Perfect riffle shuffles Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?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
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Perfect riffle shuffles
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?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 Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
$endgroup$
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 Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
$endgroup$
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 Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
$endgroup$
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 Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
asked Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
asked Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
asked Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
asked Mar 25 at 18:46
Dr XorileDr Xorile
14.1k33083
14.1k33083
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 Mar 25 at 19:12
noednenoedne
9,25012566
$endgroup$
1
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
2
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
add a comment |
Your Answer
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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 Mar 25 at 19:12
noednenoedne
9,25012566
$endgroup$
1
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
2
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
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 Mar 25 at 19:12
noednenoedne
9,25012566
$endgroup$
1
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
2
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
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 Mar 25 at 19:12
noednenoedne
9,25012566
$endgroup$
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 Mar 25 at 19:12
noednenoedne
9,25012566
answered Mar 25 at 19:12
noednenoedne
9,25012566
answered Mar 25 at 19:12
noednenoedne
9,25012566
answered Mar 25 at 19:12
noednenoedne
9,25012566
9,25012566
1
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
2
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
add a comment |
1
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
2
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
1
1
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
$begingroup$
These answers are the reason i feel guilty for not going back to continues learning.
$endgroup$
– Alex
Mar 25 at 19:32
2
2
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
$begingroup$
I got a PhD and I still can't put together answers like these! xD
$endgroup$
– Somebody
Mar 26 at 16:30
add a comment |
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