## Fool Me Once

I recently started watching Penn and Teller’s Fool Me, a show in which magicians try to fool Penn and Teller for a shot to perform in Vegas. After one of the episodes, there was a trick that I couldn’t even begin to understand, so I went looking for it online. Instead of getting a definitive answer, I found a web page with a bunch of guesses that had degenerated into a flame war. The flame war itself reminded me of an xkcd comic, but one of the comments mentioned a separate trick that had fooled Penn and Teller, a supposed mathematical trick performed by Graham Jolley.

The comment didn’t describe Graham Jolley’s trick, but it was pretty easy to find and watch. After experimenting with a deck of cards along with a pencil and some scratch paper, I’ve come up with one possible implementation.

The Effect

The magician takes out a deck of cards and has two participants sit on each side. Participant A is asked to cut about a third of the way into the deck and look at the bottom card. Participant B is asked to cut about half way into the remaining deck and look at the bottom card. The magician then recollects the card stack from Participant A and Participant B, placing them on top of the deck, respectively.

The magician then spreads the deck face down to reveal two jokers facing up. The jokers are taken out, and the magician holds each one close, as if it were transmitting where in the deck to look. In fact, the magician states that the first card will be found $A$ cards into the deck, and the other will be found $B$ cards in. When the magician counts through the deck and sets aside the cards at positions $A$ and $B$, they turn out to be the ones for Participant A and Participant B, respectively.

One Possible Method

The method I came up with assumes that Participant A will draw a card roughly 1/3 of the way into the deck, and Participant B draws a card roughly halfway into the remaining deck or roughly 2/3 of the way into the original deck.

Let’s start with a deck containing $N$ cards (not counting the jokers) with the first joker between cards $i$ and $i+1$, and the second between cards $j$ and $j+1$:

$1, \ldots, i, \|JOKER\|, i+1, \ldots, j, \|JOKER\|, j+1, \ldots, N$

We’ll now proceed with some definitions and assumptions

$N_A$ – the initial position of the card drawn by Participant A
$N_B$ – the initial position of the card drawn by Participant B
$i < N_A < j < N_B$

This would mean that the arrangement of the deck would be as follows:

$1, \ldots, i, \|JOKER\|, i+1, \ldots, N_A, \ldots, j, \|JOKER\|, j+1, \ldots, N_B, \ldots, N$

The magician achieves this by having Participant A get a card from near 1/3 of the way into the deck and placing the first joker well before 1/3 of the way in. Likewise for the second joker ($j = N/2$ would be the safest choice).

When Participants A and B replace their stacks on the top of the deck, Participant A goes first, so the order upon this reshuffle looks as follows:

$N_{A}+1, \ldots, j$, $\|JOKER\|, j+1, \ldots, N_B$, $1, \ldots, i$, $\|JOKER\|, i+1, \ldots, N_A$, $N_{B}+1, \ldots, N$

Note that the first group of cards before the joker

$N_{A}+1, \ldots, j$

contains $j - N_A$ cards. The second group of cards (between the first and second joker)

$j+1, \ldots, N_B, 1, \ldots, i$,

contains Participant B’s card $i$ positions from the end, and the third group

$i+1, \ldots, N_A, N_{B}+1, \ldots, N$

contains Participant A’s card $N_A - i$ cards from the beginning.

Thus, by moving the third group between the first and third (this happens when removing the jokers), Participant A’s card lands $j - i$ positions from the start, and Participant B’s lands $i$ cards from the end.

For a standard deck, $N = 52$, let $j = 26$, $i = 6$, and Participant A’s card will land at position 20, and Participant B’s will land at position 46.