Half In The Bag (trick)
The 'trick' that makes this work is simple mathematics, and doesn't seem new, but it's wrapped up with new obfuscations. Being purely mathematical, it requires no specialist manipulation skills, and can never fail. Because of this, many of the operations can be performed by the guests themselves, assuming simple card-handling skills. Use of the passive voice implies anyone may perform the action.
What You See
- Performer P invites two guests, A and B, on stage to participate and observe, sitting facing each other to P's left and right. A and B should be good friends so that they can use "telepathic empathy" or some other psychobabble to assist each other.
- P takes a pack of cards out of a brown paper bag, which he puts to one side.
- P opens the pack, places them square on the table, invites A to cut, complete and square. P then cuts, completes, squares, and then invites B to cut, complete, and square.
- Assuming guests have some handling skills, P invites one of them to spread the cards out and quickly check it is an ordinary deck. With absense of such skills, P can do this himself. Patter includes it being an ordinary deck. The deck appears random.
- P gathers the cards together, and offers everyone the opportunity to cut, complete, and square, as before.
- P fans the deck out in his hands, face down, and offers it to A, asking him to select a card.
- P holds the cards above the selected one in one hand, and asks B if he's happy with the position of the selection.
- If B's unhappy, which should be encouraged, P asks him to take the 8-sided or 4-sided die from the bag and roll one of them.
- P then moves the rolled number of cards forwards or backwards, at A and/or B's whim, through the deck.
- P tells A to take the top card from the bottom of the deck, and then tells B to take the new top card from it too.
- Patter includes telling them to look at their own card, remember it, but keep it secret from everyone else, and that the point of the trick is for them to find each others' cards.
- P tells A to put his card back on the top of the bottom part of the deck, and then instructs B likewise.
- P then reunites the two halves, squares them on the table, and offers everyone the opportunity to cut, complete, and square the deck, as before.
- Patter about how the cards will be dealts into piles, it being impossible for them to find each others' cards, how it would be easier if he allowed them to give hints where their card is, but that would be too easy, so he'll only allow them to give clues as to where their card isn't. All choices made by the guests below are explicitly of piles that do not contain their own card. When the piles are cleared away, they are discarded into the brown paper bag with patter along the lines of "let's get that half in the bag".
- The cards are dealt sequentially into 4 piles of 13 cards. Probably best to overlap this boring stage with the above patter.
- As an afterthought, P suggests that the 4 piles should be each shuffled, to make things more fun/hard.
- The 4 piles are then revealed. Again, they look random. Averting his gaze, P tells A and B to each ensure they know where their card is.
- Patter about selecting piles to keep or throw away. P choses one of the 4 piles, and asks A to chose a pile that doesn't contain his card. P then may select another pile and assign the final pile to B, or may chose not to do so. Either way, two piles will be discarded into the brown paper bag.
- The remaining 26 cards are shuffled together, and dealt into 4 piles of 6 or 7.
- Patter about doing it differently (doing it differently being important for obfuscation, see below), either a 4-sided die or A will chose pile first, rather than P. 12-14 cards will end up discarded.
- The remaining 12-14 cards are shuffled together and dealt into 4 piles of 3 or 4.
- Patter about doing it differently, whichever wasn't done previously is suggested. 6-8 cards will end up discarded.
- The remaining 6-8 cards are shuffled together and dealt into 4 piles of 1 or 2.
- Patter about doing it slightly differently again, and having them chose who should chose first. 2-4 cards will end up discarded.
- The remaining 2-4 cards are shuffled together and dealt into piles of 1. The players end up selecting each other's cards through whatever endgame is appropriate for the number of cards left.
HERE BE SPOILERS
Cards are divided into 2 classes, the ruly, and the unruly. P must be able to identify what class a card is in trivially, but it should be hard for anyone else to divine the rule dividing them.
Example rules could be:
- All reds are ruly, all blacks are unruly.
- All evens, and black kings, are ruly. Odds apart from black kings are unruly.
- All even reds and all odd blacks are ruly.
- All low reds and all high blacks are ruly.
- All even reds, high clubs, and low spades are ruly.
- Even diamonds, high hearts, odd spades and low clubs are ruly.
The deck is set up by separating the ruly cards from the unruly ones, shuffling each half thoroughly, so there's a genuine appearance of randomness, and then interleaving the two halves precisely. E.g. using the hard rule above, one might get:
S D D D S C D S S C C S H C C H S D H H C D D C C H H C H C S S H H C S H H D S H S D C S D S H C D D D
K 9 6 K A T 2 T 5 J 4 2 8 8 2 5 9 5 J 3 A J 8 K 6 2 K 7 Q Q 3 8 9 6 3 Q 7 4 4 6 T 4 Q 9 7 A J A 5 7 T 3
(Generated using a simple perl script.)
Identifying the 2 Chosen Cards
When the 4 piles are first revealed, they will be:
- one with all ruly
- one with all unruly
- one with all but one ruly
- one with all but one unruly
The two chosen cards are the two that are alone amongst the other type.0>
Here the guests are arbitrarily renamed A and B where A will be the first guest to be asked to do something. P is called "blind" if which of the 2 cards belongs to which of the players isn't known at this point, and "enlightened" if it is known. If one is enlightened, one may use a blind sequence. A's and B's cards are 'keeper' cards. Obviously, all of these steps are pure equivocation, which is why the protocol for each step is "done differently", so that the arbitrariness of mapping selected onto either throwing or keeping isn't so obvious.
P chooses first
Variation 1 - Choose to Throw
- P selects a pile to throw.
- If A selects a pile for throwing, P throws his pile and tells A to throw his pile. P remains blind.
- If A selects a pile for keeping, P selects the other pile to throw, forcing B to select the remaining pile, and then throws both of his piles. The card for keeping in A's pile is obviously B's, and so A's is in B's pile, thus P becomes enlightened.
Variation 2a - Blind, and Choose to Keep
- P selects a pile containing a keeper card.
- If A selects a pile for throwing, P selects the other pile to keep, forcing B to select the remaining pile, and then throw both of the guest piles. P remains blind.
- If A selects a pile for keeping, P instructs B to throw the remaining 2 piles. The card for keeping in A's pile is now obviously B's, and the card in P's pile is A's, thus P becomes enlightened.
Variation 2b - Enlightened, and Choose to Keep
- P selects the pile containing B's keeper card.
- A must select a pile for throwing, P selects the other pile to keep, forcing B to select the remaining pile, and then throw both of the guest piles.
4-Sided Die Chooses First
Get the guests to number the piles and roll. Accept that as your choice made for you, and follow whatever "P chooses first" algorithm matches the chosen pile.
A choses first
Path 1 - Choose to Throw
- A choses a thrower pile.
- P choses the other thrower pile, and throws both piles.
Path 2 - Choose to Keep
- A choses a keeper pile. This must contain B's card, enlightening P.
- P choses the other pile for keeping, and instructs B to throw both other piles.
When Both Cards are in a Single Pile
Similar to the above, except it's easier for everyone to pick a
thrower, there being only one pile that needs to be kept, but alas you
can't go from blind to enlightened.
- With 2 cards remaining, just get them to chose the card which isn't theirs.
- With 3 cards remaining, chose the thrower first, which reduces it to the 2 card problem.
- With 4 cards use one of the techniques above to get down to 2 cards. Variation 1 has the best chance of them ending the trick this round.
There's no force, the trick works no matter what cards are
chosen, all cards are equivalent. Similarly all cuts are irrelevant,
the merely rotate the cards, and if you consider the top and the
bottom to be adjacent in a loop, nothing ever changes. And there's
definitely no need for the perfomer to take a peek at the top
half of the deck whilst the two guests examine their cards.
The taking of the two cards, and their replacement, swaps their
parity. As the two classes were interleaved precisely, dealing 4 piles
would group cards of the same parity together, 2 piles of each parity,
so the cards who've had their parity swapped in the selection stage will
end up in the wrong pile.
To see the simplicity of the mathematical principle of parity being
used, use one of the terrible rules suggested above, one where it's
abundantly clear which card is in which class.
The core seems to be the same as "Neither Blind Nor Stupid"
by Juan Tamariz, from a quick check of am old grainy youtube video of
him performing it (absolutely brilliantly, I might add), and it works
for the same mathematical reason - a parity swap of 2 cards.
The difference is that the trick is couched in the deceit that the
guests will be using telepathy to guide each other to finding their
cards, that the dealing is into 4 piles rather than 2, and that
dealing into piles is performed repeatedly. The fact that the cards
can be detected upon first reveal means that additional full shuffles
of remaining cards can take place without affecting the trick, the
performer can always contrive that unwanted piles are discarded, no
matter the outcome of the shuffle and deal. This is all obfuscation,
the only actual trick is identifying to two cards with out-of-place
parity - the ruly card amongst the unruly, and the unruly card amongst
the ruly - and hoping that the audience don't detect the stage in the
performance where this happens.
I'm working on a version which uses "modulo 3", and 3 piles, rather
than "modulo 2" (parity) and 4 piles …