This is a game I thought of to explore ideas. I haven’t tried it yet, so it might be horribly cumbersome in practice.

You will need:

• A question which you wish to explore competing answers to
• A partner to play with. I think this works better online than in person, but you can probably do it in person too
• Some sort of versioned collaborative document system. As a programmer, I’d probably just use git, but google docs or something like etherpad should work too
• Some means of communicating with your partner. If you’re face to face, that’s easy. If not, use an instant messenging client

You and your partner should take two sides of the question. They don’t have to be sides you agree with. Indeed, if you have a difference of opinion on the subject then you might want to consider taking the side that is the opposite of your opinion.

Your goal is not to debate this topic. This is important. You are not the ones debating.

Your goal is to create a written document in which two fictional characters debate this topic.

In order to achieve this goal, you will play a game. Decide which one of you goes first. They then play their opening move, which is to state their character’s name and belief. e.g. “My name is Dr Cornelius Frugalmuffin. I believe that people have a moral obligation to wear hats”. “My name is Joe Colinsworth. I think people should be free to leave their heads bare if they want to”.

Play now proceeds as follows. You take it in turns. On a player’s turn they may do the following:

1. They may submit a draft response
2. They may edit their current draft response
3. They may commit to their current draft response, at which point play passes to the other player
4. They may request to wind back time. They pick a previous response from the other player and request to reset to that point. The other player may accept or refuse. If they accept, all responses past that point are deleted, and play proceeds with the other player with that past response as their current draft (which they may just immediately commit to if they want)
5. They may request to end the session. If the other player agrees, they make their closing statement. The other player then also has the opportunity to make their closing statement. The game now ends

During this play, the two players should be talking on the back channel a lot. In particular the period between initial draft response and submission should be spent discussing the draft, and providing constructive criticism on your partner’s character’s argument.

Care should be taken to make sure that the back channel discussion is kept collaborative and the debate does not move there. This is very much a co-op game. The goal is not to “win” the debate, and a request to end the session is not a concession. The goal is to produce a debate that is as informative as possible and explores the ideas as thoroughly as possible. The back channel is a place for saying things like “I don’t understand this particular point” or “I think there might be a flaw in this argument, but it could be fixed in this way”, not “You’re wrong because”.

The request to wind back time is to give the impression that the characters in your dialogue have thought through the issues much more thoroughly than you have: Anything you learn over the course of the discussion can be sent backwards in time to your characters so that they had already thought of it. This ensures they can give the strongest argument possible for their case.

The back channel can also simply be used to ask your partner for advice. “What’s the argument your character is hoping I’m not going to notice here?” is a legitimate question. As is “You actually know more about the position I’m arguing for than I do. What should I read before making my response?”. Remember: You are not your characters. The goal is not to win. The goal is to produce an interesting and informative fictional dialogue.

This is a bit of pure mathematics. I was originally interested in it for decision theoretic reasons, but it turned out to be the wrong structure for what I was looking for. Still, I found the result interesting enough in its own right that I thought I would write it up.

Let \(S\) be some non-empty finite set. Let \(\mathcal{C}\) be the set of non-empty subsets of \(S\). Let \(\mathcal{R}\) be the set of random variables taking values in \(S\). A strategy is a function \(s : \mathcal{C} \to \mathcal{R}\) such that \(P(s(U) \in U) = 1\).

A strategy is said to be subset consistent if when \(U \subseteq V\) and \(x \in U\) then \(P(s(U) = x) = P(s(V) = x) P(s(U) \in V)\). (The intuition here is “It doesn’t matter what order we add pieces of information in”).

It turns out that there’s a very nice characterisation of all subset-consistent decision strategies.

Theorem:

Let \(L_1, \ldots, L_n\) be a partition of \(S\) into non-empty sets. Let \(w : S \to (0, \infty)\). Define a strategy as follows:

Given \(U \in \mathcal{C}\), define \(U’ = L_i \cap U\), where i is the largest such that this set is non-empty. Now pick \(x \in U’\) with probability proportional to \(w(x)\).

This strategy is subset-consistent.

Proof:

Note that \(s(U) = s(U’)\).

Let \(V \subseteq U\), and let \(L_i\) be as above. If \(L_i \cap V = \emptyset\) then \(P(s(U) \in V) = 0\), and in particular for \(x \in V\) we have \(P(s(U) = x) = 0\), so the result is proved in this case as both sides of the equation are 0.

In the case where \(V \cap L_i \neq 0\), we have \(V’ = V \cap L_i\), and \(s(V)\) is chosen from \(V’\). Note that \(s(U) \in V\) iff \(s(U) \in V’\).

\[\begin{align*}
P(s(U) = x) &= \frac{w(x)}{\sum\limits_{y \in U’} w(y)} \\
&= \frac{w(x)}{\sum\limits_{y \in V’} w(y)} \frac{\sum\limits_{y \in V’} w(y)}{\sum\limits_{y \in U’} w(y)} \\
&= P(s(V’) = x) P(s(U) \in V’) \\
&= p(s(V) = x) P(s(U) \in V)\\
\end{align*}\]

As desired.

QED

We will spend the rest of this proving that in fact every subset-consistent strategy arises this way.

From now on let \(s\) be some fixed strategy.

We’ll need some notational convenience. Let \(\rho_U(x) = P(s(U) = x)\). Let \(\tau(x,y) = \rho_{\{x,y\}}(x)\) if \(x \neq y\) and \(\tau(x,x) = \frac{1}{2}\).

The following lemma is practically just definition shuffling, but is surprisingly useful:

Lemma: Let \(x, y \in U\). Then \(\rho_U(x) = (\rho_U(x) + \rho_U(y)) \tau(x, y)\).

Proof: If \(x = y\) then this just boils down to \(\rho_U(x) = 2 \rho_U(x) \frac{1}{2}\). So suppose \(x \neq y\).

Let \(V = \{x, y\}\). By subset consistency we have \(\rho_U(x) = \rho_V(x) P(s(U) \in V)\). But \(\rho_V(x) = \tau(x,y)\) by definition, and \(s(U) \in V\) iff \(s(U) = x\) or \(s(U) = y\), so \(P(s(U) \in V) = \rho_U(x) + \rho_U(y)\). Combining these gives the desired result.

QED

Say \(x \prec y\) if \(\tau(x,y) = 0\). Say \(x \sim y\) if neither \(x \prec y\) nor \(y \prec x\). Note that because \(\tau(x,y) = 1 – \tau(y, x)\) it’s not possible to have both \(x \prec y\) and \(y \prec x\).

Lemma: If \(x, y \in U\) with \(x \prec y\) then \(\rho_U(x) = 0\).

Proof: \(\rho_U(x) = (\rho_U(x) + \rho_U(y)) \tau(x, y) = 0\).

This has a partial converse:

Lemma: If \(x, y \in U\) with \(\rho_U(x) = 0\) and \(rho_U(y) \neq 0\) then \(x \prec y\).

Proof:

By the preceding lemma we have \(\tau(x,y) = \frac{\rho_U(x)}{\rho_U(x) + \rho_U(y)} = \frac{0}{0 + \rho_U(y)} = 0\). This is well defined because \(\rho_U(y) \neq 0\).

QED

Lemma: If \(x \prec y\) and \(y \sim z\) then \(x \prec z\). Similarly if \(x \sim z\) then \(z \prec y\).

Proof:

We’ll only prove the first version. The second will follow similarly.

Let \(V = \{x,y,z\}\). Then because \(y \prec z\) we have \(\rho_U(y) = 0\). Then \(\rho_U(x) = \rho_U(x) \tau(x, y)\). But because \(x \sim y\) we have that \(tau(x, y)\) is neither \(0\) nor \(1\), so the only possible solution to this is that \(\rho_U(x) = 0\).

QED

Lemma: If \(x \prec y\) and \(y \prec z\) then \(x \prec z\).

Proof: Again consider \(V = \{x,y,z\}\). We must have \(\rho_V(x) = \rho_V(y) = 0\) because \(x \prec y\) and \(y \prec z\) respectively. Therefore \(\rho_V(x) \neq 0\). Therefore by a previous lemma we have \(x \prec z\).

Lemma: \(\sim\) is an equivalence relationship.

We have \(\tau(x,x) = \frac{1}{2} \neq 0, 1\) by definition, so \(\sim\) is reflexive.

It is symmetric because \(\tau(x, y) = 1 – \tau(y, x)\), so if \(\tau(x,y) \neq 0, 1\) then \(\tau(y,x) \neq 0, 1\).

We now must show transitivity. Suppose \(x \sim y\) and \(y \sim z\). If it is not the case that \(x \sim z\) then we must have either \(x \prec z\) or \(z \prec x\). Suppose without loss of generality that \(x \prec z\). Then by a previous lemma we must have \(y \prec z\), contradicting our assumption that \(y \sim z\).

QED

Theorem: There exists a partition of \(S\) into finitely many sets \(L_1, \ldots, L_n\) such that for any \(U \in \mathcal{C}\) we have \(P(s(U) \in U \cap L_i) = 1\) where \(i\) is maximal such that this set is non-empty.

Proof:

We will take as our partition the set of \(\sim\) equivalence classes. These equivalence classes are given a total ordering by \(\prec\), because if \(w \sim x\) and \(y \sim z\) then \(x \prec y\) implies \(w \prec z\). There are finitely many of them, so we can put them in an order \(L_1, \ldots, L_n\) such that if \(x \in L_i\) and \(y \in L_j\) with \(i < j\) then \(x \prec y\). If \(L_i\) is maximal such that \(L_i \cap U \neq \emptyset\), then for any \(x \in L_j\) with \(j < i\) we must have \(\rho_U(x) = 0\), because for some \(y \in U\) we have \(y \in L_i\) and thus \(x \prec y\). Therefore \(P(s(U) \not\in L_i \cap U) = 0\) as desired. QED We now need only define our weight function. Define \(w(x) = \rho_{L_i}(x)\) where \(x \in L_i\). Note that \(w(x) \neq 0\) because if it were 0 then we'd have \(x \prec y\) for some \(y \in L_i\), contradicting that \(x \in L_i\). Theorem: Let \(s(U)\) is chosen by choosing from \(U'\) with probability proportional to \(w(x)\). Proof: We have \(P(s(U) = x) = P(s(U') = x)\) by the previous theorem. Now, \(U' \subseteq L_i\). Therefore by subset consistency we must have \(P(s(U') = x) = \frac{P(s(L_i) = x)}{P(s(L_i) \in U'\)} = \frac{w(x)}{\sum\limits_{y \in U'} w(y) }\). This was the desired result. QED We have now shown the converse to our original theorem: Every subset-consistent strategy can be constructed from a partition and a weight function. As an additional note, I've thought a bit about how this works if you lift the restriction that \(S\) must be finite. A lot of this carries through naturally, but some of it doesn't. I think if you replace the subset consistency hypothesis with the natural infinitary equivalent that if \(U \subseteq V \subseteq W\) then \(P(s(W) \in U) = P(s(W) \in V) P(s(V) \in U)\) and add the additional restriction that for every \(U\) there exists an \(x \in U\) with \(P(s(U) = x) \neq 0\) then a lot of this carries over naturally with the partition replaced by one whose reverse is a well-ordered set and each partition is countable. But I haven’t worked through the details yet, and I’m not sure I’m interested in doing so.

A question I occasionally get asked is “What do you have against the Von Neumann Morgenstern Utility Theorem, David?”. Or the more specific variant “Which of its premises do you disagree with?”.

Well, it would be faster for me to list what aspects of the theorem I do agree with.

Which is, err.

Um.

Well I haven’t come up with a convincing objection to the independence hypothesis. That’s not to say that I necessarily think it’s right, only that I haven’t got a strong opinion on it either way.

On top of its explicit premises, I also reject the implicit ones that this is a sensible way to reason about peoples’ behaviours, and the conclusion would say anything useful even if I were to grant all of the other premises.

So, yeah, I have a lot of objections. Far too many to write only one article about it. So instead I’m going to break this up and write it as and when I feel like it. Advance warning: When I write series of articles, they tend not to be finished. I will try to make each of these self contained so you don’t need to hold your breath waiting for the next one.

I will start with the most obvious place to begin this series: Somewhere in the middle picked almost at random.

Preferences over lotteries are discontinuous

Amongst the “reasonable hypotheses” of the VNM theorem is a continuity assumption. It can be formulated as this:

Suppose we have lotteries A, B, C with A < B < C. There is some probability p such that the lottery \(L_p = pA + (1-p)C\) is equivalently preferable to B. There are other, weaker, formulations of this but that doesn't matter very much: If continuity is not a hypothesis of your variant of the VNM theorem, it's still a conclusion of it, so its truth value is relevant.

I think continuity is simply false. Consider the following example: C is “I have a sandwich”, B is “I have a kitten to play with” and A is “The world is destroyed”.

We have \(L_0 = A < B < C = L_p\), certainly. We also have that for \(p < p'\), \(L_p > L_{p’}\) (I will always prefer to increase the probability of my getting a sandwich versus destroying the world).

There is no probability with which I will accept a sandwich as a fair exchange for some probability of destroying the world. No matter how tasty the sandwich is, or how cute the kitten I’d get in the non world destroying chance.

But I certainly prefer kittens to sandwiches (not to eat of course. To cuddle). If the probability of destroying the world is functionally indistinct from 0, I will pick \(L_p\) – give me a kitten with probability 1. If it is functionally distinct from 0, I will always pick \(B\) – the possibility of the kitten isn’t worth it, give me the sandwich instead.

You might argue that there is some true tiny probability with which I would consider them equal, but I’m skeptical. Further, I think that even if there were such a probability in theory (which I am not convinced of), it is so small as to not be plausible to estimate or work with.

Android apparently thinks I have a potty mouth.

To be fair, it’s right. I do have a potty mouth. I also regard this as none of my phone’s goddamn business.

Specifically the problem is that Android does not auto complete or correct profanity. It’s in the dictionary, but it will never correct typos for profanity or allow me to swipe type it. Apparently it’s really convinced that I like to talk about ducking, and that when I complain about this fact I am hitching.

I did a bunch of searching on this and couldn’t find a solution. Some mild lateral thinking provided me with one though. Here’s some Google bait to share it.

Basically, the problem is that the word is in the dictionary with a special flag that says “You can’t use this word”. It’s recognised but disallowed.

So the solution is to get it to ignore the fact that it’s in that dictionary. As well as the system dictionary you have your personal dictionary (where it remembers words you’ve said “Remember this word”). You can manually edit that dictionary. This feature is mostly to remove existing words that you’ve accidentally added, but it works for adding new words too.

How to do this:

1. Go to settings
2. Select “Language and Input”
3. Select “Personal Dictionary”
4. Select “+”
5. Select the Phrase field
6. Add the specific profane word you want to use in that field
7. Select “Next” on your keyboard
8. Select “Done” on your keyboard
9. Select “Back” to go back to the dictionary
10. That word is now in your personal dictionary and should be usable.

You have to manually add every word that you want to use this way, but once you add any given word it will be picked up correctly. About fucking time.