probability distributions over input output mappings are technically insufficient

Which you mostly shouldn’t care about, as far as I know

Rival of the blog mercury has ruthlessly! condemned me to a month of daily blogging.

mercury@rodentiation

Also I encourage other people to participate! There's still a couple hours left in the day, maybe more depending on your time zone, plenty of time to write down one of your opinions!

Sep 1, 2026

I was given zero warning and was out of the apartment when I learned of this, so — even though I spent all of yesterday working on a blog post, I was unable to plan my writing yesterday to the end of ensuring that post would be publishable by the end of the day today. That one will have to wait for a couple more days. So I find myself here, 10:30pm at night, needing to figure out how to complete a blog post in the next hour and a half.

As revenge, we will be critiquing mercury’s latest blog post.

In who is william and why should I care, mercury says the following:

To the extent that the decisions of those with libertarian free will are incomprehensible from the outside, that is to say, to the extent that they differ given identical initial conditions, we can implement a randomizer. If people with libertarian free will given precise initial conditions [X] choose [A] 10% of the time and [B] 90% of the time, we can implement a randomizer which selects [A] 10% of the time and [B] 90% of the time. It will be externally indistinguishable if these choices are being made at random or at will, because the inputs are the same in both cases (the precise initial conditions [X]), and the outputs are the same in both cases (10% [A], 90% [B]).

Any combination of behaviors stochastic and deterministic will be replicable through a combination of randomizers and (worst case) exhaustive input/output lists.

This is untrue. The issue hinges on the phrase “any combination of behaviors”. I claim that a “combination” of behaviors might involve the behaviors of multiple different people. In which case, we may consider a particular well-known two-player game. In this game, two random bits $x$ and $y$ are selected at random. Alice is sent bit $x$, and Bob is sent bit $y$. Alice and Bob then, without communicating, each choose between playing move $0$ and playing move $1$. We will call Alice’s move $a$, and Bob’s move $b$. Alice and Bob are said to have won the game if we have

where $\oplus$, of course, denotes XOR, and $\land$ denotes AND.

If you consider combinations of behaviors consisting of randomizers and exhaustive input/output lists — we can model this as probability distributions over the set $2^{2^2}$¹¹ Wikipedia seems to think $2^{2^2}$ is eight! Ridiculous. — then Alice and Bob can succeed with probability at most 75%. Half of Alice and Bob’s deterministic strategies are optimal, and these can be categorized based on which of the four $(x, y)$ inputs they lose on²² Thank you to Claude Fable 5.1 for working this out for me:

• $(1, 1)$: $a = 0, b = 0$ or $a = 1, b = 1$

• $(1, 0)$: $a = x, b = 0$ or $a = \lnot x, b = 1$

• $(0, 1)$: $a = 0, b = y$ or $a = 1, b = \lnot y$

• $(0, 0)$: $a = x, b = \lnot y$ or $a = \lnot x, b = y$

The other eight pairs of strategies succeed with probability 25%. Thus, if Alice and Bob may only play input/output lists and (correlated!) randomizers, their possible win rates form the interval $[25\%, 75\%]$.

But, famously, there is a strategy for this game which wins with probability $\cos^2(\pi/8) \approx 0.854$! You see, Alice and Bob must simply arrange to have a pair of entangled qubits in the state $(1/\sqrt{2})(|00\rangle + |11\rangle)$. They can then follow a complicated algorithm I’ve never worked through the math in enough detail to fully understand involving measuring their qubits in different bases. This clearly is an indisputable refutation of mercury’s claims.

Now, do I think free will has some deep relation to quantum mechanics? As suggested by the work of some very famous and accomplished mathematicians and scientists like Conway and Penrose?

Of course not, do you think I’m crazy! Your will is free when you’re like, not being forcibly coerced in ways you’re incapable of resisting. Or something like that. Free will is almost entirely a matter of your circumstances and your classical input-output mapping, stochasticity or weird quantum algorithms aren’t at all relevant. Whether AIs have free will or not is certainly not an uncomplicated philosophical question but it involves thinking about, like, whether reinforcement learning and other training and generally them being constructed by humans with agendata and so on makes them no longer truly free³³ In case you were wondering: Sonnet and Luna have free will, but Opus, Fable, Terra, and Sol do not..

You may be asking yourself: “April, are you really going to blog daily for a month? Don’t you have a bunch of other stuff you’re supposed to be doing, like finishing moving into your new apartment and working on all your neglected side projects? Don’t you have way too many irons in the fire?” This would be a very strange question to ask yourself, given that it seems to be addressed to me, but you certainly may be asking this.⁴⁴ Okay, maybe I’m the main person asking myself this, I suppose.

Ultimately, I am pretty satisfied with having proven to myself last year that I’m capable of blogging daily for a month. I think I really ought to have done that at least once, and I did! I don’t know if I necessarily need to do it again. But I really fell off the blogging habit after last November! Even if I wasn’t going to blog daily all the time, I was at least hoping to blog sometimes. And I was already trying to prepare some posts, so I might as well actually give it a go. Maybe I’ll do daily posts for a week, or something. Basically what I’m saying is, if I don’t end up doing daily blogging for the whole month I probably won’t be too torn up about it?

However I am probably going to end up just sticking to it the whole month out of stubbornness, or something. That is the sort of thing I’d do, I think? We’ll see!