-Payoff(today) = I/6 + J/6 - D/2 - H/2.

Since I and J are positive infinite numbers, and D and H are finite (or even if infinite, infinitely smaller than J and I, respectively), the difference is positive, and so you should procrastinate. Which is paradoxical, since the same analysis works on any day.

"But then consider the payoffs in A and B. A = I/2 = +oo. B = (1-p)I = (at least) +oo, for any positive p you like. So there is no chance that Payoff(believe tomorrow) - Payoff(believe today) is negative."

This makes the mistake of considering all infinite numbers to be the same.

Actually, B = (1-p)I/2.

So, A-B = pI/2, and you should go for A. In a decent mathematical model of infinite utilities, they can be manipulated according to exactly the same arithmetical rules as ordinary numbers. (It is a basic meta theorem of non-standard arithmetic that any first order arithmetical claim remains true given a non-standard extension.)

"There is also a non-zero chance that you die before you utter today, 'I believe'. There is always a non-zero chance in Pascal's wager of finding yourself not believing, no matter what you do."

We can model all that. Let p1 be the probability that I will die before today's time for uttering "I believe". Let p2 be the probability that I will die before tomorrow's time for uttering "I believe". Let p3 the probability that I couldn't get myself to believe no matter what I did. If that's all we take into account, then:

Payoff(today) = (1-p1)(1-p3)(1/2)I - (1-(1-p1)(1-p3))(1/2)J
Payoff(tomorrow) = (1-p2)(1-p3)(1/2)I - (1-(1-p2)(1-p3))(1/2)J + E

Then (by subtracting and collecting terms--I cheated and used Wolfram Alpha):
Payoff(tomorrow)-Payoff(today) = E - (J+I)(1-p3)(p2-p1).

Since p2 > p1, and since both J and I are positive infinite numbers, (J+I)(1-p3)(p2-p1) is a positive infinite number. Since E is a finite numer, E - (J+I)(1-p3)(p2-p1) is a negative infinite number. And so you should play today.

You can model lots of other small perturbations of the probabilities, but the lesson is that as long as the infinite prize is more likely to be obtained and the infinite punishment is more likely to be avoided if you plan to play today than if you plan to play tomorrow, then, regardless of what finite addends there might be in the equations, you should play today.

Well, the exact decision we need to make now is to either:
A: Plan to believe today.
B: Plan to believe tomorrow.

We get the result that we should go for A instead of B as long as:

(*) P(eventually believe | A) > P(eventually believe | B).

Is (*) true? Well, it's pretty plausible. Planning to do something typically makes it more likely that we do it. However, we more often deviate from plans for tomorrow than for plans for today, because (a) we have more time to change our minds, and (b) we may have an accident before tomorrow that renders the fulfillment of the plan impossible.

Anyway, it seems the question whether (*) is true has nothing to do with infinities in the payoff.

The question isn't exactly whether to believe today, but whether to try to believe today. But then the relevant probability query is whether:
(**) P(eventually believe | try to believe today) > P(eventually believe | don't try to believe today).

I take "eventually believe" to mean "believes at some time or another". And for simplicity (and contrary to fact), I take that to imply certainty of salvation.

Now (**) is pretty plausible.

Cantorian infinities are infinite cardinalities. There is some arithmetic one can do with cardinalities, but only some. For instance, the negation of an infinite cardinality is nonsense. Likewise, the product of an infinite cardinality and a real number is nonsense.

Now I don't understand your earlier statement "But the expected value (EV) of my believing today depends in part on the probabiliyt that I succeed in believing today." The expected value of an action is a conditional expectation given that action's success. Otherwise we are talking of the expected value of a trying.

Hi, I know I'm late to this party, but there are two cents here in my pocket. From a gambler's perspective, you've misstated the terms of Pascal's wager. You are not gambling an eternity in hell vs. an eternity in heaven, or even a finite time on earth vs. an eternity in heaven, but rather the extra pleasure available on earth vs. the existence of heaven. An analogy is if someone were to offer you a trade, which you could accept at any time. One the one hand, you receive a pension of a penny every week for life. On the other, you can trade it for a lottery ticket with today's winning numbers. You don't have to make the choice now. The only way you would ever choose the pennies is if you believed the lottery did not exist, or did not award prizes. If you believe the lottery is real, or even suspect it's real, you will take the ticket immediately, because the pennies are negligible by comparison, and you could miss your chance at the ticket if you delay. Nobody rational would gamble $100 million to win a penny, or even handfuls of pennies.

Likewise, Gracely's wager is one of foolish risk vs. finite reward. Every day you wait costs you one day in hell, but a single loss costs you infinite days in hell. Only a fool would play the first day, but even after a year or a hundred years, the risk is unacceptably high. The point of Gracely's problem, I believe, is that the wisest course, the most profitable course, is to spend an infinite number of days in hell waiting for your win to become assured. This is for two reasons - the amount you risk losing is so much greater than what you are assured of paying, and because the Satan of the fable will most assuredly rig the game. Every fool who plays will certainly wind up condemned to hell by his own choice, whereas the wise man stays in hell for an equal eternity, preserving his hope. There is never a point at which playing the game becomes a wise move, because you are still gambling millions to earn pennies. This is Zeno's Race - you never quite get there.

Going back to Pascal's wager, you state that a rational person could justify waiting minutes, or days, or some amount of time, believing heaven was available but wanting to accumulate a little more sinful pleasure. But our nearsightedness as humans obscures the ridiculous nature of this argument. In my analogy, suppose the person waits several years successfully prior to accepting the ticket. Will he then say proudly to other lottery winners that "I won $100 million , and also $1.73 in pennies!" He will never mention the pennies again, and if you ask him his net worth, he will tell you $100 million even. No amount of finite reward adds to eternity - it is always subsumed.

So the only rational time to accept Pascal's wager is immediately upon the advent of faith, "the evidence of things unseen". And the best time to play Satan's game is never.