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Perfect riffle shuffles
Rigged casino that prevents pairsSolitaire PuzzleHow to cheat at cardsThree Cards TrickFive-hand pokerRandom Shuffled Deck of CardsOptimal game of Bluff (not the wikipedia one)6 Cards, Top to Bottom98 Cards: Optimal Strategy with Rule of Ten'sThe magic trick
$begingroup$
Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.
Amazingly, if you perform 8 such riffle shuffles you will return to where you started.
Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.
But two cards will simply swap position back and forth each shuffle.
What are they?
Bonus question:
If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?
cards
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
$endgroup$
add a comment |
$begingroup$
Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.
Amazingly, if you perform 8 such riffle shuffles you will return to where you started.
Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.
But two cards will simply swap position back and forth each shuffle.
What are they?
Bonus question:
If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?
cards
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
$endgroup$
add a comment |
$begingroup$
Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.
Amazingly, if you perform 8 such riffle shuffles you will return to where you started.
Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.
But two cards will simply swap position back and forth each shuffle.
What are they?
Bonus question:
If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?
cards
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
$endgroup$
Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.
Amazingly, if you perform 8 such riffle shuffles you will return to where you started.
Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.
But two cards will simply swap position back and forth each shuffle.
What are they?
Bonus question:
If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?
cards
cards
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
asked 1 hour ago
Dr XorileDr Xorile
13.8k32975
13.8k32975
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.
If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed18text and 35$.
Bonus:
It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed52text shuffles$.
answered 1 hour ago
noednenoedne
7,55212159
$endgroup$
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.
If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed18text and 35$.
Bonus:
It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed52text shuffles$.
answered 1 hour ago
noednenoedne
7,55212159
$endgroup$
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
add a comment |
$begingroup$
Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.
If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed18text and 35$.
Bonus:
It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed52text shuffles$.
answered 1 hour ago
noednenoedne
7,55212159
$endgroup$
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
add a comment |
$begingroup$
Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.
If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed18text and 35$.
Bonus:
It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed52text shuffles$.
answered 1 hour ago
noednenoedne
7,55212159
$endgroup$
Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.
If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed18text and 35$.
Bonus:
It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed52text shuffles$.
answered 1 hour ago
noednenoedne
7,55212159
answered 1 hour ago
noednenoedne
7,55212159
answered 1 hour ago
noednenoedne
7,55212159
answered 1 hour ago
noednenoedne
7,55212159
7,55212159
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
add a comment |
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
42 mins ago
add a comment |
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