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Kaj_Sotala comments on Drawing Two Aces - Less Wrong
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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'.