Est. 1998 — 27 years of woodworking knowledge
A correction and a Gimme
Alex Y
In the second hand, I said:
I goofed. It should have been9H
AS
9D
AHYou are holding the 2 of Hearts
And the gimme:
In bridge, the suits are ordered alphabetically from low to high:
C D H S. So in the above example, the last three are HML.
Also, my assistant and I are treating Aces as high. Would be fine to treat them as low, as long as we were both on the same page.
Another important fact about Fifth set
Alex Y
With this set of cards, the Jack of Hearts is the only card the audience member could hold, and the "encoding" with the other four cards is unique.
This is not the case with the other four hands; in those cases there were multiple cards that could have been given to the audience member, and in some cases there were multiple possible encodings to reveal the same card.
Last hint until after the 4th
Alex Y
Will be too busy with visiting family to keep up with this next week. I'll try to check in periodically, to see if I can crown a winner, but else we will resume this in a week.
I used a web page to deal five pseudo random hands of 5 cards each (and these are pseudo-random in more than the mathematical sense--I modified the last one to get one hand that had only one pair of cards of the same suit).
Here is a table showing the five hands and every possible way my assistant could have designated the card to be hidden and the stack of cards revealed to me.
And a reminder: Aces high, and suit order from low to high is Clubs, Diamonds, Hearts, and Spades. So 2C is the lowest card in the deck, while AS is the highest.
Happy 4th to all, but in particular to fellow Americans.
Re: Hints
John Horst
Bill Earls, that was my father in laws name. Which state are you in?
Re: Hints
bill earl
That's Earl with no 's'. I'm originally from Western NY.
Re: Last hint until after the 4th
bill earl
"And a reminder: Aces high"
The last three cards
Alex Y
We use the following system:
LMH = 1
LHM = 2
MLH = 3
MHL = 4
HLM = 5
HML = 6
This seems the most natural conversion to me, since substituting the DIGITS 1,2,3 for the letters L,M,H, the NUMBERS formed by the permutations above are in numerical order.
Re: The last three cards
Larry Barrett
The first card turned over provides the clue for the suit. This order of the last three cards provides a way to identify a specific 6 cards in the suit, but we need a way to identify 13. So perhaps the way the assistant lays the stack on the table could be what is needed. If the stack is laid on the table in a vertical position it would mean that the order of the last three cards would determine the 2 - 7 card; if the stack is laid on the table in a horizontal position it would mean the 8 - K, and if the stack is laid at an angle it would mean the A.
Re: The last three cards
Alex Y
Nothing but the values of the four cards is significant. they could be read out to me without my even seeing them.
But you are close. Go back and reread the hint I left before the 4th.
Summary and last hint
Alex Y
5 cards picked by an audience member
Assistant gives one back to the audience member, and arranges the other four to be revealed in a particular order.
First thought: There are 48 possibilities for that other card, but only 4! = 24 possible orderings of the four remaining cards, so this looks impossible.
First AHA!: The assistant picked the card the audience member holds, and repeated instances of the trick led to correct speculation that the first card in the stack of 4 indicates the suit of the hidden card.
(BTW, the popularizer of this trick suggests "rotating" the suit key to prevent this induction. i.e., do the trick the first time with the first card as the suit of the hidden card. The second time doing the trick for the same audience, make the second card the one showing the suit, etc.)
Second try: We know the suit of the hidden card, and it is not the one revealed, so there could be 12 possibilities for the hidden card. But there are only 3! = 6 ways to arrange the three cards following the suit key, so it still appears we have only 1/2 the information we need.
What's missing? The key is in the hint titled "Last hint until after the 4th"
Re: Summary and last hint
bill earl
As I mentioned in my response to your last hint before the 4th, there was some semi-misinformation in your hint. Aces are only high for the purpose of ranking the cards.
Re: Summary and last hint
Alex Y
In a sense, they are high and low for other purposes.
Are you saying you have it? Out with it!
Re: Summary and last hint
bill earl
The sticking point of the pure combinatorial approach is that you have 3! (6) combinations of cards to identify one of 12 possible values. And the first card only tells you the suit - or does it?
If you forget about "Aces high" and view card values 'modulo Ace' (mod 13) you get a circular arrangement that has no "high card". Counting up from the Ace takes you back to the 2.
With this arrangement, no two card values are more than 6 cards apart. Once you determine the suit, pick the card that is no more than 6 less than the other card (mod 13) to reveal.
The first card revealed now gives you 2 pieces of information: The suit and starting point for the count. The other 3 cards tell you how far to count up (mod 13) to determine the answer.
Thanks Alex (I think) - I lost a bit of sleep on that one 
Good job, Bill!! *LINK*
Alex Y
That is exactly right.
Here's a nice article describing the trick and some of its history (it has been around a while). He uses a different ordering of HML, which I find less intuitive, but any order that the assistant and performer can remember works. After discussing this trick, he goes into some pretty heavy math, and shows that the trick can work with up to 124 cards (using a standard 52-card deck, there is a lot of "wasted" information, e.g., where there are more than two cards of the same suit, and you never use the values of the counting cards to eliminate them from contention).
courses.csail.mit.edu/6.042/spring10/cardTrick.pdf