Say you’re dealt 13 out of 52 standard playing cards. Call the chance of getting two aces A. Now imagine a second round, in which I tell you that I know that you already have at least one ace in your hand. The chance of holding two aces in this scenario is B. Lastly, I tell you that the ace I know you’re holding happens to be the ace of spades. The chance of holding two aces is now C. Can you sort A, B and C?
Part of the reason this is so counter-intuitive is that this setup is actually ambiguous and the answer depends on how you interpret it. “The ace I know you’re holding” is worded poorly—I may be holding more than one ace! Consider the following options:
C1: I check whether you hold the ace of spades specifically and tell you either that you have it or that you do not have it.
C2: I check whether you hold at least one ace, and if so, I tell you the suit of a randomly chosen one of those aces, otherwise I tell you that you have no aces.
The probabilities of getting at least two aces, conditional on being told you have the ace of spades, is different in these two scenarios! In fact, B < C1 and B = C2. I get p(C1) = 56%, p(C2) = 36.9% = p(B). Maybe the intuition here is a little clearer, since we can see that winning hands that contain an ace of spades are all reported by C1 but some are not reported by C2, while all losing hands that contain an ace of spades are reported by both C1 and C2 (since there’s only one ace for C2 to choose from). So C2 is “enriches” for losing states when conditioning on being told that we have an ace of spades.
This is somewhat like the “Ignorant Monty” variant of the Monty Hall problem where Monty chooses a door (other than the contestant’s door) at random, potentially revealing either a goat or a car. Should you switch when he reveals a goat? If you haven’t seen this before, solve it yourself first—I found it as unintuitive as the original Monty Hall problem.
Maybe the intuition here is a little clearer, since we can see that winning hands that contain an ace of spades are all reported by C1 but some are not reported by C2, while all losing hands that contain an ace of spades are reported by both C1 and C2 (since there’s only one ace for C2 to choose from)
If I am not mistaken, this would at first only say that “in the situations where I have the ace of spades, then being told C1 implies higher chances than being told C2”?
Each time I try to go from this to C1 > C2, I get stuck in a mental knot.
[Edited to add:] With the diagrams below, I think I now get it: If we are in C2 and are told “You have the ace of spades”, we do have the same grey/loosing area as in C1, but the winning worlds only had a random 1⁄2 to 1⁄4 (one over the number of actual aces) chance of telling us about the ace of spades. Thus we should correspondingly reduce the belief that we are in these winning worlds. I hope this is finally correct reasoning. [end of edit]
I can only find an intuitive argument why B≠C is possible: If we initially imagine to be with equal probability in any of the possible worlds, when we are told “your cards contain an ace” we can rule out a bunch of them. If we are instead told “your cards contain this ace”, we have learned something different, and also something more specific. From this perspective it seems quite plausible that C > B
Okay, I think I managed to make at least the case C1-C2 intuitive with a Venn-type drawing:
(edit: originally did not use spades for C1)
The left half is C1, the right one is C2. In C1 we actually exclude both some winning ‘worlds’ and some losing worlds, while C2 only excludes losing worlds. However due to symmetry reasons that I find hard to describe in words, but which are obvious in the diagrams, C1 is clearly advantageous and has a much better winning/loosing ratio.
(note that the ‘true’ Venn diagram would need to be higher dimensional so that one can have e.g. aces of hearts and clubs without also having the other two. But thanks to the symmetry, the drawing should still lead to the right conclusions.)
I think your left diagram is correct but the one for C2 is off somewhat. In both, we’re conditioning on the statement that “you have an ace of spades”, so we’re exclusively looking in that top circle. Both C1 and C2 have the same exact grey shaded area. But in C2, some of the green shaded region inside that circle is also missing: the cases where you have an ace of spades but I happened to tell you about one of the other aces instead. So C2 is a subset of C1 (condition on being told you have the ace of spades) where only a randomly selected subset of the winning hands are chosen (1/2 of the ones with two aces, 1⁄3 of the ones with three, etc).
But that correction doesn’t really change much since your diagram is just the combination of four disjoint diagrams, one for each of the suits. So the ratio of grey to green is right, but I find it harder to compare to C1.
Either way, my main point was that C2 might have been driving our intuition that C=B, and in fact, C2=B, so our intuitions isn‘t doing too bad.
Part of the reason this is so counter-intuitive is that this setup is actually ambiguous and the answer depends on how you interpret it. “The ace I know you’re holding” is worded poorly—I may be holding more than one ace! Consider the following options:
C1: I check whether you hold the ace of spades specifically and tell you either that you have it or that you do not have it.
C2: I check whether you hold at least one ace, and if so, I tell you the suit of a randomly chosen one of those aces, otherwise I tell you that you have no aces.
The probabilities of getting at least two aces, conditional on being told you have the ace of spades, is different in these two scenarios! In fact, B < C1 and B = C2. I get p(C1) = 56%, p(C2) = 36.9% = p(B). Maybe the intuition here is a little clearer, since we can see that winning hands that contain an ace of spades are all reported by C1 but some are not reported by C2, while all losing hands that contain an ace of spades are reported by both C1 and C2 (since there’s only one ace for C2 to choose from). So C2 is “enriches” for losing states when conditioning on being told that we have an ace of spades.
This is somewhat like the “Ignorant Monty” variant of the Monty Hall problem where Monty chooses a door (other than the contestant’s door) at random, potentially revealing either a goat or a car. Should you switch when he reveals a goat? If you haven’t seen this before, solve it yourself first—I found it as unintuitive as the original Monty Hall problem.
Thanks for the attempt at giving an intuition!
I am not sure I follow your reasoning:
If I am not mistaken, this would at first only say that “in the situations where I have the ace of spades, then being told C1 implies higher chances than being told C2”? Each time I try to go from this to C1 > C2, I get stuck in a mental knot. [Edited to add:] With the diagrams below, I think I now get it: If we are in C2 and are told “You have the ace of spades”, we do have the same grey/loosing area as in C1, but the winning worlds only had a random 1⁄2 to 1⁄4 (one over the number of actual aces) chance of telling us about the ace of spades. Thus we should correspondingly reduce the belief that we are in these winning worlds. I hope this is finally correct reasoning. [end of edit]
I can only find an intuitive argument why B≠C is possible: If we initially imagine to be with equal probability in any of the possible worlds, when we are told “your cards contain an ace” we can rule out a bunch of them. If we are instead told “your cards contain this ace”, we have learned something different, and also something more specific. From this perspective it seems quite plausible that C > B
Okay, I think I managed to make at least the case C1-C2 intuitive with a Venn-type drawing:
(edit: originally did not use spades for C1)
The left half is C1, the right one is C2. In C1 we actually exclude both some winning ‘worlds’ and some losing worlds, while C2 only excludes losing worlds.
However due to symmetry reasons that I find hard to describe in words, but which are obvious in the diagrams, C1 is clearly advantageous and has a much better winning/loosing ratio.
(note that the ‘true’ Venn diagram would need to be higher dimensional so that one can have e.g. aces of hearts and clubs without also having the other two. But thanks to the symmetry, the drawing should still lead to the right conclusions.)
I think your left diagram is correct but the one for C2 is off somewhat. In both, we’re conditioning on the statement that “you have an ace of spades”, so we’re exclusively looking in that top circle. Both C1 and C2 have the same exact grey shaded area. But in C2, some of the green shaded region inside that circle is also missing: the cases where you have an ace of spades but I happened to tell you about one of the other aces instead. So C2 is a subset of C1 (condition on being told you have the ace of spades) where only a randomly selected subset of the winning hands are chosen (1/2 of the ones with two aces, 1⁄3 of the ones with three, etc).
But that correction doesn’t really change much since your diagram is just the combination of four disjoint diagrams, one for each of the suits. So the ratio of grey to green is right, but I find it harder to compare to C1.
Either way, my main point was that C2 might have been driving our intuition that C=B, and in fact, C2=B, so our intuitions isn‘t doing too bad.
oh.., right—it seems I actually drew B instead of C2. Here is the corrected C2 diagram:
Beautiful! That’s also a nice demonstration of B=C2.