Here's code to compute the probability empirically (I got an answer of 0.2051384 for 10000 draws. It's written in C# but can be readily converted to other functional languages such as Haskell).

var cards = new[] { "AH", "AS", "2C", "2D" };  
var pairs = cards.Combinations (2);  
var pairsWithAce = pairs.Where (p => p.Any (c => c.Contains ("A")));  

var drawsWithAce =  
(  
    from i in Enumerable.Range (1, 10000)  
    let randomPairWithAce = pairsWithAce.ElementAt (rand.Next (pairsWithAce.Count()))

You have three sets of 100 cards - either all aces, no aces, or exactly one ace; all three are equally likely. You lay the cards out in front of you.

I either (a) ask you whether you've got an ace, and you say yes, or (b) turn over one card chosen at random, and find that it's an ace.

In both cases I now know that you've got at least one ace, but the posterior probability that you have all aces is 1/2 in case "a" , and (I think) 100/101 in case "b".

Argument 2 isn't expressed clearly enough for me to understand it; it seems to be missing some steps. But the answer is 1/5.

Argument 2 is correct. There are lots of ways to show this, since it has a numeric conclusion, and there are lots of correct ways to arrive at that number.

Where precisely does Argument 1 fail?

Argument 1 says "which is just the same state of knowledge..." but it is flat-out lying. One direction is fine: if you answer "yes, yes" to Scenario 3, then you would answer "yes" in Scenario 1. But if you would answer "yes" to Scenario 1, then you would not answer "yes, yes" to Scenario 3. This is not the same state o... (read more)

The question I think is interesting is: When you throw away probability information and keep only possibility information, like when you go from "a random ace was a spade" in Scenario 2 to "there was an ace which was a spade" in Scenario 1, when does that cause bias? How do you have to think about the information you have left? When can you just condition on your information being true, and when do you have to think about the probability that you would have been left with this information and not some other information?

The probability is 1/5 (as independently calculated by me). I've no idea if argument 2 is correct, because I don't understand it. My reasoning:

There are 6 combinations of 2 cards: AsAh, As2c, As2d, Ah2c, Ah2d, 2c2d.

Of these, only the first 3 (AsAh, As2c, As2d) could I have answered yes to both questions (assuming I'm not lying, wihch is outside the context).

But if I have AsAh, only 1/2 the time would I have answered yes to the second question. So AsAh needs 1/2 the weight of the other 2 possibilities.

So the probability is (1/2)/(2+1/2) = 1/5.

I'm sorry to confess that I fell for the equivalent of Argument 1 in a similar puzzle in the past.

In terms of the current puzzle, I realized my mistake only when I realized that my state of knowledge isn't just that you're holding the Ace of Spades. I also know that you told me that you're holding the Ace of Spades, instead of some other card that you might have told me about. This knowledge introduces a new conditional probability into the Bayesian computation, which then yields the correct answer.

Argument 2, I figure. Finding out if a randomly chosen ace is a AS or not tells you nothing about p(2A) - since the chance of choosing AS is 50-50 for both 2A and across all 4 of the !2A cases.

That's assuming a genuine random choice. If there's a bias that causes the choice to become non-random once hearing the AS question, that might mess things up.

I have to agree that both arguments seem to be lacking, and the correct method for finding the answer is one of (enumerate the possibilities | do some math).

ETA: I should stop using regex-style disjunctions given the syntax for "given" used extensively here for probability.

I didn't read either argument before cranking up Bayes' Theorem. After the first question, the five remaining possibilities are equiprobable, so the posterior probabilities are proportional to the likelihoods of each possibility. The two non-pairs containing the ace of spades have likelihood 1 and the pair of aces has likelihood 0.5. Normalizing the likelihoods gives 1/5 chances for the pair of aces.

Argument 2 is correct. Showing an ace provides no relevant information. If you want to do the math, I like Blueberry's solution. Or: if you have 2 aces, then your chances of saying "yes, it's AS" are 50-50 since you were equally likely to have chosen either ace (you picked one at random), and if you have 1 ace, then your chances of saying "yes, it's AS" are 50-50 since you are equally likely to have either ace, so your "yes" answer does not provide any information about how many aces you have.

Argument 1 is wrong because in ... (read more)

Before drawing the cards, I decided randomly whether to "prefer" hearts or spades, so that if I had both cards I would tell you about the preferred one. That gives me twelve scenarios, of which five result in the answers that I gave you, of which I hold both aces in only one. Therefore 1/5.

If your second question was instead "Are you holding the Ace of Spades?" as in Scenario 1, then I'm twice as likely to answer "Yes" in the instance that I really have both aces as I was before - ie there are now six scenarios allowed by m... (read more)

Heh! After the first 4 answers, there is an even split!

This is the way that I see the problem (please correct me if I'm wrong).

ANSWER: 1/5

We know that the person has an Ace of spades in their hand which means the following are the only combinations the person could have is AH and AS, 2C and AS, 2D and AS. The probability of the person answering the question with "I have an Ace of spades" when the person has the combination of AH and AS is 1/2. The probability with 2C and AS is 1 and the probability of 2D and AS is 1 as there is only one ace in these combinations meaning the person would be forc... (read more)

Argument 2 is right. Here is how I think about it. If I am holding one ace the probability it is a spade is 1/2 (there are only two aces). If I am holding two aces the probability that one selected randomly is the spade is also 1/2. The cases aren't distinguished by the expected response to the question "Do you hold the ace of spades?" so that information cannot possibly be used to update the prior answer of 1/5.

There's an easy way to figure out the probability: say that the person holding the cards flips a coin. If he has two Aces, when asked to pick one, heads means he picks the Ace of Spades, and tails means he picks the Ace of Hearts.

There are twelve possible outcomes: 6 possible two-card hands times 2 possibilities for the coin flip. The person's responses have ruled out all but five: 1) heads, AS, AH; 2) heads, AS, 2C; 3) tails, AS, 2C; 4) heads, AS, 2D; 5) tails, AS, 2D. Each is equally likely and he has two Aces in 1), so the probability must be 1/5.

(We don't have the same information as in Scenario 1 because the coin flip made it less likely that he has two Aces, as Unknowns explained.)

The conclusion of the second is correct, but to arrive at that conclusion I had to write out all the possibilities and observe how the sequence of answers pruned them. I only understood the second argument when I realised that the symmetry of spades and hearts is what makes it work.

By that symmetry, the posterior probability of having two aces after answering yes to the final question about the ace of spades -- P(2A|AS=yes) for short -- must equal P(2A|AH=yes). But P(2A|AH=yes) = P(2A|AS=no). So the posterior after asking about the ace of spades is indepe... (read more)

I see another way to show that 1/5 is the correct solution:

P(2 Aces | Ace of Spades revealed)= P(2 Aces AND Ace of Spades revealed)/P(Ace of Spades revealed)

(note: for further calculations, I'm assuming that there are 5 possible hands and the probability for each hand is 1/5, since it already has been revealed that there is at least one Ace. The end result would be the same if you would also set aside a random card in case you have no Ace,but the probabilities in the steps before the end results would have to change accordingly)

P(2 Aces AND Ace of Spades reveled)=P(2 Aces)*1/2 = 1/5 * 1/2 =1/10

P(Ace of Spades revealed)= 2/5 * 1 + 1/5 * 1/2 = 5/10

(1/10)/(5/10)=1/5

Even further: the probability that I can guess your hand is 1/5. However, the probability that you have 2 aces is 1/3. No?

Further: write out your 12 possible trial pulls. Raw odds of ace-ace =2/12. Once an ace is pulled, do two things: cross out the two null-null pulls. The odds of ace-ace appear to be 2/10, but... We didn't draw out Schrodinger's ace; it is either ah or as. pick one (it doesnt matter which, they are symmetrical in distribution) and cross out the combinations that do not have this ace. As the waveform collapses the true odds of ace-ace appear- 2/6. Do you see that? We didnt draw an equally hearty or spadey ace, it had to be one of them or the other, which mad... (read more)

I find the responses funny and interesting. Should I write a post explaining this problem, since I don't think it's possible to achieve decent signal to noise by writing another comment? Naturally, I know where the error is.

Combination vs. permutation, right? I don't care which ace is where; I just care if, among the three cards that the other card COULD be, that card is the one ace left.

But how about this: I deal each of us two cards out of a deck of AS, AH, 2C, and 2D. I ask you if you have an ace and you say yes. Surely the odds of me having both 2C and 2D are the same as above.

I figured this before reading your arguments:

If you have two aces, the probability that the one you pick is the ace of spades is 0.5. If you have only one ace, the probability that it's the ace of spades is also 0.5 (and if so the probability that you pick it is 1).

The conditional probabilities are 50/50 in both cases. Therefore the evidence has no influence on the prior probability. The probability of two aces remains 1/5.

As for the arguments:

I think argument 1 is oversimplifying. We'd have the same information as in scenario 1 if we'd just ask "... (read more)

The key to this is to understand that in half of all cases where two aces were drawn, the second random step will pick the heart 1/2 the time.

Thus - the first step selects 5 out of 6 possible outcomes. In the second step, 4 of the 5 cases have fixed outcomes; the other (Aa) has two posssible outcomes.

Thus there are six possible final states:

1 a D showing a (1/5) 2 a d showing a (1/5) 3 A D showing A (1/5) 4 A d showing A (1/5) 6 A a showing a (1/10) 6 A a showing A (1/10)

The answer to the second question is yes in cases 2, 3, and 6. The collective probability of these cases is 1/5 + 1/5 + 1/10 = 1/2.

1/10 over 1/2 is 1/5.

I didn't quite understand the verbal arguments well enough to confidently evaluate them directly (though argument 2 seems to hit the correct applause buttons), but I diagrammed out the options on a spreadsheet and got the same answer as argument 2.

Essentially the reason argument 1 is wrong is because in enumerating the different possible outcomes compatible with the response given, the hand with 2 aces should receive half the weight of the other hands containing the ace of spades, since it will only yield a "yes" half the time, and therefore has ... (read more)

In some detail:

edit: I haven't looked at the post in detail, but I think Morendil did basically the same thing I do here.

Let S count the number of aces of spades you draw, and H the number of aces of hearts. Select one ace randomly from your hand, with equal probability of selection each of the aces you hold. Let R = 1 if the ace of spades is selected, and 0 otherwise. By Baye's Rule we have:

P(S+H=2 | S+H >= 1 and R=1)= [ P(S+H=2) * P(S+H >= 1 and R=1 | S+H=2) ] / [ P(S+H >= 1 and R=1) ]

P(S+H=2) = 1/6 since there are 4 choose 2 = 6 possibl... (read more)

(EDITED) Computing this the long way, by straightforward application of the product rule, yields probability 1/5.

There is a shortcut corresponding to argument 2. Call H the hypothesis that you have two aces, and E the evidence constituted by my confirming the ace I pick at random is spades. It turns out that P(E|H)=P(E|!H), so the evidence can't change my probability assignment. P(E|H) = P(spades | 2 aces) = 1/2. P(E|!H) = P (spades | 1 ace) = 1/2.

Now do it the long way: I start from P(2A,SR|A), that is, the probability that I hold two aces AND that I chos... (read more)

If the player has the SPADE:

• 1/3 of the time, he also has the HEART.
• 2/3 of the time he doesn't, and so must choose the SPADE.
• 1/6 of the time he chooses the SPADE though he did have the HEART.

So 5/6 of the time he chooses the SPADE, but only 1/6 of the time does he choose the SPADE while having the HEART.

Thus, the chance of him having the HEART when he has chosen the SPADE is 1/5.

V guvax gur reebe vf urer.

V nfx lbh "Qb lbh unir na npr?" Lbh erfcbaq "Lrf." Gur cebonovyvgl lbh ubyq obgu nprf vf 1/5:

Engure gur cebonovyvgl lbh unir gur n fcrpvsvp npr vf 1/2 naq gur cbffvovyvgl gerr sbe rnpu npr frcnengryl tvirf n 1/3 punapr bs univat obgu nprf. Tvivat n 2/6 be 1/3 punapr bs univat obgu nprf va gbgny.

Va fubeg, vg vf abg rdhvyvxryl gb unir rnpu bs gur 5 pbzovangvbaf.

Edit: This is wrong feel free to ignore.