Independed of the operations he does on the cards, the order of cards is always:
14 cards + A1 + 15 cards + A2 + 15 cards + A3 + 4 Cards
(He is cutting the decks here and there, but they are useless transformations, they just go ontop of each other anyway.)
And that is the perfect organisation to do the even/odd throwing away.
Aha! I finally figured out a simple way to explain it to a layman:
The cuts don’t matter (see for yourself). So the cards always end up in positions 15, 31, 47. In other words, there are 15 cards between each of the chosen cards.
By removing every other card, we keep halving the deck. So after the 4 divisions, we have reduced the deck to be 1/16 of the original size. So basically we remove the 15 cards between each of the chosen cards.
I think this explanation is wrong in every particular, but it will make a non-mathematician feel like they understand what’s going on. In my book, that makes it a good Lie-to-children 🙂
You put 9 cards on top of the 3rd Ace. Thus, the 3rd Ace is card #10 counting from the top.
You split the leftmost 15 card pile into two parts, one part is moved to cover the 2nd ace, the other part remains. But then you stack that other part on top of the 2nd ace too, so no matter how you split it, there are 15 cards between the 3rd and 2nd ace. In other words, the “cut” is a ruse, you are putting part of the 15 on top of pile #2, but then you put the rest of them on top of pile #2 as well! Thus, there are always 15 cards between the 3rd ace and the 2nd. The same is true of the 1st ace, it is covered by part of pile #2 when cut, and the rest of pile #2 in the final stack, hence there are 15 more cards between the second and first ace.
Thus, numbered from the top, the 3rd ace will be card #10 always. Then there will be 15 more cards and then the 2nd ace, thus 10+16 = 26, it will be card #26 always. And the final ace will have another 15 plus the ace itself, 16 more from 26 is 26+16 = 42.
Your 3 cards, numbered from the top, will be numbers:
10, 26, 42.
Always! This is easy enuf to verify, try it with a deck, count them, your three target cards will be #10, #26, and #42 from the top, ALWAYS.
Now, take 4 off the top, we subtract 4 from each number and they become numbers:
6, 22, 38 (numbered from the top)
Now pile them up/down/up/down,
#1 -> to up pile
#2 -> to down pile #1 (first card in the down pile)
#3 -> to up pile
#4 -> to down pile #2 (second card in the down pile)
#5 -> up
#6 -> down #3 (third card in the down pile)
In other words, each card becomes half its number in the down pile. All three of our cards are even numbers, they will all go down, and will become half their number. That is, #6, 22, and 38, become #3, 11, and 19 in the down pile, BUT THAT IS COUNTING FROM THE BOTTOM!
That is, what I just called #1 in the down pile, ie, the first card that went to the down pile, what was numbered #2 in the original pile, counting from the top, has become #1 in the down pile, but that is counting from the bottom!
Let’s convert that to “from the top” numbering.The down pile now has half of 52 or 26 cards. Card #1 counting from the bottom, ie, the bottom card, is #26 from the top. #2 counting from the bottom is #25 from the top. #3 from the bottom is #24 from the top. That is,
#1 from bottom -> #26 from top
#2 from bottom -> #25 from top
#3 from bottom -> #24 from top.
In general, card #X from the bottom is # 27-X from the top. So our three cards, #3, 11, 18 from the bottom become 27-3, 27-11, and 27-19 from the top, ie, the correct numbers, from the top, are:
24, 16, and 8.
Now, repeat the procedure. #8 will be the 4th card in the down pile, 16 the 8th, 24 the 12th, so our cards become
12, 8, 4 AGAIN COUNTING FROM THE BOTTOM.
The pile has half of 26 or 13 cards. So, the n’th card from the bottom is 14-n from the top, and so our cards are really:
14-12, 14-8, 14-4 =
2, 6, 10 COUNTING FROM THE TOP.
Repeat the procedure, they become:
1, 3, 5 COUNTING FROM THE BOTTOM OF THE DOWN PILE. Which contains now 6 cards.
(This is a bit trickier because we had 13 cards, so the down pile will have 6 and the up 7. Imagine 3 cards, they would go: up, down, up = 2 in the up pile, 1 in the down. Thus, since we always start with up, the up pile has the extra card. So, starting with 13 cards, the up pile will have 7, the down pile 6.)
So our down pile has 6 cards, and our cards are 1, 3, 5 from bottom, so from top they are:
7-1, 7-3, and 7-5 from the top which is
6, 4, and 2 COUNTING FROM THE TOP.
Repeat one more time and our down pile will be just 3 cards numbered 3, 2, and 1 from the bottom. At this point, we hardly need to convert that, 3,2,1 from the bottom becomes 4-1, 4-2, AND 4-3 from the top, ie, 1,2,3 from the top, but we are done. We have our 3 original cards!!!
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is this guy serious?
Independed of the operations he does on the cards, the order of cards is always:
14 cards + A1 + 15 cards + A2 + 15 cards + A3 + 4 Cards
(He is cutting the decks here and there, but they are useless transformations, they just go ontop of each other anyway.)
And that is the perfect organisation to do the even/odd throwing away.
Aha! I finally figured out a simple way to explain it to a layman:
I think this explanation is wrong in every particular, but it will make a non-mathematician feel like they understand what’s going on. In my book, that makes it a good Lie-to-children 🙂
Here u go! I worked this out myself!
You put 9 cards on top of the 3rd Ace. Thus, the 3rd Ace is card #10 counting from the top.
You split the leftmost 15 card pile into two parts, one part is moved to cover the 2nd ace, the other part remains. But then you stack that other part on top of the 2nd ace too, so no matter how you split it, there are 15 cards between the 3rd and 2nd ace. In other words, the “cut” is a ruse, you are putting part of the 15 on top of pile #2, but then you put the rest of them on top of pile #2 as well! Thus, there are always 15 cards between the 3rd ace and the 2nd. The same is true of the 1st ace, it is covered by part of pile #2 when cut, and the rest of pile #2 in the final stack, hence there are 15 more cards between the second and first ace.
Thus, numbered from the top, the 3rd ace will be card #10 always. Then there will be 15 more cards and then the 2nd ace, thus 10+16 = 26, it will be card #26 always. And the final ace will have another 15 plus the ace itself, 16 more from 26 is 26+16 = 42.
Your 3 cards, numbered from the top, will be numbers:
10, 26, 42.
Always! This is easy enuf to verify, try it with a deck, count them, your three target cards will be #10, #26, and #42 from the top, ALWAYS.
Now, take 4 off the top, we subtract 4 from each number and they become numbers:
6, 22, 38 (numbered from the top)
Now pile them up/down/up/down,
#1 -> to up pile
#2 -> to down pile #1 (first card in the down pile)
#3 -> to up pile
#4 -> to down pile #2 (second card in the down pile)
#5 -> up
#6 -> down #3 (third card in the down pile)
In other words, each card becomes half its number in the down pile. All three of our cards are even numbers, they will all go down, and will become half their number. That is, #6, 22, and 38, become #3, 11, and 19 in the down pile, BUT THAT IS COUNTING FROM THE BOTTOM!
That is, what I just called #1 in the down pile, ie, the first card that went to the down pile, what was numbered #2 in the original pile, counting from the top, has become #1 in the down pile, but that is counting from the bottom!
Let’s convert that to “from the top” numbering.The down pile now has half of 52 or 26 cards. Card #1 counting from the bottom, ie, the bottom card, is #26 from the top. #2 counting from the bottom is #25 from the top. #3 from the bottom is #24 from the top. That is,
#1 from bottom -> #26 from top
#2 from bottom -> #25 from top
#3 from bottom -> #24 from top.
In general, card #X from the bottom is # 27-X from the top. So our three cards, #3, 11, 18 from the bottom become 27-3, 27-11, and 27-19 from the top, ie, the correct numbers, from the top, are:
24, 16, and 8.
Now, repeat the procedure. #8 will be the 4th card in the down pile, 16 the 8th, 24 the 12th, so our cards become
12, 8, 4 AGAIN COUNTING FROM THE BOTTOM.
The pile has half of 26 or 13 cards. So, the n’th card from the bottom is 14-n from the top, and so our cards are really:
14-12, 14-8, 14-4 =
2, 6, 10 COUNTING FROM THE TOP.
Repeat the procedure, they become:
1, 3, 5 COUNTING FROM THE BOTTOM OF THE DOWN PILE. Which contains now 6 cards.
(This is a bit trickier because we had 13 cards, so the down pile will have 6 and the up 7. Imagine 3 cards, they would go: up, down, up = 2 in the up pile, 1 in the down. Thus, since we always start with up, the up pile has the extra card. So, starting with 13 cards, the up pile will have 7, the down pile 6.)
So our down pile has 6 cards, and our cards are 1, 3, 5 from bottom, so from top they are:
7-1, 7-3, and 7-5 from the top which is
6, 4, and 2 COUNTING FROM THE TOP.
Repeat one more time and our down pile will be just 3 cards numbered 3, 2, and 1 from the bottom. At this point, we hardly need to convert that, 3,2,1 from the bottom becomes 4-1, 4-2, AND 4-3 from the top, ie, 1,2,3 from the top, but we are done. We have our 3 original cards!!!
QED!!!
This trick can be done in class, which is a good thing.