Argument 2 is correct, but I’m having considerably difficulty putting the reason into succinct terms in a way that’d feel satisfying for me, even after reading all the comments here.
However, here’s the thoroughly worked out answer for anyone who wants it:
We start with no questions asked. All of the six possible sets are equiprobable.
(AS 2C): 1⁄6
(AS 2D): 1⁄6
(AS AH): 1⁄6
(AH 2C): 1⁄6
(AH 2D): 1⁄6
(2C 2D): 1⁄6
Next you ask the first question, “do you have an ace?” I respond “yes”. This eliminates the (2C 2D) possibility. This leaves the following sets:
(AS 2C) or (AS 2D): 2⁄5
(AH 2C) or (AH 2D): 2⁄5
(AS AH): 1⁄5
Now the probability of me holding both aces is 1⁄5, the probability that I have (only) the Ace of Spades is 2⁄5 (3/5), and the probability that I have (only) the Ace of Hearts is 2⁄5 (3/5).
Suppose that after this, you’re about to ask me “is a randomly picked ace the ace of spades?” The prior probability that I’ll answer “yes” is, for the different scenarios:
(AS 2C) or (AS 2D): 2⁄5 * 1 = 2⁄5
(AH 2C) or (AH 2D): 2⁄5 * 0 = 0
(AS AH): 1⁄5 * 1⁄2 = 1⁄10
Totaling 5⁄10, or 1⁄2.
Assuming that I have answered “yes”, that leaves three possible sets:
(AS 2C)
(AS 2D)
(AS AH)
However, they are not equiprobable, as there was previously a random element involved. Now that we know I have the Ace of Spades, let’s take another look at the probabilities for “is a randomly picked ace the ace of spades?”
(AS 2C): 1⁄3 * 1 = 1⁄3
(AS 2D): 1⁄3 * 1 = 1⁄3
(AS AH): 1⁄3 * 1⁄2 = 1⁄6
Totaling 5⁄6. So now, suppose you only know that I have an ace. What is the probability that I have both aces, given that a randomly chosen ace from my hand produced the Ace of Spades?
By Bayes’ rule, P(A|B) = P(B|A)P(A) / P(B). In this case, A = I have both aces, B = A randomly chosen ace from my hand produces the Ace of Spades. Naturally, that makes B|A = A randomly chosen ace from my hand produces the Ace of Spades, given that I have both aces.
P(A) = P(I have both aces) = 1⁄5
P(B) = P(A randomly chosen ace from my hand produces the Ace of Spades) = 1⁄2
P(B|A) = P(A randomly chosen ace from my hand produces the Ace of Spades, given that I have both aces) = 1⁄2
Next you ask the first question, “do you have an ace?” I respond “yes”. This eliminates the (2C 2D) possibility. This leaves the following sets:
(AS 2C) or (AS 2D): 2⁄5
(AH 2C) or (AH 2D): 2⁄5
(AS AH): 1⁄5
I disagree with this step (my rot13d explanation hasn’t garnered much attention).
I don’t think the sets are equiprobable. Consider the following tree
The first question represents asking what the first ace drawn was, the second question what the other card was. As the first question is 50:50 either way and the second each card has a equal probability. However as AHAS comes up twice on the tree it has twice the weighting and 1⁄3 probability from the start.
Or to think of it another way. You know they have one ace, what are the options for the other card. They are equally probably 2C, 2D and the other Ace. So I say it is 1⁄3 the chance of getting two aces, once you know they have one ace.
I believe AdeleneDawner is right: yes, there are three options each in the last branch, but they aren’t all equally likely. Though I’m uncertain of how to show it from the top off my head: give me a while.
The colors of the squares in the grids show how you’d answer the question ‘Is your preferred ace the ace of spades?’ and whether you have 1 or 2 aces. The ‘P=’ notation in the corner of each grid shows what you’re preferring; in the first case you always prefer the first ace drawn; the latter two are meant to be read together and assume that you’re picking which ace you prefer ahead of time with a coin toss. The red and green squares to the side show how many of each response you could see in each case.
Thanks that cleared it up for me. I’ve been trying analyse where I went wrong. I reformulated the question in a way that I didn’t notice lost information
Also, simply asking “Do you have the ace of spades?′ returns a chart that looks like the P=AS one; red (and peach, if ‘Do you have an ace?’ isn’t asked first) squares are instances where you answer ‘No’, and the remaining 4 light green and 2 dark green squares show the 1 in 3 chance that you have 2 aces given that you answered ‘Yes’.
My first question is phrased as “first ace drawn”, the second question is “other card drawn”. This card could have been drawn before or after, it doesn’t matter which (unless it is the other ace in which case it couldn’t have been drawn before).
Picking the first ace is really just a way to fix what one definite unknown ace is in some way, so you can ask what the other cards are.
The flaw in this argument is that you don’t have an equal probability of drawing a 2 as an ace alongside your first ace, since there are two possibilities for drawing the twos—drawing them before or after—but only one for drawing the second ace, since it must be drawn after.
Missed that on my first read-through, but it still kind of points in the direction of the problem with your chart. Assume that the first ace is AS. There’s two instances where the other card could be 2C (AS then 2C, or 2C then AS), two instances where it could be 2D (AS then 2D, or 2D then AS), and one instance where it could be AH (AS then AC, but not AC then AS). The three branches for ‘other card’ are not equally likely.
Okay I’ll dispense with draw order entirely. Imagine if instead of asking them if they had an ace, ask them if they had an ace and mentally select one of their aces to be the primary ace at random.
They don’t tell you what it is or give any other information. So the first question on my tree is what is their primary ace, and the second question is what is their other card.
Their primary ace still has a 50:50 chance of being either (if they only have one ace, it could have been either drawn from the deck, and if they have two then it is selected randomly by the person with the cards). If you guess that their primary ace is one of the aces then the other cards are drawn from a pool of three possibilities.
I see what you’re doing, but I still think you’re making a mistake: Just because there are three possibilities, doesn’t mean that those possibilities are equally likely. It’s similar to flipping a fair coin twice; you could get two heads, two tails, or one of each. There are three possible outcomes, but the ‘one of each’ option is twice as likely as either of the other two.
Argument 2 is correct, but I’m having considerably difficulty putting the reason into succinct terms in a way that’d feel satisfying for me, even after reading all the comments here.
However, here’s the thoroughly worked out answer for anyone who wants it:
We start with no questions asked. All of the six possible sets are equiprobable.
(AS 2C): 1⁄6
(AS 2D): 1⁄6
(AS AH): 1⁄6
(AH 2C): 1⁄6
(AH 2D): 1⁄6
(2C 2D): 1⁄6
Next you ask the first question, “do you have an ace?” I respond “yes”. This eliminates the (2C 2D) possibility. This leaves the following sets:
(AS 2C) or (AS 2D): 2⁄5
(AH 2C) or (AH 2D): 2⁄5
(AS AH): 1⁄5
Now the probability of me holding both aces is 1⁄5, the probability that I have (only) the Ace of Spades is 2⁄5 (3/5), and the probability that I have (only) the Ace of Hearts is 2⁄5 (3/5).
Suppose that after this, you’re about to ask me “is a randomly picked ace the ace of spades?” The prior probability that I’ll answer “yes” is, for the different scenarios:
(AS 2C) or (AS 2D): 2⁄5 * 1 = 2⁄5
(AH 2C) or (AH 2D): 2⁄5 * 0 = 0
(AS AH): 1⁄5 * 1⁄2 = 1⁄10
Totaling 5⁄10, or 1⁄2.
Assuming that I have answered “yes”, that leaves three possible sets:
(AS 2C)
(AS 2D)
(AS AH)
However, they are not equiprobable, as there was previously a random element involved. Now that we know I have the Ace of Spades, let’s take another look at the probabilities for “is a randomly picked ace the ace of spades?”
(AS 2C): 1⁄3 * 1 = 1⁄3
(AS 2D): 1⁄3 * 1 = 1⁄3
(AS AH): 1⁄3 * 1⁄2 = 1⁄6
Totaling 5⁄6. So now, suppose you only know that I have an ace. What is the probability that I have both aces, given that a randomly chosen ace from my hand produced the Ace of Spades?
By Bayes’ rule, P(A|B) = P(B|A)P(A) / P(B). In this case, A = I have both aces, B = A randomly chosen ace from my hand produces the Ace of Spades. Naturally, that makes B|A = A randomly chosen ace from my hand produces the Ace of Spades, given that I have both aces.
P(A) = P(I have both aces) = 1⁄5
P(B) = P(A randomly chosen ace from my hand produces the Ace of Spades) = 1⁄2
P(B|A) = P(A randomly chosen ace from my hand produces the Ace of Spades, given that I have both aces) = 1⁄2
Thus P(A|B) = P(B|A)P(A) / P(B) = (1/2)(1/5) / (1/2) = (1/10) / (1/2) = 1⁄5.
I disagree with this step (my rot13d explanation hasn’t garnered much attention).
I don’t think the sets are equiprobable. Consider the following tree
The first question represents asking what the first ace drawn was, the second question what the other card was. As the first question is 50:50 either way and the second each card has a equal probability. However as AHAS comes up twice on the tree it has twice the weighting and 1⁄3 probability from the start.
Or to think of it another way. You know they have one ace, what are the options for the other card. They are equally probably 2C, 2D and the other Ace. So I say it is 1⁄3 the chance of getting two aces, once you know they have one ace.
I believe AdeleneDawner is right: yes, there are three options each in the last branch, but they aren’t all equally likely. Though I’m uncertain of how to show it from the top off my head: give me a while.
I’m almost done putting a diagram together, if you want it.
Please do show it.
The colors of the squares in the grids show how you’d answer the question ‘Is your preferred ace the ace of spades?’ and whether you have 1 or 2 aces. The ‘P=’ notation in the corner of each grid shows what you’re preferring; in the first case you always prefer the first ace drawn; the latter two are meant to be read together and assume that you’re picking which ace you prefer ahead of time with a coin toss. The red and green squares to the side show how many of each response you could see in each case.
Thanks that cleared it up for me. I’ve been trying analyse where I went wrong. I reformulated the question in a way that I didn’t notice lost information
Also, simply asking “Do you have the ace of spades?′ returns a chart that looks like the P=AS one; red (and peach, if ‘Do you have an ace?’ isn’t asked first) squares are instances where you answer ‘No’, and the remaining 4 light green and 2 dark green squares show the 1 in 3 chance that you have 2 aces given that you answered ‘Yes’.
What about the instances where you get 2C or 2D first, and then one of the aces?
My first question is phrased as “first ace drawn”, the second question is “other card drawn”. This card could have been drawn before or after, it doesn’t matter which (unless it is the other ace in which case it couldn’t have been drawn before).
Picking the first ace is really just a way to fix what one definite unknown ace is in some way, so you can ask what the other cards are.
The flaw in this argument is that you don’t have an equal probability of drawing a 2 as an ace alongside your first ace, since there are two possibilities for drawing the twos—drawing them before or after—but only one for drawing the second ace, since it must be drawn after.
Missed that on my first read-through, but it still kind of points in the direction of the problem with your chart. Assume that the first ace is AS. There’s two instances where the other card could be 2C (AS then 2C, or 2C then AS), two instances where it could be 2D (AS then 2D, or 2D then AS), and one instance where it could be AH (AS then AC, but not AC then AS). The three branches for ‘other card’ are not equally likely.
Okay I’ll dispense with draw order entirely. Imagine if instead of asking them if they had an ace, ask them if they had an ace and mentally select one of their aces to be the primary ace at random.
They don’t tell you what it is or give any other information. So the first question on my tree is what is their primary ace, and the second question is what is their other card.
Their primary ace still has a 50:50 chance of being either (if they only have one ace, it could have been either drawn from the deck, and if they have two then it is selected randomly by the person with the cards). If you guess that their primary ace is one of the aces then the other cards are drawn from a pool of three possibilities.
Does this clear what I am getting at up for you?
I see what you’re doing, but I still think you’re making a mistake: Just because there are three possibilities, doesn’t mean that those possibilities are equally likely. It’s similar to flipping a fair coin twice; you could get two heads, two tails, or one of each. There are three possible outcomes, but the ‘one of each’ option is twice as likely as either of the other two.
That’s how I did it too, thanks for saving my the typing.