You Too Can be a Mathematician Magician

John Bonomo

Westminster College

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The Basic Trick

• Volunteer picks one of 15 cards (call this the “key” card)

• Cards dealt in three piles, volunteer identifies “key” pile

• Pick up piles with key pile in the middle

• After three passes, key card is in the middle of the deck

Two Modifications

• Place key card in any location of the deck

• Allow users to pick up the decks

Analysis

03

69

12

14

710

13

25

811

14

← Row 0← Row 1

← Row 2← Row 3

← Row 4

f(p) = 5 + p/3

• Let p = position prior to deal

• Then new position is given by

p3 = f(p2)

• Let p0 = initial position; p1, p2, p3, positions after deals 1, 2 and 3

= f(f(p1)) = f(f(f(p0)))

p3 =

• Let p0 = initial position; p1, p2, p3, positions after deals 1, 2 and 3

7 + 6 + p0

27

• Since 0 ≤ p0 ≤ 14, we have

p3 = 7 + 6 + p0

27

p3 = 7

= 5 + 2 + 6 + p0

27

p3 = 7 + 6 + p0

27

base offset

• Generalized position function:

• Where i = 0 (bottom pile),

1 (middle pile),

2 (top pile)

fi(p) = 5i + p/3

0

i1 1

2

0

i2

1 2

12 + p04 + 27

6 + p02 + 27

24 + p03 + 27

18 + p01 + 27

21 + p02 + 27

15 + p0 27

3 + p01 + 27

9 + p03 + 27

p0 27

0 ≤ p0 ≤ 14

0

i1 1

2

0

i2

1 2

24 + p03 + 27

18 + p01 + 27

21 + p02 + 27

15 + p0 27

0 ≤ p0 ≤ 14

0

1

2

3

4

What’s your favorite number? Nine

Nine

8 = 9 - 18 = 5 + 3

i1=0, i2=2, i3=1Bottom, top, middle

9 = 8 - 18 = 5 + 3

i1=0, i2=2, i3=1Bottom, top, middle

Why is your face so sweaty?

And pale?

Zzzzzzzzz…

0

i1 1

2

0

i2

1 2

24 + p03 + 27

18 + p01 + 27

21 + p02 + 27

15 + p0 27

0 ≤ p0 ≤ 14

0

1

2

3

4

0

i1 1

2

0

i2

1 2

24 + p03 + 27

18 + p01 + 27

21 + p02 + 27

0 ≤ p0 ≤ 14

0

1

2

3

4

0, 0 ≤ p0 ≤ 11

1, 12 ≤ p0 ≤ 14

0

i1 1

2

0

i2

1 2

0 ≤ p0 ≤ 14

0

1

2

3

4

0, 0 ≤ p0 ≤ 11

1, 12 ≤ p0 ≤ 14

1, 0 ≤ p0 ≤ 8

2, 9 ≤ p0 ≤ 14

2, 0 ≤ p0 ≤ 5

3, 6 ≤ p0 ≤ 14

3, 0 ≤ p0 ≤ 2

4, 3 ≤ p0 ≤ 14

0

i1 1

2

0

i2

1 2

0 ≤ p0 ≤ 14

0: 100%

1: 100%

2: 100%

3: 100%

4: 100%

0, 0 ≤ p0 ≤ 11

1, 12 ≤ p0 ≤ 14

1, 0 ≤ p0 ≤ 8

2, 9 ≤ p0 ≤ 14

2, 0 ≤ p0 ≤ 5

3, 6 ≤ p0 ≤ 14

3, 0 ≤ p0 ≤ 2

4, 3 ≤ p0 ≤ 14

0

i1 1

2

0

i2

1 2

0 ≤ p0 ≤ 14

0: 100%

1: 100%

2: 100%

3: 100%

4: 100%

1, 0 ≤ p0 ≤ 8

2, 9 ≤ p0 ≤ 14

2, 0 ≤ p0 ≤ 5

3, 6 ≤ p0 ≤ 14

3, 0 ≤ p0 ≤ 2

4, 3 ≤ p0 ≤ 14

0: 80%

1: 20%

0

i1 1

2

0

i2

1 2

0 ≤ p0 ≤ 14

0: 100%

1: 100%

2: 100%

3: 100%

4: 100%

0: 80%

1: 20%

1: 60%

2: 40%

2: 40%

3: 60%

3: 20%

4: 80%

Always pick the “best” card

• 87% chance of selecting key card on first pick

• 100% chance of selecting key card on second pick (if necessary)

Generalize “Any Position” Trick

• n piles of m cards each

• still use only three deals

• Two questions:

– What values of n and m work?

– How do we determine i1, i2 and i3?

m ≤n2 + gcd(n2,m)

2

n

(piles)

m

(cards in pile)

3 1,…,6,9

5 1,…,13,15,25

6 1,…,18,20,24,36

m ≤n2 + 1

2(n,m relatively prime)

Valid n,m pairs

Determine i1,i2 and i3 for a given location L

Let s = (L mod m) n2

m

Then i1 = s mod n

i2 = s/n

i3 = L/m

Example: n=5, m=11 L = 40-1 = 39

i1 = 14 mod 5 = 4

i2 = 14/5 = 2

i3 = 39/11 = 3

s =(39 mod 11) 52

11= 14