

keeblem Inner circle Essex, UK 1167 Posts 
Hello
I need some help with the down under deal. If a spectator deals any number of cards, is there a formula which will tell me where to place the chosen card so it will be the one left in their hands after going through the down under deal? Thanks! Mark 
Nick Pudar Veteran user 369 Posts 
Mark,
Not exactly a formula, but my StackView software has the Down Under/Australian Deal as part of the simulation. It can be found in the Piles module. Essentially, you establish a Pile of cards, and then you set the appropriate parameters for the deal. When you run the deal, the final card will be on top of the Pile. There is also an "Inverse" checkbox, which will cause the deal to be done in reverse. Therefore, whatever card starts as the top card of the Pile, will end up in the correct position. You can simulate different numbers of cards, as well as deal parameters. Nick
Let me explain. No, there is too much. Let me sum up.
www.stackview.com Version 5.0 is available! 
uri New user Israel 95 Posts 
If you have "Scarne on card tricks", the formula is described there at some point.

edh Inner circle 4698 Posts 
Can someone tell me where I can find this deal? I think I may have it but forgot where to look.
Thanks edh
Magic is a vanishing art.

Hideo Kato Inner circle Tokyo 5649 Posts 
N is number of cards you use.
Find the largest 2^B which is smaller than N. Let's assume N=19. Such 2^B for 19 is 16. subtract 16 from 19, the result is 3. Multiply it with 2, the result is 6. If N=6, 2^B=4, 64=2, 2*2=4, the result is 4. If N=13, 2^B=8, 138=5, 5*2=10, the result is 10. If N=2^B, the last card is always the bottom card. Hideo Kato 
Gary Plants Special user 543 Posts 
I published a routine (with the formula needed) where groups of people cut off small packets of cards from a shuffled deck and pass the deck to another person, etc, etc. Each person now selects a card from the packet and it is "replaced" to the proper position for a down/under deal to work. Now everyone trades packets and does the down and under deals......the last cards are all of the selections.
This is in one of Steve Beam's Semi Automatic books. 
edh Inner circle 4698 Posts 
Hideosan, is 2^B translated equals 2 to the power of B?
What does B equal. I'm sorry but I don't understand the equation of N=19(I know what that means) but 2^B for 19 is 16. I don't know how you arrived at 16. Thanks for your help. edh
Magic is a vanishing art.

uri New user Israel 95 Posts 
I found the formula, and here is how I think about it which might be simpler for some:
You have four key numbers 4, 8, 16, and 32. These are easy to remember as each one is the previous one times two. (These are the only numbers needed since no numbers smaller than 4 will be used, and as the next number in the series is 62 it is not relevant either to regular 52 cards deck applications. These are the 2^B Katosan was talking about.) You have your packet whose total number of cards is N. Subtract the nearest lower key number from N and multiply the result by two. The card in this position from the top of the packet will turn up after the Down Under deal. (Notice it's Down Under deal, not Duck and Deal...) So, if you have 19 cards in your packet you subtract 16 from 19, leaving you with 3, which you then multiply by 2 which gives you 6, which is the card that will end up in the hand after the Down Under deal. 
edh Inner circle 4698 Posts 
Uri, thanks for the clarification for where B was coming from.
Magic is a vanishing art.

Hideo Kato Inner circle Tokyo 5649 Posts 
Urisan kindly explained what I posted in common language. Thanks.
Hideo Kato 
keeblem Inner circle Essex, UK 1167 Posts 
Thanks guys
I thought this may have all been lost when the magic Café crashed! I'd better quickly print it! Mark 
Bill Hallahan Inner circle New Hampshire 3220 Posts 
Hideo Kato, I got the algorithm from your post. It was quite clear to me what 'B' meant from this part of your post:
Quote:
Find the largest 2^B which is smaller than N. Thank you for posting the algorithm to figure out the position of the last card that remains after a Down Under deal. I think this might be a useful thing to use in a routine such as, "The Trick That Can't Be Explained", or in general Jazz magic with cards. I have one issue with the down under deal, and I've never been sure if it's relevant. To me, such a deal makes it obvious that only 1 prepositioned card will be the result. I feel this must also be obvious to a layperson. However, not having any experience using it, I don't know if that's true.
Humans make life so interesting. Do you know that in a universe so full of wonders, they have managed to create boredom. Quite astonishing.
 The character of ‘Death’ in the movie "Hogswatch" 
Hideo Kato Inner circle Tokyo 5649 Posts 
Bill Hallahansan, Down Under Deal or Under Down deal never produces a magical effect by itself. Compare the following two ways.
1. After receiving the selected card, you cut the packet several times and do DU. 2. After receiving the selected card, spectator shuffle the packet and you do DU. Spectators must know that you don't know the position of the selected card. Of course there can be exceptions to this rule. But most DU/UD tricks are based on this rule. Hideo Kato 
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