In The Logic of Decision, second edition, Richard Jeffrey endorsed a version of evidential decision theory supplemented with a principle of ratification. A decision is ratifiable iff, conditional on making that decision, the expected utility of carrying out the decision is at least as high as than the expected utility of doing something else. In Newcomb’s Problem, although one-boxing has highest expected utility, only two-boxing is ratifiable. In zero sum games with each agent knowing the other agent’s strategy and no Nash equilibrium, only mixed strategies are ratifiable. So there might be some cases where the principle looks attractive. But it isn’t defensible in general. Here I’ll argue against a relatively weak version of the ratification principle.

Weak Ratificationism: In a decision problem with only one ratifiable strategy, that strategy is uniquely rational.

In our puzzle, the Agent has to name some non-negative integer. Call this number a. After the Agent names the integer, another number will be displayed on a scoreboard. The other number, call it s, will be generated by a process that is causally independent of the Agent’s choice. The Agent’s payout is as follows.

• If a = s + 1, and s > 0, then the Agent gets 2 utils
• If either a = 0 or s = 0, the Agent gets 0 utils
• Otherwise, the Agent gets 1 util

The number on the scoreboard will be the number named by a psychological Twin of the agent (called Twin) who is playing the same game. Agent believes, with probability 1, that Twin will follow the same strategy as Agent in producing the number. This doesn’t mean Twin will produce the same number – both players may be following a mixed strategy of some kind. If Agent is following the mixed strategy that leads to them saying 1 with probability 0.4, and saying 2 with probability 0.6, then Twin is following the same mixed strategy. Since the probability of Agent saying 2 and Twin saying 1 is 0.24, Agent’s expected return is 1.24.

The only ratifiable strategy for the Agent is to say 0. This is ratifiable, because given the Agent says 0, then with probability 1 the Twin says 0, so the expected return of any strategy is 0, so saying 0 is as good as any other strategy.

Clearly this is the only ratifiable pure strategy for the Agent. If they say any number greater than 0, say x, then saying x+1 has an expected return of 2 (conditional, that is, on having said x), while saying x has an expected return of 1.

But there are no other ratifiable mixed strategies in the game. Or at least there aren’t if we assume, as I do, countable additivity for probabilities. Given countable additivity, any mixed strategy must be representable by a function f from non-negative integers to [0, 1] such that the various values of f(x) sum to 1 as x ranges over the non-negative integers. When no ambiguity results, we’ll let f name both a function and the strategy it represents. Let F be the sum, as x ranges over the positive integers, of f(x)f(x+1). Then the expected return of f is (1 – f(0))² + F. (The first term is the probability, given that both players are playing f, of Agent’s return being at least 1. The second term is the probability, given that same assumption, of Agent’s return being 2.)

If Agent decides to play f, the Twin will play f with probability 1. We want to prove that no f, other than the f that makes f(0) = 1, is ratifiable. To this end, we aim to construct, for any other f, a rival strategy g such that the expected return of g is greater than the expected return of f, conditional on Twin playing f.

Note that if Twin plays f, and Agent plays g, then Agent’s expected return is (1 – f(0))(1 – g(0)) plus the sum, as x ranges over positive integers, of f(x)g(x+1). The first term is the probability of Agent’s return being at least 1, and the second is the probability of it being 2.

Given that, it is easy to construct g. Let k be the smallest number such that f(k) is positive. Let g agree with f over every value except k and k+1. In those cases, g(k) = 0 and g(k+1) = f(k) + f(k+1). It’s easy to see that making this change doesn’t lower the value of any term in the sum that is the expected return of g, and raises at least one term. (Either the first term, if k = 0, or the value for x=k otherwise.) So f is not ratifiable; given it is being played, g has a higher expected return.

So the only ratifiable strategy is to simply play 0. But this strategy is guarantees getting the worst possible result in the game, and is weakly dominated by every other strategy. (That is, there are some possibilities where the alternative strategy does better, and none where it does worse.) Perhaps, if you know that things will go maximally wrong for you come what may, a strategy that guarantees the worst possible outcome might be rationally permissible, as it doesn’t make things worse. But it is impossible to see how a strategy that guarantees the worst outcome could be uniquely rational. So Weak Ratificationism fails.

Posted by Brian Weatherson in Uncategorized

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