Shuffling Card Trick - Numberphile
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Edit: I just Googled it, and apparently it comes from a book from 1902 and may even be older. Mind blown.
Date: 2022-04-08
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Comments and reviews: 9
ido
i was riffleshuffling the other day and thought, how many perfect riffleshuffles can you perform on a deck of cards until it returns to the starting point. So i wrote a small program to calculate it, and found some interesting results:
first of all, for 52 cards, the answer is 8.
next, the numbers had no clear consistency. i tried to find any sort of formula to calculate it but the results seemed pretty random to me, Except for powers of 2.
for any number 2-n, the result is n. I've tried for a couple of minutes to figure out why but didn't seem to get anywhere. And so I turn to you Numberphile, oh lords of the mathematics, I have results, yet no conclusions, and it would be amazing if you could get to this topic because I'm really interested.
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i was riffleshuffling the other day and thought, how many perfect riffleshuffles can you perform on a deck of cards until it returns to the starting point. So i wrote a small program to calculate it, and found some interesting results:
first of all, for 52 cards, the answer is 8.
next, the numbers had no clear consistency. i tried to find any sort of formula to calculate it but the results seemed pretty random to me, Except for powers of 2.
for any number 2-n, the result is n. I've tried for a couple of minutes to figure out why but didn't seem to get anywhere. And so I turn to you Numberphile, oh lords of the mathematics, I have results, yet no conclusions, and it would be amazing if you could get to this topic because I'm really interested.
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Jason
+Numberphile, I have a question which I have wondered for a while now, and I was wondering if you could calculate the total number of possible key combinations that are possible on a piano if a human had infinite hands. (So of the 88 keys how many ways are there to play the notes together (and singularly) in any combinations up to all 88 keys at once. (e. g all 88 keys, all keys excluding g7 and b3, no keys, and so forth for all the other combinations which are possible)
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+Numberphile, I have a question which I have wondered for a while now, and I was wondering if you could calculate the total number of possible key combinations that are possible on a piano if a human had infinite hands. (So of the 88 keys how many ways are there to play the notes together (and singularly) in any combinations up to all 88 keys at once. (e. g all 88 keys, all keys excluding g7 and b3, no keys, and so forth for all the other combinations which are possible)
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Healey
Someone please correct me if I'm wrong, but to my understanding:
The trick requires the deck to be prepared such that each pair has R & B; each foursome has D, C, H, & S; and each 13-some has A-K.
The cut, count, and riffle shuffle will always rearrange the cards in an order that also conforms to those patterns.
So theoretically, can't the trick immediately be performed again without having to re-prepare the deck, since it's still in an acceptable order?
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Someone please correct me if I'm wrong, but to my understanding:
The trick requires the deck to be prepared such that each pair has R & B; each foursome has D, C, H, & S; and each 13-some has A-K.
The cut, count, and riffle shuffle will always rearrange the cards in an order that also conforms to those patterns.
So theoretically, can't the trick immediately be performed again without having to re-prepare the deck, since it's still in an acceptable order?
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K0ggy
Am I right that this isn't possible with a 32-card set (7, 8, 9, 10, J, Q, K, A?
Because every 8th card had to be the same number and also the same color (because 8 is a multible of 2) which is not possible.
Or in general: It isn't possible if the different kind of cards (numbers) are a multible of 2 (or 4, which is included in 2.
Or is there any other permutation for these cases to make the trick work?
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Am I right that this isn't possible with a 32-card set (7, 8, 9, 10, J, Q, K, A?
Because every 8th card had to be the same number and also the same color (because 8 is a multible of 2) which is not possible.
Or in general: It isn't possible if the different kind of cards (numbers) are a multible of 2 (or 4, which is included in 2.
Or is there any other permutation for these cases to make the trick work?
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JaySay63
You know guys. this is numberphile. every card trick will be math based and meant to just -trick- you, not amaze with smoke and juggling blades.
I mean, for christ's sake, the man doing the trick is a mathematician, not a magician!
He's no mathemagician! He's a MATHEMATICIAN! (Sorry, Just realized that mathematician and magician rhyme)
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You know guys. this is numberphile. every card trick will be math based and meant to just -trick- you, not amaze with smoke and juggling blades.
I mean, for christ's sake, the man doing the trick is a mathematician, not a magician!
He's no mathemagician! He's a MATHEMATICIAN! (Sorry, Just realized that mathematician and magician rhyme)
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funny
All this while i sat thinking that there's some inherent mathematical property among the cards that allows them to be perfectly aligned from any order we start, and then you reveal some cheap magician parlor trick used to set the cards up beforehand!
That's not math, that's a waste of time.
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All this while i sat thinking that there's some inherent mathematical property among the cards that allows them to be perfectly aligned from any order we start, and then you reveal some cheap magician parlor trick used to set the cards up beforehand!
That's not math, that's a waste of time.
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Robin
If those modulo-properties exist before the -Brady-shuffle- and persists after it couldn't Brady have done it repeatedly?
And if cutting doesn't disturb the order isn't any sequence of cuts and -Brady-shuffles- doing the trick?
Would make it more impressive. (Even more as it already is)
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If those modulo-properties exist before the -Brady-shuffle- and persists after it couldn't Brady have done it repeatedly?
And if cutting doesn't disturb the order isn't any sequence of cuts and -Brady-shuffles- doing the trick?
Would make it more impressive. (Even more as it already is)
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Jaykrushna
To the people who are wondering about the Upside down writing. Jason is left handed. Few of them write everything upside down. bending their hand other way unlike right handed. so, he can easily write upside down with straight hand.
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To the people who are wondering about the Upside down writing. Jason is left handed. Few of them write everything upside down. bending their hand other way unlike right handed. so, he can easily write upside down with straight hand.
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Tetsujin
This trick depends on how the person would pick out any number of cards. Say if someone just cut the deck and jumbled them on the table, if the same conditions were true at the end, THEN it would be a card trick; )
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This trick depends on how the person would pick out any number of cards. Say if someone just cut the deck and jumbled them on the table, if the same conditions were true at the end, THEN it would be a card trick; )
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