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.
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.