yofxThe Unofficial Blog of the Tunxis Math Department

One of my old students asked me to explain this card trick. If you go to the comments on the dailymotion page, you’ll see a couple of other explanations. The trick breaks down in to two sections: stacking the deck and sifting cards.

Stacking the deck
Though the aces look semi-randomly inserted into the deck, it is only an illusion. The aces will always wind up in positions 16, 32, and 48 in the stack no matter how one cuts the cards. Follow the bouncing cards…
Here’s the original setup:
(pile 1)(pile 2)(pile 3) (left overs)
Let’s insert the aces. We put the first ace on pile 1:
(1st ace)
(pile 1)     (pile 2)(pile 3) (left overs)
Now we cut pile 2 and make a space for the 2nd ace in pile 2:
(top of pile 2)
(1st ace)           (2nd ace)
(pile 1)              (bottom of pile 2)(pile 3) (left overs)
Now we do the same thing to pile 3, cut it and insert the third ace.
(top of pile 2)    (top of pile 3)
(1st ace)              (2nd ace)                (3rd ace)
(pile 1)                 (bottom of pile 2)(bottom of pile 3) (left overs)
Now they all get stacked on top of each other in this way:
(left overs)
(3rd ace)
(bottom of pile 3)
(top of pile 3)
(2nd ace)
(bottom of pile 2)
(top of pile 2)
(1st ace)
Then the first 4 left overs are put on the bottom (remembering there are 9 left overs all together):
(last 5 left overs)
(3rd ace)
(bottom of pile 3)
(top of pile 3)
(2nd ace)
(bottom of pile 2)
(top of pile 2)
(1st ace)
(pile 1)
(1st four left overs)
It is easy to see that the 3rd ace is the 6th card in the stack. It’s a little harder to see that the 2nd ace is 22nd and the 1st ace is 38. But remember, pile 2 and pile 3 each had 15 cards, so if the bottom of pile 3 sits on top of the top of pile 3 that would still 15 cards. The same is true for pile 2. We also know that pile 1 had 10 cards in it. So we can simplify the stack.
(last 5 left overs)
(3rd ace)
(15 cards)
(2nd ace)
(15 cards)
(1st ace)
(10 cards)
(1st four left overs)
Now it is easy to see our aces are in the 6th, 22nd, and 38th position. So the cutting of decks is really just slight of hand.

Sifting the cards
In the second part of the trick, the man goes through the deck and throws out every other card four times. We need to show the result of the process will always yield the 6th, 22nd, and 38th cards.
The first sifting out is easy to follow. The man turns the first card up, the second card down, the third card up… The result is all the odd cards are thrown out. We have just the even cards left, but they’re stacked from high to low, so…
52
50…
4
2
Now the process is repeated, 52 is up, 50 is down…. So that we end up throwing out all the multiples of 4 and reordering the deck from low to high.
2
6…
46
50
Repeating the process again will throw out all those numbers that have a remainder of 2 when divided by 8. (BTW, there is an interesting thing going on with modular arithmetic here.)
46
38…
14
6
We do it once more and throw out everything that has a remainder of 14 when divided by 16. We get the desired:
6
22
38.

Sorry it took so long, Jay!