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Kaj_Sotala comments on Drawing Two Aces - Less Wrong
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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.
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:
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:
Totaling 5/10, or 1/2.
Assuming that I have answered "yes", that leaves three possible sets:
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?"
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'.