Any idea how this works?

Incredible math card trick. [VIDEO]

its pretty cool...is it anything to do with odd and even numbers?

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- Jul 2nd 2011, 11:21 AMtjnorthamMath Card Trick
Any idea how this works?

Incredible math card trick. [VIDEO]

its pretty cool...is it anything to do with odd and even numbers? - Jul 2nd 2011, 12:09 PMMacstersUndeadRe: Math Card Trick
I believe it is indeed the positioning of the 3 cards into 'odd slots'

Pile 1: 10

Pile 2: 15

Pile 3: 15

Pile 4: 9

--

Let A1, A2, A3 be the chosen cards

Step 1: A1 on Pile 1. Pile 1 has 11 cards, A1 in 'odd' position

Step 2: add any number of cards from Pile 2 to Pile 1. if the number of cards brought to Pile 1 is even, number of cards on Pile 2 is odd. if the number of cards brought to Pile 1 is odd, number of cards on Pile 2 is even. (because even + even =! 15 and neither can odd + odd =! 15, since 15 itself is odd)

Step 3: A2 on Pile 2. If # of cards on Pile 2 even, now # is odd. If # of cards on Pile 2 odd, now # is even.

...

now I'm having trouble tracking. if the trick works all the time, it for sure is independent of the number of cards chosen from each pile, and dependent on the number of cards in each pile in the beginning. - Jul 3rd 2011, 10:02 PMWilmerRe: Math Card Trick
Since (at end) 4 cards are moved from top to bottom, and since these 4 cards are from the

leftover pile of 9 cards (which had been placed on top), then this can be done with

pile#1 having 14 cards instead of 10; the leftover will then be 5 cards.

Easy to see that since these 5 cards are on top of last spectator card, then that spectator

card will be initially in the "down" pile, since the down pile contains even-turned cards.

And since "cutting" can be any number of cards, we can take the whole pile!

So, after placing 1st spectator card on top of pile#1, we can take all of pile#2 and place it

on top of the spectator card; similarly with pile#3.

So, numbering the cards 1 to 52, making the spectator ones 50, 51 and 52, we end up with

the pile of 52 cards being:

1,2, ... 13,14,[50],15,16, ... 28,29,[51],30,31, ... 43,44,[52], 45,46,47,48,49

The first "up-down" job will leave this pile as the down pile:

48,46,[52],43,41,39,37,35,33,31,[51],28,26,24,22,20,18,16,[50],13,11,9,7,5,3,1

The next: 3,7,11,[50],18,22,26,[51],33,37,41,[52],48

The next: [52],37,[51],22,[50],7 and finally: [50],[51], [52]

No matter what number of cards are "cut" (none, some, all), it'll turn out the same general way.

BUT I can't prove it...not yet anyway... - Jul 4th 2011, 10:03 AMWilmerRe: Math Card Trick
Pile#1: 14 ; Pile#2: 15 ; Pile#3: 15 ; Pile#4: 5 (the leftover) : baptized as P1,P2,P3,P4 !

My previous post may be a little "too simplistic" as explanation; but to explain fully

would require way too much typing...and I'm a one-finger-typer!

The main thing to remember (when you try and "see" what's happening) is that the

"cuts" made (like some cards from P2 go to P1) really are irrelevant: the final process

of putting the piles together (P3 on top, P2 next, P1 at bottom) reinstates the

original 15 in P2 and P3, since, as example, bottom of P2 goes to top of P1, thus

the 15 originals from P2 are together: not quite in same order, but what's important

is the "15" in between the 2 spectator cards.

This is the result of the final "piling"

Code:`5 P4`

1 spectator card

15 original P3 (not in order, but irrelevant)

1 spectator card

15 original P2 (not in order, but irrelevant)

1 spectator card

14 exact original P1

ends with 6-22-38 (try it!) as last 3 cards down; and quite clear here:

5 + 1 = 6

6 + 15 + 1 = 22

22 + 15 + 1 = 38

In other words, the whole thing is complicatedly simple!! - Jul 12th 2011, 11:48 PMWilmerRe: Math Card Trick
Just figured out how to do this trick with 26 cards instead of 52; takes less time!

Try it; the 3 piles are 2-7-7 cards. - Aug 22nd 2011, 03:37 PMDevenoRe: Math Card Trick
the "cuts" are just for show, they don't actually change the positioning of the chosen cards. for the first cut we add m cards to the 1st pile, then we place the 2nd chosen card on top of the 15-m remaining. this puts the 2nd chosen card 10 + 1 + m + 15-m + 1 = 27th from the bottom, no matter what "m" is. the same logic for the 2nd cut.

after the final stacking, the chosen cards are 11th, 27th, and 43rd from the bottom, after the 4 cards are put from the top to the bottom, this puts them at 15th, 31st, and 47th from the bottom, which is the same as 6th, 22nd and 38th (in reverse order, now) from the top. since we do the final deals (up-down) from the top, the even-numbered cards (including our chosen 3) are all "down".

dealing reverses the order (first down goes to the bottom of the stack), so our chosen 3 are now 8th, 16th and 24th from the top of the "down" stack. these are all even numbers, so they will all wind up in the "down" pile. at this point, we have 26 cards left.

after the 2nd deal, our chosen cards are now the 2nd, 6th, and 10th from the top (again, all even numbers). we now have just 13 cards left. dealing alternately up-down will eliminate 7, leaving 6 cards remaining: our chosen 3 wind up 2nd, 4th and 6th from the top.

it is now clear that the final deal will eliminate all but our chosen 3.

why this works: the "spacing" between our 3 chosen cards is 16 (this is why the 2 "middle piles" in the beginning have to have 15 cards). the placing of 4 cards from the top to the bottom is again, mostly for show: the real purpose is to make the top stack have 5 (one more than 4, which is important).

14 * 15 * 15 * 5 <--- initial spacing (bottom to top)

7 * 7 * 7 * 2 <--- spacing after 1st deal (top to bottom)

3 * 3 * 3 * 1 <--- spacing after 2nd deal (bottom to top)

1 * 1 * 1 * <--- spacing after 3rd deal (top to bottom)

* * * <--- final 3 cards. - Aug 22nd 2011, 06:45 PMWilmerRe: Math Card Trick
Right on Deveno!