How many perfect shuffles will return a full deck
of cards to their original order?




Divide 52 cards into 2 equal piles
Shuffle by interlacing cards
Keep top card fixed (Out Shuffle)
8 shuffles => original order


Label the positions 0-51
Then




0->0 and 26 ->1
1->2 and 27 ->3
2->4 and 28 ->5
… in general?
0  x  25
 2x
f ( x)  
2 x  51 26  x  51
f(x) = 2x mod 51

Minimum integer k such that 2 k x = x mod 51
for all x in {0,1,…,51}

In particular, this has to hold when x = 1

Minimum integer k such that 2 k - 1= 0 mod 51

Thus, 51 divides 2 k - 1



k= 6, 2 k - 1 = 63 = (3)(3)(7)
k= 7, 2 k - 1 = 127 (prime)
k= 8, 2 k - 1 = 255 = (5)(51) = 0 mod 51
0  p  n 1

p = position
In Shuffles
mod n  1, n even
I ( p)  2 p  1 
n odd
 mod n,

Out Shuffles
mod n  1, n even and 0  p  n  1
O( p )  2 p 
n odd and 0  p  n  1
 mod n,


8 Out Shuffles for 52 Cards
In General?


o (O,2n-1) = o (O,2n)
o (I,2n-1) = o (O,2n)
 => o (O,2n-1) = o (I,2n-1)


o (I,2n-2) = o (O,2n)
Therefore, only need o (O,2n)

1 shuffle: O(p) = 2p mod (2n-1), 0<p<N-1

2 shuffles:
O2(p) = 2 O(p) mod (2n-1) = 22 p mod (2n-1)

k shuffles:

Order: o (O,2n) = smallest k such that
Ok(p) = p mod (2n-1) for all p between 0 and 2n

Which means 2k = 1 mod (2n-1) => (2n – 1) | (2k – 1)
Ok(p) = 2kp mod (2n-1)
n
o(O,n) o(I,n)
n
o(O,n) o(I,n)
2
1
2
13
12
12
3
2
2
14
12
4
4
2
4
15
4
4
5
4
4
16
4
8
6
4
3
17
8
8
7
3
3
18
8
18
8
3
6
50
21
8
9
6
6
51
8
8
10
6
10
52
8
52
11
10
10
53
52
52
12
10
12
54
52
20