A Disproof of the SIA
The proof presented below I believe is novel, but builds on a hole in the SIA that’s been known for at least 14 years, if not longer.
For those unfamiliar with what the Self-Indication Assumption (SIA) tries to address, here is an explainer:
The SSA is essentially the logic that drives the “Doomsday Argument”, which I think the majority of people agree is self-evidently false. The SIA is an attempt to rectify where the SSA goes wrong. In the above post, I’ve argued that you don’t need the SIA to reject the SSA, which simply rests on a self-begging assumption. But I’d never figured out how to disprove the SIA (except indirectly, i.e. arguing what it implies about Sleeping Beauty) until now.
The Proof
God decides to play a little game. He creates one person in a room, then gives that person a vision of Him flipping a coin. If it comes up Heads, He stops the game. If it comes up Tails, He creates two indistinguishable rooms each with one person and gives them the same vision of another coin flip. The procedure repeats: Heads stops; Tails leads to the creation of twice as many rooms as last created (2 then 4 then 8 and so on).
Let’s call the current round of your creation (before God’s next coin flip) “C”, and the final round where God finally flips Heads “F”.
Let’s also say that God looks into the future and tells us when he’ll eventually flip heads: F = j. The chance that the next coin will flip Heads can be written as P(C = j | F = j). The chance that we’re in any particular batch of creations can be written as P(C = k | F = j).
Under the SIA, the probability of being in the final batch of creations is just the fraction of observers who are in that batch (since nobody knows their rank, all else is equal, and you can be anybody).
- The count of observers in batch k is
2^(k - 1) - The count of total observers is
2^j - 1 - Lemma 1:
P(C = k | F = j) = 2^(k - 1) / (2^j - 1) P(C = k | F = k) > 2^(k - 1) / (2^k)- Lemma 1.5:
P(C = k | F = k) > 1/2
Normally, we think of SIA as favoring Tails: The more observers that get created, the more likely you are to have been from that larger group of observers.
However, when we fix F, suddenly the SIA will start slightly favoring Heads. This might seem counter-intuitive, but it’s just the SIA again weighing towards theories with greater numbers of observers: You’re marginally more likely to be in the final batch (size 2^(j - 1)) than everything preceding the final batch (size 2^(j−1) − 1).
Now instead consider if God tells us our current round number C = k.
As puts it, with the SIA “your existence is rendered more probable according to theories on which there are more people who you might be”. When our batch becomes known, then this stops becoming relevant because we can no longer consider ourselves possible members of future k+m batches.
Without the SIA to modify anything, the chance of God’s next coin flip being heads is a simple 50/50:
- Lemma 2:
P(F = k | C = k) = 1/2
If we like, we can derive this like so:
P(C = k | F = j) / P(C = k | F = k) = (2^k - 1)/(2^j - 1)- By Bayes,
P(F = j | C = k) / P(F = k | C = k) = (P(C = k | F = j) / P(C = k | F = k))*(P(F = j) / P(F = k)) - By the SIA,
P(F = j)/P(F = k) = (2^(k-j))*(2^j - 1)/(2^k - 1)- (I’m omitting the math for this step for brevity, but can share it on request)
- Cancel out terms,
P(F = j | C = k) / P(F = k | C = k) = 2^(k - j) P(F = k | C = k) summed over every j ≥ k = 1- The geometric series of 2^(k - j) sums to 2, leaving:
P(F = k | C = k) = exactly 1/2
Let’s call “H” the event that God’s next coin flip will be Heads, just another name for “C = F”. When we’re given C = k, H is the same event as F = k, and also vice versa.
The contradiction arises when we calculate P(H).
We can do so via Lemma 1.5:
P(H | F = k) > 1/2, for all k- Therefore by the law of total probability…
- Lemma 3:
P(H) > 1/2
Similarly:
- From Lemma 2,
P(H | C = k) = 1/2 - This is true for all k, therefore…
- Lemma 4:
P(H) = 1/2
P(H) cannot be both 50% and greater than 50%, hence the SIA must be false.
Reality
Here’s the real math for the Lemma 3 section:
P(H | F = k) = P(C = k | F = k) = P(F = k | C = k)*P(C = k)/P(F = k)P(H | F = k) = (1/2)*P(C = k)/P(F = k)P(H) = sum over P(H | F = i)*P(F = i)- Substituting in
P(H | F = k)and canceling out terms: P(H) = (1/2)*(sum over P(C = i))P(H) = 1/2
Ah, but Lemma 3 was built off of Lemma 1. Surely the mistake must have been earlier, when we derived a P(C = k | F = j)?
Yep! We can use more complicated math to get here, but I’ll skip it because this part is intuitive:
P(C = k | F = j) = 1/j
It’s just applying the principle of indifference to the possible current rounds.
None of this however gets to the heart of the issue, which I believe is easier to see with simpler examples—such as the one that Bentham’s Bulldog gives here:
Clone coin toss: There are five empty rooms. A coin is tossed. If it comes up heads, someone is created in the first room. If it comes up tails, someone is created in all five rooms. Upon being created, what should your credence be that the coin came up heads?
The SIA would have you believe that P(E | H) = 1/6.
Actually:
P(E | H) = 1P(E | T) = 1P(T | E) = P(E | T)*P(T)/P(E)P(T | E) = (1)*(1/2)/(1) = 1/2
The definition of “E” I borrow from , who in the comments:
P(someone with my experiences exists | heads) = 1P(someone with my experiences exists | tails) = 1
When we define an event like “I exist”, that “I” will either refer to the only person it can refer to in Heads, or it will refer to any one of the five people it can refer to in Tails. We assign probabilities of 1 here, but they’re not actually probabilities (a rationalist should avoid 0% and 100% probabilities); they’re just expressions of the identity function. The setup of the problem defined “I” as someone already selected, so there’s no way that “I” couldn’t have been selected.
This can feel unsatisfying. Surely there’s some sense in which my existence is not a guaranteed outcome? Can we not take a further step back and consider that uncertainty?
We can. For instance: What could someone have believed about you before you existed?
Let’s say we’re eagerly awaiting for God to create Igor, our only friend from a past life.
Igor is only one soul out of trillions, so the chance he happens to get picked after any flip is low. If we learn after a particular flip that he has indeed been picked, then we will update towards believing God more likely created more people.
This, by the way, is perfectly analogous to a classic problem that was solved back in 1959 by Martin Gardner, the Girl Boy Paradox. Asking about Igor’s existence matches up with asking “Of the two dogs you’ve put up for adoption, ma’am, is either one a male?” then increasing your credence that both dogs are male.
On the other hand, say we approach God and ask him to provide the name of ANY one person created this round, and he provides “Ivan”. That info then neither implies Heads nor Tails and we keep our 50/50 credence. This matches the Girl Boy Paradox variant of meeting the eldest child of a two-child family, which tells you nothing of the other child’s gender and therefore remains at a 50/50.
Igor was pre-marked by our attention (what I like to think of as “premarkable”), so a random draw of Igor represents either 1/N or 5/N chance depending on the coin flip. Being told “Ivan”, however, tells us nothing beyond the fact of his name.
The right way to think about waking up inside of one of God’s rooms is to see that it’s an event that’s not been at all pre-marked. It’s a fact we learn, but not a fact that narrows; my identity tells me nothing beyond the fact of my identity.
This time I’m putting a Subscribe button just before the final paragraph (not counting the post script “Prior work” section). Is this distracting? Should I swear to never do this again? Please let me know in comments. Or don’t, I’m not the boss of you!
In some setups, our existences can be viewed as probabilistic events. In other setups, they’ll only be viewed as factual properties already established. The SIA over-generalizes, trying to assume that we are random draws from all matching possible observers in all situations (and the SSA does the same thing). The only correct move is to reject this assumption (thereby avoiding both the SIA’s presumptuous philosopher and the SSA’s Adam and Eve problem), under a theory I’m not sure has a name already, but if I could give it one, I’d call it the Self-Identity Non-assumption: Don’t make ungrounded assumptions about your identity; not every piece of info is a probabilistic statement; some are best expressed as identity functions.
Prior work
This disproof does not work if we limit ourselves to finite scenarios. The fact that the SIA doesn’t hold up with infinities was already recognized by here:
https://www.lesswrong.com/posts/99isjR8W4JGZdro7x/sia-fears-expected-infinity
I am more convinced by this disproof than I am by Armstrong’s (though maybe I shouldn’t be??) since his relies on the inability to renormalize (arguably easy to dismiss with a shrug and “math is weird”) where this one leads to a contradiction (and has no specific points where I would want to get off the train with just a shrug).
I understand if this won’t be convincing to those who only believe in the existence of finite scenarios, and I would be interested in investigating that angle further. I do hope that this will be convincing for anyone who happens to believe that the universe is infinite (a sentiment I don’t believe anyone will ever be able to prove, but which I also think is more likely than not!).
Also see ‘s response to the above-linked Bentham’s post here:
Lastly, please share in the comments any corrections, typos, objections, opinions or errata!
You should share this post. That’s because, though as we’ve already established I’m not the boss of you, I did give your boss a call and let them know that this was very important and they agreed, and thank you for understanding.
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