This is following in the line of this question asking what the difference between free-will and randomness/indeterminism is:

What is the difference between free-will and randomness and or non-determinism?

Now, I can describe both determinism and randomness in a mathematically precise manner. Yet, free-will is supposedly neither deterministic nor random.

Past is a past state.
Curr is the current state.
Futr is the future state.

Determinism: P(Curr|Past) = 1
Randomness: P(Curr|Past) = f(Curr, Past) < 1, for all Curr, Past, for some f
Free-will?: P(Curr|Past) < P(Curr|Future)

So, for instance, the digits of pi are not random, because each new digit calculated is fully determined by the prior steps of the calculation.

What is a mathematically precise description of free-will that distinguishes it from determinism and randomness? Is my definition adequate?

And if you say there is no such description possible, you must also prove that in a mathematically precise way :)

Past is a past state.
Curr is the current state.

Determinism: P(Curr|Past) = 1

This appears to potentially leave room for free will in the following form:

Free will: P(Curr|Past) ∈ F(Curr, Past), for all Curr, Past, for some multivalued function F

Note that a multivalued function is not a proper function.

This is precisely how game theory describes a mixed extension of a strategy set. (In game theory you usually may choose the probabilities with which an action is performed.)

Viewed this way, the mathematics of free will is the mathematics of game theory.

And indeed we also find such underdeterminedness in game theory 'proper': it has some (or great) difficulty dealing with equilibrium selection.


Take the following normal-form game:

             Player 2
              L   R 
Player 1  T  9,9 0,8
          B  8,0 7,7

This game has three Nash equilibria. Without further assumptions there is absolutely no way to determine play, not even in terms of probabilities. You might say that the choice of strategy in this game is 'free'.

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    Would you mind explaining a bit more how this differs from the definition of randomness you also recommended? – yters May 2 '14 at 7:21
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    I can try. f is a proper function. The coin falls with probabilities .5 and .5. There's stochasticity, but no choice. F, on the other hand, is a multivalued function. You may choose how you weight the coin and decide upon any p and 1-p you like. The multivaluedness in the formulation introduces indeterminedness in the solution beyond mere stochastics. – user3164 May 2 '14 at 7:28

This answer is superseded by my other answer. I'll leave this one for reference. (Or for voting, e.g. if you like this one more.)

In this lovely paper, which also has a Wikipedia entry and a whole lecture series somewhere on iTunes and YouTube, we find:

The Free Will Theorem. The axioms SPIN, TWIN and MIN imply that the response of a spin 1 particle to a triple experiment is free—that is to say, is not a function of properties of that part of the universe that is earlier than this response with respect to any given inertial frame.

In there, you'll find a definition of sorts of free will. It is a rather strange definition though, because it only says what free isn't:

A free response is a response that is not a function of properties of that part of the universe that is earlier than this response with respect to any given inertial frame.

With a bit of LaTeX, you may turn this into an actual formula.

  • Yes, I've seen that paper before. Unfortunately, that definition fits within what I'd consider random. I'm looking for a third formulation that is neither deterministic nor random. – yters May 2 '14 at 4:50
  • @yters I don't quite remember, but I think that the authors don't think of their 'free' as random (stochastic). I would have to read the paper again to make sure. But perhaps it would be more fun if you had a closer look. :) – user3164 May 2 '14 at 4:53
  • The pertinent point the paper proves is the particle is not determined by a prior state. That is necessary, but not sufficient to define free-will as specified above. – yters May 2 '14 at 5:08
  • @yters I assume that always happens if you define something by stating what it isn't. :) I can't really improve on it, but I did make a suggestion for the question. – user3164 May 2 '14 at 5:18
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    Also, here's a question regarding Conway's paper: philosophy.stackexchange.com/questions/624/… – yters May 2 '14 at 5:56

Before being mathematical or logical, the issue might be philosophical. Are we not facing an aporia?

If we make a conjecture that absolute free will exists, then the outcome of any test experiment is free and could be anything. It could be random, pseudo-random (as in pi) or deterministic, or have any property that the free agent cares to give to that result. Any test might or might not be defeated, there is no way to tell, especially if the agent is given a feedback.

In practice (at least, e.g. in business process modeling, where human intervention is considered), considering free will is actually a useful simplification, providing there are constraints that limit the number of possible choices. This reason why is Occam's razor: by postulating free will, we don't need to get into the details of how a decision is taken; we only need to know it is taken at a specific point.

For example, if we consider an application for a loan from a public administration, we can safely assume that the officer receiving the loan could a) grant it, b) be reject it or c) suspend it pending receipt of more information. The detailed decision process is assumed to be devolved to "free will". The approach of merely providing guidelines (rather than complete and consistent decision charts) greatly simplifies the administrative modeling and might be better suited for practical needs.

In those cases, whereas we will not be able to predict the individual outcomes, we will be able to take statistical measurements, e.g. a) 35%, b) 49% and c) 16%), to give us an appreciation of whether the process goes as we want. Here we have would have a case of relative free will.

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