Subject: card puzzle
Date: Tue, 21 Nov 2000 15:11:48 -0600 (CST)
From: Weiping Shi <wshi@ee.tamu.edu>
To: shai@stonehill.edu

Shai,

Here is how you can do it with 52+1 cards.
Name the extra card as spade 14.
Then the 5 cards
can be in one of the following cases:

case 1. at least 2 cards in suit not spade.
case 2. 3 or more spades.
case 3. 2 spades, 1 each heart, diamond, and club.

For case 1, use the same algorithm.

For case 2, use the same algorithm except that
we shoose two spades of distance not 7 to show.
This is always possible with 3 spades.

For case 3, we can use the same algorithm except
the two spades are distance 7 apart, say A and 8.
If we have spade A and 8, then
show both spades, with one being the first.
If my acomplice sees two spades of distance 7,
then it has to be case 3. So he'll know the suit
of the missing card is the missing suit.

The number of ways we have at this time is
3*2*3!=36, (3 for choosing the missing suit,
2 for picking the first spade, and 3! for ordering
the rest), if we don't fix the missing suit.
Here is one way to do it:
Assume for the worse case at least two non-spade
cards are in [A,...,7], since the other interval
[8,...,K] is smaller.
Then we show spade A as the first card.

Now min{X-Y, Y-X} (mod 7) has to be less
than or equal to 3. Use one card in the
permutation to identify the offset.
If all three non-spade cards are in [A,...,7],
choose the highest two.

Weiping