(In reply to please read
Absolutely correct. This ends up being equivalent to saying that B's choice choice must tell A what the order of all 5 cards is.
Thus, to do this trick with 4 audience members requires at least cards 1 through 124, giving B 120 choices (enough to encode 5!).
If we plug into your analysis, right side is P(124,4) = 124!/120! = 124!/(119!*5!) = C(124,5) is left side.
So, the question now is do we have a method that works when there are 124 cards?
Posted by Joel
on 2007-05-12 21:47:05