Nearly 15 years ago, I wrote about the following trick attributed to William “Fitch” Cheney, Jr., in [1]:

Effect 1: While the magician turns their back or even leaves the room, a spectator shuffles a standard deck of cards, and selects an (unordered) subset of any cards and hands them to the magician’s assistant. For example:

The assistant arranges of these cards in a row left to right, as shown below, keeping the remaining selected card hidden.

Upon observing this arrangement, the magician announces the fifth selected card, in this case the ace of diamonds.

The article linked above describes how the trick works in practice (including Python source code demonstrating both the assistant and the magician), as well as a proof that the following condition on is necessary and sufficient for the trick to be feasible, at least in principle:

Intuitively, the LHS binomial coefficient counts the number of spectator selections that may be presented to the assistant, and the RHS falling factorial counts the number of arrangements that may be presented to the magician.

The motivation for this post is to observe that this basic argument, applying the König-Egerváry theorem to a bi-regular graph to find a saturating matching, is more generally applicable: it should be possible to perform other “Fitch-like” tricks, whose feasibility rests on similar counting arguments. For example:

Effect 2: A spectator rolls dice each with sides (maybe from their Farkle set), and arranges the dice in a row left to right, in any desired order. The assistant covers one of the dice by placing a cup over it, leaving the ordered arrangement otherwise undisturbed. The magician observes this arrangement with the values shown, and announces the value on the covered di(c)e.

This trick is feasible if and only if:

For another example of this same form, corresponds to writing down a 10-digit number, the assistant covering one of the digits, and the magician announcing the hidden digit.

The linked proof shows that both of these tricks are feasible in principle— but it’s a separately interesting challenge to come up with a practical scheme for performing them. In this case, there is a relatively simple solution left as a puzzle for the reader; but things can get complicated quickly, as in this puzzling.stackexchange.com post that discusses the case.

Effect 3: A spectator selects any cards from a deck of cards, and arranges them face-up in a row left to right, in any desired order. (More practically, imagine shuffling and cutting strictly more than half the deck, and fanning the cards face-up in a row.) The assistant flips one of the cards face-down, leaving the ordered arrangement otherwise undisturbed. The magician observes this arrangement with the face-up cards shown, and announces the face-down card(s).

In this case, the trick is feasible if and only if:

Of these two variants, I think this one would be the more powerful effect. Note that unlike Fitch’s original trick, the spectator gets to not only choose the cards, but to arrange them in any chosen order as well. The assistant’s only freedom is in which card(s) to flip face-down.

Here again, the obligatory puzzle is to determine a practical scheme for performing this trick. I’m not sure how much it helps as a hint to note that, in this particular case of where the assistant flips exactly one card face-down, for the trick to be possible we need the spectator to select and arrange strictly more than half of the cards in the deck. (So that, for example, it is perhaps useful that the spectator is guaranteed to select at least one pair of cards of the same rank and color in their arrangement?)

Reference:

1. Kleber, M., The Best Card Trick. Mathematical Intelligencer, 24 (2002). [PDF]