# MathsWatch

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#1
Take a pack of 53 cards and deal them into 13 piles placing one card on each pile in turn, so that the first pile has one extra card. Now pick up the piles in order placing the second pile on top of the first, then the third on top of this, and so on. will repeating this process enough times ever return all the cards at once to their original positions in the pile? if so how many repetitions are needed?
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4 weeks ago
#2
(Original post by Mathshm)
Take a pack of 53 cards and deal them into 13 piles placing one card on each pile in turn, so that the first pile has one extra card. Now pick up the piles in order placing the second pile on top of the first, then the third on top of this, and so on. will repeating this process enough times ever return all the cards at once to their original positions in the pile? if so how many repetitions are needed?
There may be a slick way to do this, but doesn't spring to mind.

The whole process of dealing and restacking the cards can be thought of as a permutation. That alone tells you the cards will eventually return to their original positions if you repeat it enough times.

You "just" need to work out the order of that permutation. Rather tedious, but doable. About 20 min work (at least when I did it).
Last edited by ghostwalker; 4 weeks ago
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#3
Hi, I knew it was along the lines of permutations, but the bit where you have to stack the second on top of the first and so fourth is confusing. I also tried doing this literally but I was on 50 something and not reaching there. How did you go on doing it ? Thanks for your reply 🙌🏽
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4 weeks ago
#4
(Original post by Mathshm)
Hi, I knew it was along the lines of permutations, but the bit where you have to stack the second on top of the first and so fourth is confusing. I also tried doing this literally but I was on 50 something and not reaching there. How did you go on doing it ? Thanks for your reply 🙌🏽
Last edited by ThiagoBrigido; 4 weeks ago
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4 weeks ago
#5
I think it might be 212. Because you've got every 4 cards being in the same pile but because of 53 being a prime number, it takes longer to get them back than if it were 52 or 56. So I think it's the 4 x 53. I can't really give you a brilliant explanation. It's just one of those things which makes sense in my mind.
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4 weeks ago
#6
I'd have thought that an unbiased pack of cards contains 52 cards, therefore starting with 53 cards and work its way back; the first pile carries 14 cards, less the other three piles containing 13 cards each would give you the starting point again.
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4 weeks ago
#7
(Original post by Mathshm)
Hi, I knew it was along the lines of permutations, but the bit where you have to stack the second on top of the first and so fourth is confusing. I also tried doing this literally but I was on 50 something and not reaching there. How did you go on doing it ? Thanks for your reply 🙌🏽
It is literally the case of writing out the cards in 13 columns.

Cards 1 to 13 will form the first card in each pile, the bottom card.
14 to 26 go on top of them in order,
27 to 39 on top of them, and so on.

So, your first pile consists of 53, 40, 27,14, 1 (from top down)

Now restacking. Since the first pile goes on the bottom and the second on top of that, etc., we can work from the other end and say, the 13th pile will be on top, then the 12th beneath it, then the 11th, etc. Rather than write anything else we can just note that the new order is read down the 13th pile, then down the 12th, etc.

You now need to workout / write out the permutation. To start you off, the card in position 1 has gone to the very bottom of the pile, position 53.... Probably easiest to write it out in two rows to start, top row, 1,2,3,4,5,..., bottom row, 53,....

Seems to be a fair bit of interest in the question - Where is it from?
Last edited by ghostwalker; 4 weeks ago
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4 weeks ago
#8
(Original post by Possibly this)
I think it might be 212. Because you've got every 4 cards being in the same pile but because of 53 being a prime number, it takes longer to get them back than if it were 52 or 56. So I think it's the 4 x 53. I can't really give you a brilliant explanation. It's just one of those things which makes sense in my mind.
'Fraid not.
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#9
(Original post by Possibly this)
I think it might be 212. Because you've got every 4 cards being in the same pile but because of 53 being a prime number, it takes longer to get them back than if it were 52 or 56. So I think it's the 4 x 53. I can't really give you a brilliant explanation. It's just one of those things which makes sense in my mind.
But there’s 5 cards in pile 1 always, so 4x53 shouldn’t work
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#10
(Original post by ghostwalker)
It is literally the case of writing out the cards in 13 columns.

Cards 1 to 13 will form the first card in each pile, the bottom card.
14 to 26 go on top of them in order,
27 to 39 on top of them, and so on.

So, your first pile consists of 53, 40, 27,14, 1 (from top down)

Now restacking. Since the first pile goes on the bottom and the second on top of that, etc., we can work from the other end and say, the 13th pile will be on top, then the 12th beneath it, then the 11th, etc. Rather than write anything else we can just note that the new order is read down the 13th pile, then down the 12th, etc.

You now need to workout / write out the permutation. To start you off, the card in position 1 has gone to the very bottom of the pile, position 53.... Probably easiest to write it out in two rows to start, top row, 1,2,3,4,5,..., bottom row, 53,....

Seems to be a fair bit of interest in the question - Where is it from?
I will give this a go! Are you going to tell us what the answer is ?
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4 weeks ago
#11
(Original post by Mathshm)
I will give this a go! Are you going to tell us what the answer is ?
No. But I'll confirm/deny any answer you come up with. You can post the permutation you come up with if you wish it checked. I presume you're familiar with converting it to disjoint cycle form and working out the order; if not, check your group theory text books / notes.
Last edited by ghostwalker; 4 weeks ago
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4 weeks ago
#12
(Original post by ghostwalker)
No. But I'll confirm/deny any answer you come up with. You can post the permutation you come up with if you wish it checked. I presume you're familiar with converting it to disjoint cycle form and working out the order; if not, check your group theory text books / notes.
Ha! This is a fascinating question, and fun to solve over lunchtime. I must admit I can't find a simpler way of doing it than writing the permutation down, and then breaking it into disjoint cycles. I am wondering if there's a symmetry argument that simplifies this process - along the lines of certain elements must have the same length of orbit...
Last edited by Gregorius; 4 weeks ago
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4 weeks ago
#13
(Original post by Gregorius)
Ha! This is a fascinating question, and fun to solve over lunchtime. I must admit I can't find a simpler way of doing it than writing the permutation down, and then breaking it into disjoint cycles. I am wondering if there's a symmetry argument that simplifies this process - along the lines of certain elements must have the same length of orbit...
Nothing leaps out at me regarding symmetry. Might try plotting the cycles over the card piles when I have time, but just doing the first couple of "average" sized ones doesn't look promising - being a bit cryptic as I don't want to mention any specific values.

I'd not appreciated, until I'd thought about it a bit, that any repeated reordering of the cards must of necessity bring you back to the original order at some stage. Yea, for finite groups!
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4 weeks ago
#14
(Original post by Mathshm)
But there’s 5 cards in pile 1 always, so 4x53 shouldn’t work
Ah. I read the question wrong, thinking it was 4 piles of 13 (+1)
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#15
Okay, I think I got it. Is it 280?!
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4 weeks ago
#16
(Original post by Mathshm)
Okay, I think I got it. Is it 280?!
'Fraid not.

You can post your working, if you'd like someone to check it.
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#17
(Original post by ghostwalker)
'Fraid not.

You can post your working, if you'd like someone to check it.
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4 weeks ago
#18
(Original post by Mathshm)

Well you've gone for simulation, rather than the method I suggested, and you seem to have used 4 piles, rather than 13. And your piles are upside down. Top card (1), goes to the bottom of pile 1, which is position 53.
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#19
(Original post by ghostwalker)
Well you've gone for simulation, rather than the method I suggested, and you seem to have used 4 piles, rather than 13. And your piles are upside down. Top card (1), goes to the bottom of pile 1, which is position 53.
You might understand better
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4 weeks ago
#20
(Original post by Mathshm)
You might understand better

You have card1 going to position1. Card2 going to position5, etc.

Whereas card1 actually goes to position53, and card 2 to position48, etc.
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