After shuffling 52-card deck 8 times, how many cards are in place?
This problem’s question: After shuffling 52-card deck 8 times, how many cards are in place?
The phrase "in place" means the the card is in the same relative place in the deck, e.g., if it were in the fourteenth place from the top originally, then it is in place if that card is in the fourteenth place from the top presently.
It is necessary to define exactly what is meant by a shuffle. If the cards were numbered from 1 to 52, then the output of a single shuffle 1 , 2 7 , 2 , 2 8 , 3 , 2 9 , 4 , 3 0 , 5 , 3 1 , 6 , 3 2 , 7 , 3 3 , 8 , 3 4 , 9 , 3 5 , 1 0 , 3 6 , 1 1 , 3 7 , 1 2 , 3 8 , 1 3 , 3 9 , 1 4 , 4 0 , 1 5 , 4 1 , 1 6 , 4 2 , 1 7 , 4 3 , 1 8 , 4 4 , 1 9 , 4 5 , 2 0 , 4 6 , 2 1 , 4 7 , 2 2 , 4 8 , 2 3 , 4 9 , 2 4 , 5 0 , 2 5 , 5 1 , 2 6 , 5 2
The deck is divided exactly in half. The cards are then merged such that the top card of each pair comes from the original top half of the deck.
By all means, you may treat this as a computer science problem.
The answer is 52.
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2 solutions
See Out Shuffle .
⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ 1 1 1 1 1 1 1 1 2 7 1 4 3 3 1 7 9 5 3 2 2 2 7 1 4 3 3 1 7 9 5 3 2 8 4 0 4 6 4 9 2 5 1 3 7 4 3 2 2 7 1 4 3 3 1 7 9 5 2 9 1 5 8 3 0 4 1 2 1 1 1 6 4 2 8 4 0 4 6 4 9 2 5 1 3 7 3 0 4 1 2 1 1 1 6 2 9 1 5 8 5 3 2 2 7 1 4 3 3 1 7 9 3 1 1 6 3 4 4 3 2 2 3 7 1 9 1 0 6 2 9 1 5 8 3 0 4 1 2 1 1 1 3 2 4 2 4 7 2 4 3 8 4 5 2 3 1 2 7 4 2 8 4 0 4 6 4 9 2 5 1 3 3 3 1 7 9 5 3 2 2 7 1 4 8 3 0 4 1 2 1 1 1 6 2 9 1 5 3 4 4 3 2 2 3 7 1 9 1 0 3 1 1 6 9 5 3 2 2 7 1 4 3 3 1 7 3 5 1 8 3 5 1 8 3 5 1 8 3 5 1 8 1 0 3 1 1 6 3 4 4 3 2 2 3 7 1 9 3 6 4 4 4 8 5 0 5 1 2 6 3 9 2 0 1 1 6 2 9 1 5 8 3 0 4 1 2 1 3 7 1 9 1 0 3 1 1 6 3 4 4 3 2 2 1 2 3 2 4 2 4 7 2 4 3 8 4 5 2 3 3 8 4 5 2 3 1 2 3 2 4 2 4 7 2 4 1 3 7 4 2 8 4 0 4 6 4 9 2 5 3 9 2 0 3 6 4 4 4 8 5 0 5 1 2 6 1 4 3 3 1 7 9 5 3 2 2 7 4 0 4 6 4 9 2 5 1 3 7 4 2 8 1 5 8 3 0 4 1 2 1 1 1 6 2 9 4 1 2 1 1 1 6 2 9 1 5 8 3 0 1 6 3 4 4 3 2 2 3 7 1 9 1 0 3 1 4 2 4 7 2 4 3 8 4 5 2 3 1 2 3 2 1 7 9 5 3 2 2 7 1 4 3 3 4 3 2 2 3 7 1 9 1 0 3 1 1 6 3 4 1 8 3 5 1 8 3 5 1 8 3 5 1 8 3 5 4 4 4 8 5 0 5 1 2 6 3 9 2 0 3 6 1 9 1 0 3 1 1 6 3 4 4 3 2 2 3 7 4 5 2 3 1 2 3 2 4 2 4 7 2 4 3 8 2 0 3 6 4 4 4 8 5 0 5 1 2 6 3 9 4 6 4 9 2 5 1 3 7 4 2 8 4 0 2 1 1 1 6 2 9 1 5 8 3 0 4 1 4 7 2 4 3 8 4 5 2 3 1 2 3 2 4 2 2 2 3 7 1 9 1 0 3 1 1 6 3 4 4 3 4 8 5 0 5 1 2 6 3 9 2 0 3 6 4 4 2 3 1 2 3 2 4 2 4 7 2 4 3 8 4 5 4 9 2 5 1 3 7 4 2 8 4 0 4 6 2 4 3 8 4 5 2 3 1 2 3 2 4 2 4 7 5 0 5 1 2 6 3 9 2 0 3 6 4 4 4 8 2 5 1 3 7 4 2 8 4 0 4 6 4 9 5 1 2 6 3 9 2 0 3 6 4 4 4 8 5 0 2 6 3 9 2 0 3 6 4 4 4 8 5 0 5 1 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
here is the mapping that shows the new position of card n .
f ( n ) = { 2 n − 1 2 ( n − 2 6 ) i f n ≤ 2 6 i f 2 6 < n ≤ 5 2
using the map, we can work out the algebraic cycles that are created by recursively applying shuffling. I have included the cycles below.
( 1 ) , ( 5 2 ) , ( 1 8 , 3 5 ) , ( 2 , 3 , 5 , 9 , 1 7 , 3 3 , 1 4 , 2 7 ) , ( 4 , 7 , 1 3 , 2 5 , 4 9 , 4 6 , 4 0 , 2 8 )
( 6 , 1 1 , 2 1 , 4 1 , 3 0 , 8 , 1 5 , 2 9 ) , ( 1 0 , 1 9 , 3 7 , 2 2 , 4 3 , 3 4 , 1 6 , 3 1 ) , ( 1 2 , 2 3 , 4 5 , 3 8 , 2 4 , 4 7 , 4 2 , 3 2 ) , ( 2 0 , 3 9 , 2 6 , 5 1 , 5 0 , 4 8 , 4 4 , 3 6 )
all cycles are of a length that divides 8 ,so, all cards would return to their original position.