Consider a choice between two wagers on a fair coin:
W1: on heads, you get $1 if you are perfectly rational and $3 if you are not
W2: on tails, you get $2 if you are perfectly rational and $1 if you are not.
Suppose you are perfectly rational, and that it’s a part of perfect rationality that you know for sure you’re perfectly rational. It’s obvious you should go for W2. But let’s calculate. We immediately run into the zero-probability problem that I’ve lately been thinking about. For if you’re perfectly rational, the probability that you go for W1 is zero, so E(U|W1) seems to be undefined. Of course, E(U|W2) is unproblematically half of $2, or $1, but you can’t say whether that beats “undefined” or not.
Suppose you think: Maybe E(U|W1) is undefined in classical probability, but maybe I can use some other way of defining it, say using Popper functions.
Well, let’s think about what E(U|W1) “should be”. So imagine that you actually go for W1. Now, only an imperfectly rational agent would go for W1. So, if you were to go for W1, you would get $3 on heads, so your expected payoff would be $1.50, which beats anybody’s expected payoff for W2. So, formally, E(U|W1) is undefined, but if you close your eyes to that and think intuitively, you get E(U|W1) equally $1.50, which yields the wrong result that as a perfectly rational agent you should go for W1.
What if we say that a perfectly rational agent need not know for sure that they are perfectly rational? Suppose, say, you are perfectly rational agent who is 0.99 sure you are perfectly rational. Then E(U|W1) and E(U|W2) are both well-defined. But what are they? Well, it’s intuitively clear that if you are 0.99 sure that you are perfectly rational, you should go for W2. But supposing that’s right, then W1 entails you are not perfectly rational, and since P(W1) = 0.01, the expectation E(U|W1) is well-defined, and must be equal to $1.50. Oops!
This line of reasoning assumed evidential decision theory. What if you go for causal decision theory? Well, there are two causal hypotheses: R (you are perfectly rational) and Rc (you are not) with P(R) = 0.99 and P(Rc) = 0.01. So now your causal expected utility on W1 equals
- CE(U|W1) = 0.99E(U|W1∩R) + 0.01E(U|W2∩Rc).
What is this? Well, W1 ∩ R is the empty set! But conditionalizing on an empty set is not a merely technical problem in the way that conditionalizing on a specific zero-probability outcome of a continuous spinner is. Rather, it is simply nonsense. So the first summand is undefined, and hence the sum is undefined. Thus you simply cannot make a decision with causal decision theory here.
It’s obvious that if you’re nearly sure you’re perfectly rational you should go for W2. But neither evidential nor causal decision theory gives a way to that conclusion.
[By the way, the reason I set up W1 and W2 as I did, with one having the payoff on heads and the other on tails, was to ensure that we didn’t have domination. For one might reasonably say that a perfectly rational agent will try to decide on grounds of domination first, before resorting to probabilities.]
3 comments:
Maybe the fact that being perfectly rational about a choice seem to be an option for us is confusing the picture. We could restate this with a red toke for a perfectly rational agent and a blue token for an imperfectly rational agent, and assume to token is assigned is a way outside of the agent's control. Then:
I have a red token.
Choose W1 and heads: $1
Choose W1 and tails: $0
Expected with choose W1: $0.50
Choose W2 and heads: $0
Choose W2 and tails: $2
Expected with choose W2: $1.00
I have a blue token.
Choose W1 and heads: $3
Choose W1 and tails: $0
Expected with choose W1: $1.50
Choose W2 and heads: $0
Choose W2 and tails: $1
Expected with choose W2: $0.50
So red token agents choose w2, because they cannot at the time of the choice choose their token, (with red token made equivalent to whether they are perfectly rational). All numbers seem to be defined at the time of the choice of W1 versus W2. Whether the token is changed after that choice is up to the person framing the problem.
OK, let's see. So I have a red token and I am perfectly rational. But then I can't get over the problem that if I know for sure I am perfectly rational and have a red token, then P(W1)=0, and so E(U|W1) is undefined rather than $0.50. On the other hand, if I am merely confident, say 99%, that I am perfectly rational and have a red token, while I am completely certain that I am perfectly rational iff I have a red token, then E(U|W1)=1.50, because the only way I am going to choose W1 is if in fact I am NOT perfectly rational. What am I missing?
I've been thinking a bit more about the EDT version with 99% probability that I am perfectly rational.
Imagine I am one of 200 agents offered the choice, and I know that of these 200 agents all but two are perfectly rational. I have no reason to think myself special, so my credence is 99% that I am one of the perfectly rational ones.
Hypothesis 1: In a case like this, it's rational to choose W1.
If this is right, then let's think how the 200 agents will tend to do. Two of them will in fact be imperfectly rational, so when they choose W1, on average one of them will get $3. Of the remaining 198 agents, on average 99 of them will get $1.
Hypothesis 1 thus has a rather weird consequence. For if these 200 agents instead had chosen W2, then 99 of them on average would have got $2, while 1 of them would have got $1. Hypothesis 1 thus seems really implausible. In fact, it's a kind of reversal of the Newcomb situation for CDT: it is the rational EDTer who doesn't get as rich as they should!
Hypothesis 2: In a case like this, the rational thing is to choose W2.
But then choosing W1 is perfectly correlated with being perfectly rational, so everyone who chooses W1 is imperfectly rational, and hence gets on average $1.50, while choosing W2 gets one at most on average $1.00. EDT then seems to require one to choose W1.
So neither Hypothesis can be right.
What's left?
I guess:
Hypothesis 3: Both options are rational.
Hypothesis 4: Neither option is rational.
I am not attracted to either of these.
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