# How can you assign a finite value to an outcome of infinite value?

Source: 10 minutes 53 seconds juncture; Lecture 5, Video 4 (transcription);
MITx: 24.00x Introduction to Philosophy; by MIT Associate Prof Caspar Hare PhD (Princeton)

[...] where infinite rewards are at stake, we need to expand upon our account of rationality. The natural way to do it is this: So if there's some number, n, such that if we set the value of the infinite reward at n, then the expected value of a particular option is greater than the expected value of all other options. And furthermore, if we set the value of the infinite reward at any finite number higher than n, then the expected value of that same option comes out as higher than the expected value of all other options. Then we rationally ought to take that first option.

[Prof Hare doesn't verbalise the following, which is only depicted in the video]:

Rational Behavior When An Outcome of Infinite Value is at Stake

It is rational for you to take an option when there is a finite number n, such that, if we assign any value n or higher to the outcome that is of infinite value, then that option has highest expected value.

It's a bit of a complicated idea, but it gives us the right results, which is that you ought to take the certain eternity of a heaven, over the 1/6 of a chance of eternity in heaven. And it gives us a response to these Many Gods, Weird Gods, Punitive Gods, Generous Gods, objections to Pascal's Wager. Because [...], just so long as you think it's very slightly more likely that you'll get eternity in heaven if you believe in a particular kind of god, (rather than doing anything else, believing in another kind of god or not believing in God,) then that is rationally what you ought to be doing. [...]

I don't understand the above. The bold requires you to assign some finite value to an outcome of infinite value, but how is this possible? Both (1 × ∞) and (1/6 × ∞) = ∞ .

2. How does assignment of value help? Instinctively, one would just assign the highest value to whatever one desires subjectivity. So this lack of objectivity is biased and self-defeating.

What the professor would like to have happen is the following:

Let's suppose you can achieve infinite value with action A, with `p(A) = 0.7`. Let's suppose you can achieve infinite value with action B, with `p(B) = 0.4`. Now, if this infinite value were finite N, we could compare A & B:

``````  0.7*N + 0.3*(small stuff) > 0.4*N + 0.6*(small stuff)
because we can ignore the small stuff if N is really big and
0.7*N > 0.4*N
``````

So, let's just pick sufficiently big N and do it this way.

This, at first glance, seems like it might be a way to deal with infinity. (The explicit step of plugging in probabilities and ignoring the small stuff is what was missing in saying that it gives the right answer for certain vs. 1/6th-certain obtainment of infinite value.) If, for instance, the only possible infinite good is being "saved" by a (single, known) effectively omnipotent monotheistic god, it seems to work out okay.

But let's think about this a little bit more. How could you accrue infinite value, anyway? One obvious choice would be everlasting finite value: something has goodness G per unit time, and you can make it happen forever (and apply no temporal discounting). Awesome! Then, picking N is equivalent to picking some point sufficiently far in the future and only counting the first however many millennia: N = G*T for some really big T.

However, what if you can accrue goodness not at rate G but at rate 2*G if you take the riskier 0.4-chance method? Then the two accounts diverge:

Fixed value N method:

``````0.7*N > 0.4*N
``````

Fixed time T method:

``````0.7*G*T < 0.4*2*G*T = 0.8*G*T
``````

Oops. So it doesn't even work in really simple cases like this, where finite actions now lead to infinite ongoing outcomes.

More generally, if you have two competing infinite goods, you are unable to decide between them, because if you pick separate `N` and `N'` for the two things, it is no longer true that one will always beat the other: who beats whom will depend on your choice.

As if this weren't bad enough, the consequences of treating infinity this way are rather shocking: as soon as you can imagine an infinite good that is not completely, absolutely impossible, you are forced to abandon all else in pursuit of it, including consideration of other people. So, for instance, if you need to massacre and torture an entire civilization in order to save one person for eternity (something which you imagine has infinite value where all else is finite), then it is "rational" and just to do it.

So I don't think you can argue anything that comes out of this is "rational"; the analytic framework doesn't adequately capture the issues that might be at play.

(But you should understand the "make N really big" idea and why it fails mathematically despite looking initially promising.)

• Could you rewrite this so you explain the professor's idea first, and then debunk it second? The way it is presented now, it reads like you're advancing your own argument rather than answering the question, even though the pieces are the same. – Chris Sunami supports Monica Jun 23 '15 at 19:12
• @ChrisSunami - Good point. How is it now? – Rex Kerr Jun 23 '15 at 20:47
• This reads much better to me. Now it's a direct answer to the question with some additional commentary. – Chris Sunami supports Monica Jun 24 '15 at 13:21

He's addressing the issues that can arise when considering infinite rewards.

Instead of considering heaven to be "infinite" reward, consider a Pascal's wager where the benefit of correct belief is some large, but finite, amount of "gain", call this amount `n`. If you can find a finite `n` such that it is rational to believe in god on cost/benefit analysis, and that the belief remains rational for larger (but still finite) values of `n`; then one can argue that the belief continues to remain rational in the limit of infinite rewards.

It's a mathematical approach to the problem. One useful way around the difficulty of using infinity in mathematical calculations is to replace it with an arbitrarily high number.

In this case it seems like he's comparing n with 1/6 of n. As you pointed out, those are seemingly equal when n is infinity. However, if you replace n with any finite positive number, no matter how bit, n will be larger than 1/6 of n (six times larger). This pattern doesn't alter as you approach infinity --in absolute terms, the imbalance gets steadily bigger as n gets bigger.

Hare is suggesting that the above result makes n > 1/6 * n a better model than n = 1/6 * n even at n = infinity. (As Dave suggested, another way of expressing this is that the limit of n divided by [n * 1/6] is 6, as n approaches infinity.) That would be a legitimate conclusion in most mathematical contexts, but whether it is aptly applied to Pascal's Wager is open to debate.