Let's see if he can talk me round on this one .....



(I have just read the other thread and changed my opinion by the time I got to the end of it )

 

At first glance, I'll have to agree with you two. I don't think it matters, i.e., P1 = P2.

But I have a feeling magician has something up his sleeves...

 

it doesnt matter what suite it is, p1 = p2

 

"Suite"?

 

Computing P1 and P2 with 52-card decks and 13-card hands is quite a chore--I know; I've done it.

So, let's simplify the problem:

There are only three cards: the ace of spades, the ace of hearts, and the king of diamonds.

RHO gets only two cards, and nobody looks at the third card.

Case 1: RHO says "I have an ace." Quick! Calculate the probability that he has a second ace! Call this probability P1.

Case 2: RHO says "I have the ace of spades." Quick! Calculate the probability that he has a second ace! Call this probability P2.

How do they compare?

 

^ at both times that he announces his ace, nobody else has looked at their cards right?

 

That's right. The only one who looks at his cards is RHO who makes the announcement.

 

and we *know* he isn't lying right?

even though it got me into trouble the last time, I'm gonna say 50/50 again for this simplified example.

I'm also going to stick with P1=P2.

 

Do the calculations in the simplified (three-card) problem and post them, please.

(Hint: in each case, figure out how many hands are consistent with his announcement, and of those how many have a second ace.)

 

Here is how I simplify it in my mind:

AS = Ace of spades
AH = Ace of hearts
KD = King of diamonds

there are only 3 possibilities: that the third card (which he doesn't get) is either the AS, AH, or KD.

Therefore, in his first announcement, he says he has an A.
Probability of him having a 2nd A = 1 out of 3.

In his second announcement, he says he has the AS.
Probability of him having a 2nd A = 1 out of 2.

wow, that's surprising! or I'm wrong, which wouldn't be surprising.