Proof that God doesn't exist

A conclusive philosophical argument

I come with unfortunate news: in my investigations, I have stumbled on a completely decisive proof of atheism. I know, me even saying this is proof enough to convince you, but you should at least read the argument since it’s a good maxim to think for yourself. Take some time to emotionally adjust, and then proceed.

Everyone learns in grade school that it’s possible for things with probability 0 to happen. If I flip a coin infinitely many times, it could happen that I get a heads every time. But the probability of that is less than or equal to the first n tosses all being heads, the probability of which is 1/2ⁿ. So the probability of getting all heads is less than or equal to 1/2ⁿ for any n, so the probability is 0, known for being the only number in [0, 1] less than or equal to all 1/2ⁿ.

You may reply: “Well, it seems too crazy to accept that something with probability 0 can happen. So I’ll just say it’s impossible for each of the infinity tosses to land heads.” That won’t work, Joe. Whatever the sequence of flips actually ends up being—whichever random but even distribution of heads and tails comes up—there is likewise a probability of zero that that particular sequence would be the one that comes up, for exactly the same reason: the probability that the flips match a given sequence for the first n tosses is 1/2ⁿ, etc.

So, things with probability 0 can happen. Now, consider the following scenario.¹ A man, Bob, has just finished an infinite sequence of independent and fair coin flips (he started a very long time ago). After surveying his results, he tells you a shocking fact: only finitely many of the flips landed tails. Now, he asks you what the probability is that the 100^th most recent coin flip laded tails (EDIT—look at this footnote²). Easy question, right? There are infinity heads, but only N tails, and you have no evidence as to where the tails-flips occurred, so it seems like the probability that one of the N tails was on the 100^th-most-recent flip is equal to N/infinity = 0. Or, if the preceding reasoning doesn’t pan out, at least it is very slim. Okay.

Now imagine an alternative scenario. Again, Bob just finished his infinity coin flips. Except this time, he wasn’t paying attention during the 100^th most recent flip, and he forgot to write it down. Anyway, he tells you that of the coin flips he was paying attention during (that is, all of them except the 100^th most recent), only finitely many of them were tails. Now, what’s the probability that the 100^th coin flip was tails? Well, that’s just an independent coin flip; it’s the same as if you just flipped a coin now, except it just happened to be the 100^th in this series. Bob didn’t take note of the result, so nothing he tells you has any correlation to what happened on the 100^th flip, so it’s just a fair coin toss whose result you don’t know. It seems clear that the probability that flip landed tails is 0.5.

Let F denote the proposition that only finitely many tosses landed tails, F' the proposition that only finitely many of the tosses other than the 100^th landed tails, and let H be the proposition that the 100^th landed tails. Then, summarizing, we have:

\(\begin{align*} P(H | F) &< \frac{1}{2}\\ P(H | F') &= \frac{1}{2} \end{align*}\)

But wait a minute: F and F' are the same outcomes. No matter what happened on the 100^th flip, that can’t change whether only finitely many flips were tails. If infinitely many flips other than the 100^th are tails, then the number of tails is infinite either way; if only finitely many other than the 100^th are tails, then even the 100^th being tails won’t push the finite number to infinity. Therefore, we should have P(F) = P(F'). How do we get out of this paradox?

Maybe we should say that P(H | F') is not equal to 0.5. But that seems implausible. The 100^th coin flip is just a coin flip, and its inclusion in the sequence is arbitrary. If you say it doesn’t have a 0.5 chance of tails, you’d have to say that about every possible coin flip. Say you’ll flip a coin tomorrow. We can then consider that to be the 0^th flip in the sequence. Based on Bob’s testimony, you now know that only finitely many of the total amount of flips—whether his flips, or yours tomorrow—will be finite. Do you then conclude the 0^th flip is probably going to land heads? Of course not, you’re just flipping a coin tomorrow, the probability of tails is 0.5.

Should we instead say that P(H | F) = 0.5? I don’t think so. If you thought the chance that the 100^th toss was tails was 0.5 (in the case where Bob looked at the whole sequence), you’d have to say that about every particular toss. What’s more, you’d have to say that, for any finite collection of n C and any sequence s of length n, the probability that the flips in C were in accordance with the sequence s is 1/2ⁿ. For example, the probability that the 10^th, 100^th, and 1234^th flips were H, T, and T, respectively is 1 in 8. (To see why this is, just run the second case, except where Bob failed to note those three results. No finite collection of tosses can make a difference as to whether only finitely many were tails).

However, we then run into problems. Let’s create a natural number m whose binary digits come from the results of the flips, with tails being 1 and heads being 0. So, for example, if Bob’s result was H, T, H, T, T, H, H, and then infinitely many Hs, then m is (going backwards) 11010₂ = 26. Because there will only be finitely many 1s, we are guaranteed to have a well-defined m. Here’s the problem: given that there are only finitely many tails, each possible value of m is equally likely. Say that a and b are both fewer than n digits long. If there is a 1 past the n^th digit, then neither a nor b are the value of m. If there is not a 1 past the n^th digit, then the probability that m matches a (including the leading 0s) is 1/2ⁿ, and similar for b. The probability is the same in both cases, so P(m = a) = P(m = b). This reasoning holds for all possible values of m. So, m is an integer greater than or equal to 0, each value equally likely. But this is impossible, for there is no uniform distribution on a countable set.

Maybe the two probabilities in question—P(H | F) and P(H | F')—are not well-defined? Fat chance. P(H | F') is definitely equal to 0.5, as explained above. And if it’s defined, so is P(H | F), since F = F'.

The actual solution to this paradox is the same reason why this paradox will literally never affect us in our lives: the scenario is impossible. Bob can’t flip a coin infinity times, and if he did, he wouldn’t be able to be sensitive to the whole history of the sequence so as to tell you there were only finitely many tails. I would admit it’s possible for things with probability 0 to happen—nothing rules out the possibility that something like infinitely many independent coin flips could occur in some world, and whatever the particular sequence that obtains is, that sequence has probability 0. But we could never learn that something with probability 0 happened, since that would require getting infinite information, which nobody can do. Our eyes, ears, etc. can only take in a finite amount of information. Even if a dart strikes an infinitely precise dart board, we could never measure it with infinite precision, so as to learn which probability-zero event happened.

But if the paradox can only be solved by supposing agents can’t learn that probability-zero events happen, then omniscience is impossible, for God could never know what probability-zero events have obtained. And certainly it should still be possible for such events to obtain, since if God cannot do infinite coin flips, then he is not omnipotent. If you think infinite time is impossible, then God could do the infinite flips as a supertask: he’ll take 0.5 seconds to do the first flip, 0.25 to do the second, 0.125 to do the third, etc.³ If there is some nonzero amount of time such that God cannot flip a coin that fast, then again, he is not omnipotent.

You may reply that the paradox only arises if someone learns there were finitely many tails without learning which particular flips were tails. But, you say, God already knows all the results, since he is omniscient. This doesn’t fix the problem: even if God knows everything, he could still tell an angel that there were only finitely many tails, and then the paradox would arise for that angel. If God cannot tell an angel such a thing, then he is not omnipotent.

If you reply, “perhaps the angel isn’t certain that they’re really being told by God that there were finitely many tails,” you must still admit that the agent assigns nonzero probability to God having told the truth and them having understood God correctly and such, and so they assign nonzero probability to F = “there are only finitely many heads,” and the paradox re-arises with respect to the conditional probability distribution P(— | F).

Therefore, God does not exist.

(EDIT—I’m beginning to worry some of you actually think I think this is a good argument against the existence of God. Please, live a little.)

(EDIT 2—A real, fatal flaw in my reasoning was identified by Plasma Bloggin’. In retrospect, the error is a rather silly one. Identifying this, I think, leads to a satisfying solution to the puzzle, even granting that the weird cases with zero probabilities can happen. See if you can find it before looking at his comment.)

1. I got this case from Daniel Rubio, though I only just now realized it proves that God doesn’t exist.

2. What I have in mind here is that Bob has been flipping for infinitely long, and has now come to a stop. So maybe he flipped a coin a second ago, and one two seconds ago, etc. I’ll refer to the last flip as the “first,” the second-last as the “second,” etc. But the order of the infinite series shouldn’t matter.

3. If you don’t like infinite coin flips, we can replace the example with God picking a random number between 0 and 1, and treating each 0 in the binary expansion as a heads and each 1 as a tails.