Curious if anyone has the math behind this? I'm trying perhaps to put a square peg in a round hole to make two effects work together, by taking advantage of what position you are usually in at the end of 5,9,K. My problem is my next bank of cards after Garder's set-up require two of the same cards. I'm wondering if there's alternative 4 of a kinds that will work, rather than adding in dupes?

Thank you

The procedure adds 2 extra cards on top of any specific four-of-a-kind. Let's say there are N cards on top of your four-of-a-kind with value V. For the trick to work for this particular set you need V = N + 2, since that's the number of cards dealt in the end.

So you could have the Deuces with 0 cards on top.
Or the 3s with 1 indifferent card on top.
4s, 2 cards on top.
5s, 3 cards on top.
6s, 4 cards on top.
7s, 5 cards on top.
8s, 6 cards on top.
9s, 7 cards on top.
10s, 8 cards on top.
Js, 9 cards on top.
Qs, 10 cards on top.
Ks, 11 cards on top.
As (as 14), 12 cards on top (but you really have to make sure there are at least 16 cards in the packet, so this is risky)
So now you can simply pick and choose, maybe 2s, 6s, 10s (,and maybe Aces) with no cards above the stack. Or 3s, 7s, Js with 1 card above the stack. Or 4s, 8s, Qs with 2 cards above.
Then you have the classic 5 9 K with 3 cards above.

I hope any of those works with your other trick.

/Tomas

Wow! Thank you so much, appreciate this Tomas! Did you already know the math or did you just work it out? I pressume it can only work with +4 in each bank of 4.

I just worked it out. You could have some strange distribution where interlaced cards are forbidden from selection if you can hide them, such as

x3333x8888xKKKKxxx...

or

22226666xxQQQQxxx...

But none of the x cards are valid selections during the selection procedure. Maybe you can Nyquist Spread Hideout any x card you don't want as a possible selection if there is a very specific set of three values you need for this trick.



You can change the procedure of the trick a bit, telling them to deal double the selected value from each packet. The formula then becomes 2V = N + 2 (or N = 2V - 2)

Ace (counts as 1), 0 cards on top.
2s, 2 cards on top.
3s, 4 cards on top.
4s, 6 cards on top.
5s, 8 cards on top.
6s, 10 cards on top.
7s, 12 cards on top.
8s, 14 cards on top.

More than that and you have more than a third of the deck in your stack. Here's a stack based on it, but it's 16 cards which means the spectator might cut into the 7s when he cuts the deck in three packets.

AAAA333355557777

A safer might be

xx222244446666xxx....

/Tomas