I have recently written up a very short paper, in which I suggest that an old puzzle from the Martin Gardner books, known as the ‘Second Ace’ puzzle, could be used as a counterexample to van Fraassen’s reflection principles and their more recent descendants. The puzzle doesn’t seem to have been widely discussed in the philosophical literature yet, although some people in the AI community, including Jo Halpern, apparently have something to say about it, albeit not (explicitly) in connection with Reflection.
Reflection
van Fraassen famously suggested the following (where Cr is a credence function an t, t* are times):
Special Reflection (SR): it ought to be the case that 
where 
.
This says that one’s current credence in A, conditional on having a future credence of x in A ought to be x. It follows from this that it ought to be the case that your credence at t in A is the average of your future possible credences in A, weighted by the probability that you give at t to having them.
van Fraassen offered a number of reasons to endorse SR, of which:
- It yields some intuitively correct prescriptions in certain cases. Example: I am watching a programme that I believe with absolute certainty will provide conclusive evidence for x’s guilt; intuitively, I ought to be certain of his guilt at this point in time already.
- A convincing Dutch book argument can be given for SR.
- SR is entailed by Strict Conditionalisation alongside another intuitively compelling requirement on rationality.
- SR entails Strict Conditionalisation.
I’ll bracket (2)-(4) here. Weisberg has recently argued against all three; I haven’t read his paper in any depth, so I’ll have to reserve judgment for now.
Now (1) seems good prima facie evidence in favour of something like SR. However, although SR does yield good results wrt the case mentioned above in (1), it also yields pretty silly prescriptions wrt cases in which one knows that one’s future beliefs may be arrived at in epistemically dubious manners, for instance in cases in which one may have forgotten prior evidence or found oneself in some state of cognitive incompetence. So a number of people have argued for a weakened version of SR restricted to future credence functions that don’t involve information loss and the like (Elga and Weisberg both offer something along these lines). Call this ‘improved’ version of SR, ‘SR+’.
Ok, now it strikes me that SR+ yields some funny results.
The second-ace puzzle
We have a deck of cards with four cards: ace of hearts (AH), ace of spades (AS), deuce of hearts (DH), deuce of spades (DS). Two cards are picked at random and handed to Alice. The outcome space is
Bob’s initial credences in the six possible hands are 1/6 each. In particular, he starts off, at 
, with 
as his degree of belief in Alice having the two aces.
At 
, Alice then tells Bob ‘I have at least one ace’, waving the back of one of the cards at Bob. Bob conditionalises on the info and assigns 
, having eliminated 
.
At 
, Alice moves on to say ‘Tomorrow, I’ll tell you what colour that ace is’. Bob immediately reasons that either: (i) it will be announced black, the options 
and 
will be eliminated and the updated credence will be 1/3, or (ii) it will be announced red, the options 
and 
will be eliminated and the updated probability will be 1/3. So either way the probability updated on future knowledge will be 1/3. By SR+, he sets 
.
But intuitively, by merely telling Bob that she will announce the colour of the ace that she has previously revealed that she has, surely Alice leaves Bob none the wiser as to whether or not she has two aces. SR* seems to generate counterintuitive results.
A colleague of mine has made a suggestion, which may be along the lines of of something that Halpern says, but which, in my opinion, is not satisfactory. I’ll post it shortly.
Posts
Not sure I’ve got this right, but I think you are wrong in claiming that Bob’s credence in AH&AS, after learning the colour of the ace held by Alice, should be 1/3. It seems, rather, that it should remain 1/5 and that his credence in each of the other two hypotheses not ruled out by the new (colour) evidence should be 2/5. I am assuming that Bob knows Alice will say she has at least one ace when and only when she has at least one ace.
Suppose it is announced red (ARED) While this does eliminate AS&DS and AS&DH, it leaves Cr(AH&AS) unchanged because P(AH&AS/ARED) is P(ARED/AH&AS)P(AH&AS)/P(ARED) which is (1/2)(1/5)/(1/2)(1/5) + (1)(1/5) + (1)(1/5) which is (1/5).
Am I missing something?
Hi Joel, thanks for commenting! Interesting; I think you may be right!
My reasoning was as follows. Suppose the card is announced red, then Bob has just found out that Alice has (at least) the ace of hearts, i.e. Bob finds out that AH. Now
But now I see that the way that I presented the problem doesn’t vindicate this. Indeed, as the problem is presented here, if the card is announced red, Bob learns from Alice, not just the proposition that she has the ace of hearts, but also that she asserts this. Let
stand in for ‘Alice says that P‘. We get, for example:
I think that we probably get the same kind of result even if we do away with Alice’s assertion altogether, so that Bob finds things out ‘directly’, so to speak.
Hi Jake. That is what I had in mind in taking the evidence to be ‘that Red is announced’ (ARED).
Yes, I know. Sorry, I should have stuck to your original notation!
This puzzle doesn’t look like it’s really about reflection. Leave out the part of the story where Alice announces her intention to announce the color. Bob can still reason as follows: “Either the card is black, or it is red. Each has probability one-half. If it is black, the probability of two aces is 1/3; if it is red, the probability of two aces is 1/3. So the probability that the other card is an ace is P(ace2|black1)*P(black1) + P(ace2|black2)*P(black2) = 1/3*1/2 + 1/3*1/2 = 1/3.” No appeal to reflection there, just conditionalization.
It looks like the puzzle really arises from two incompatible ways of treating the state space. The official way it was given in the statement of the puzzle treated the two positions in Alice’s hand as indistinguishable. But in that case there’s no way to talk of the color of “that ace”; i.e, there’s no proposition in the state space to be expressed by “That ace is black”, as opposed to just “I’m holding a black ace.” But if Alice only tells Bob “I’m holding a black ace”, then Bob’s second inference is mistaken. As Joel Pust explained,
and 
are both still 1/5.
The alternative is that the two positions in Alice’s hand are distinguished—say she’s holding one of them in her left hand and the other in her right hand. In that case there are in fact 12 different atomic states:
, and so on. Then if Alice announces “The card in my left hand is an ace”, there are six possibilities remaining, and the second card is an ace in two of those six possibilities. So the probability after Alice’s first announcement that she is holding a second ace is 1/3. Her later announcement of its color won’t change this.
So there are two pieces of reasoning in the puzzle, one that leads to 1/5, another that leads to 1/3. Only the first is right if the cards are indistinguishable; only the second is right if the cards are distinguishable.
Hi Jeff. Good point about the state-space distinction.
This puzzle seems to me to have very similar form to the Monty Hall problem. I choose door 1, and have credence 1/3 that I’ve chosen the door with the prize. Monty says one of the other doors has no prize, and he’s about to reveal one. I reason, if he reveals door 2, then my credence will be 1/2 that door 1 has the prize, and if he reveals door 3, then my credence will be 1/2 that door 1 has the prize, so I should already have credence 1/2 that door 1 has the prize. But of course, this isn’t how things work.
As some of the others have pointed out, the problem here is that you don’t just learn that Alice has a red ace, but also that she said she has a red ace. Similarly, in the Monty Hall problem, you don’t just learn that door 3 has no prize, but that Monty has shown that door 3 has no prize. In each case, one of the situations (the case where Alice has both aces, the case where the door you selected in the first round has the prize) allows two different ways for Alice/Monty to reveal the information. Thus, on some specifications of what happens, each piece of information only has a 50% chance in this situation, and thus in some sense, half of this possibility is excluded as well when the information is received. When you work through the calculations, you see that the proper credence both before and after this extra piece of information is still 1/5 in the Alice case and 1/3 in the Monty case.
Yes indeed Kenny, well-put! The thought had occurred to me in passing.
I concur with Joel and Kenny’s diagnosis.
If you play bridge, and are serious, you learn a principle called ‘restricted choice’ that works the same way. Suppose that I am missing both the king and queen in a key suit. I do not know which opponent (LHO or RHO) has it. Suppose I play the jack, and RHO wins it with the king. Now LHO is a 2 – 1 favorite to hold the queen. The reason is that if RHO had the king only, he would always play the king — likelihood of 1. If he had both the K and Q, he would play the K only half the time — likelihood of 1/2. This skews the posteriors toward LHO holding the Q.