Argument 2 is correct. When the person replies “Yes,” after choosing randomly, you learn not only that he has the ace of spades, but also that on one trial, he selected it after choosing randomly from his aces. This makes the combinations “AS, 2C” and “AS, 2D” more probable than “AS, AH”, since the first two combinations give a 100% chance of a positive response, while the third gives only a 50% chance of a positive response. So each of the first two combinations is twice as likely as the third, so the probability of the third combination, namely two aces, goes to 1⁄5.
This looks like a riff on the Monty Hall problem, whose solution also hinges on the fact that the host opens a door randomly or non-randomly depending on your initial choice.
Argument 2 is correct. When the person replies “Yes,” after choosing randomly, you learn not only that he has the ace of spades, but also that on one trial, he selected it after choosing randomly from his aces. This makes the combinations “AS, 2C” and “AS, 2D” more probable than “AS, AH”, since the first two combinations give a 100% chance of a positive response, while the third gives only a 50% chance of a positive response. So each of the first two combinations is twice as likely as the third, so the probability of the third combination, namely two aces, goes to 1⁄5.
I reached the same solution.
This looks like a riff on the Monty Hall problem, whose solution also hinges on the fact that the host opens a door randomly or non-randomly depending on your initial choice.
This is exactly what I thought when I read the problem.
This.