Ok, so I thought about it a bit.

If I'm right he first part of the trick with the cutting doesn't really do anything at all to the position of their 3 cards. The first ace will always be the 6th card, the second the 22nd, the 3rd the 38th and then there are 14 cards behind that to make up the 52. Its pretty clear that any even number card here will be face down. On the second pass the order is reversed, and the numbers halved. So the 1st ace will be 8th, then 16th, then 24th, then 2 cards behind to make the 26. Order reversed again. 2nd 6th 12th etc.

I can't give a complex mathmatical reason why those card positions give those results, but in a practical sense its very easy to work out. Take a deck in new deck order, perform the same filtering process and look at the final 3 cards. There position in the new deck gives you the placement you need. The sorting process is ofc completely deterministic so its just a matter of getting the right cards in the right place. In this case he just...puts them there and creates the illusion that they have been shuffled when they haven't (all the cutting does is changes the order of the "filler" cards.)

I used to do a similar trick where you'd have 13 cards from Ace to King. You'd then spell out "Ace", and for each letter you'd remove a card from the top and add it to the bottom. On the last letter you'd turn over the top card, an Ace ofc, remove it and continue with the remaining 12 cards. Spell out Two, turn it over etc etc. Once you got to your last card you'd make a big joke of "shuffling it to the bottom" and turn it over to reveal a Joker. You could easily work out the order just by working backwards. I impressed a few girls with that...just not the smart ones.

Anyway, I think I got it?

Yes, you got it sir. Isn't math great.

I don't think there is any complex mathematical explanation. Given a starting point of 52 cards and as you point out the fact that the order of the remaining cards is reversed on each pass to keep the aces in even numbered positions, as long as the aces start in the 6th, 22nd and 38th positions you'll get the desired result. This will still work if the first pile has a different number of cards as long as the number moved top to bottom at the end combined with the first pile totals 14.

I think it's pretty straightforward if you examine the card buffers.

Part of the intended misdirection is that the cards are cut, but because of the way the cards are stacked, the cut stacks are always restored.

After the initial stack-up, the sequence 100% of the time is:

• 9 random cards on top of
• 1 ace on top of
• 15 random cards (which were cut then put back together when stacked) on top of
• 1 ace on top of
• 15 random cards (which were cut then put back together when stacked) on top of
• 1 ace on top of
• 10 random cards

>> ^Jinx:

I impressed a few girls with that...just not the smart ones.
Anyway, I think I got it?


Kind of like this video impressed a few viewers...just not..., well you know the rest

OMG this video is so impressive!!

Simple like a rubik's cube.

Damnit... wish I had a deck of cards now

My mind remains unblown. Disappointing

It's witchcraft!

BURN THE WITCH!!