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.
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.