I apologize if this is old stuff to everyone on this forum but me, but, here goes: while rereading "Scarne on Card Tricks," I have found that Scarne gives a simple formula for figuring out which card will be the last one in your hand when you're doing a "down-under" deal with any given number of cards in the packet; it's part of trick #115, "Scarne's Phone Miracle," and it's given in other tricks, as well. I was pleased to find that out. But, is there a formula for figuring out which card will be last in your hand if you're doing an "UNDER-DOWN" deal? There must be, but I've never seen one, I don't think. Thanks. -- Hushai

I don't know the formula for figuring out the last card for a down-under deal, but if you have that formula you have one for an under-down deal. Use your down-under formula and add 1 to the result. That'll be the last card for an under-down deal of the same number of cards. If adding 1 gets you more than the number of cards, the answer is the first card.

Hushai,
I agree with rgranville. Just add 1. By the way, is there a mathematical formula for this or are be talking about a repeating pattern that represents the outcomes? Maybe you could share with us what the formula is as per Scarne.

Count Elmsley,
Scarne's formula is: take the number of cards in the packet; subtract from that number the highest number in the binary sequence (1, 2, 4, 8, 16, 32 ...) that is lower than that number; and double the result. For example: if you have 10 cards in your hand, subtract 8 to get 2, and multiply that by 2 to get four. The fourth card from the top of the packet of 10 will be the last in your hand if you do a down-under deal. I am too dumb, I guess, to understand why ANY of this works, either Scarne's formula or your (and rgranville's) answer to my question, but thanks to both of you anyway! I assume it must be another of those binary things, right? -- Hushai

Well, explaining the under-down formula is easy. Just think about what you're doing. Take the top card and put it on the bottom of the deck. Now stop. You have the same number of cards in your hand. You're next going to put a card down on the table, followed by one to the bottom, followed by one to the table... That is, you're now about to start a down-under deal. You have a formula for figuring out what number card is going to be the last card. (And now, so do we. Thanks!) But you already transferred a card from the top to the bottom. So what is now the first card was the second card before you did anything. What is now the second card was the third card. That's why you have to add 1.

Here's an example. Take the Ace of Hearts through the 7 of Hearts in order from top to bottom. The cards are AH, 2H, 3H, 4H, 5H, 6H, 7H. Now let's do an under-down deal. We put the AH on the bottom, so we now have 2H, 3H, 4H, 5H, 6H, 7H, AH. If we keep going from here, it's really just a down-under deal. We have seven cards, so we subtract 4 from 7, getting 3. Double that, and we have 6. The sixth card in our current stack will be the last card remaining when we're done. The sixth card is the 7H. But that was the seventh card in the original stack.

Thank you, rgranville! It does seem simple when you explain it like that. I am still intrigued by down-under and under-down deals and their relation to the binary sequence (if I am using that term right). I am not a mathematician, but I would like to understand more clearly why Scarne's formula works. Since I found it I have been applying it to card tricks I know of in which down-under and under-down deals are used, and I am gaining some insight into why those tricks work. Most of the tricks are simple (but good!) ones in books by Karl Fulves and Bob Longe, but more advanced tricks in books by people like Peter Duffie also use these deals. Peter Duffie says somewhere (in another context) that you should always understand the principles behind the tricks you are doing. Well, I would like to figure this principle out. Is it written up anywhere that anyone knows of? Thanks again.
-- Hushai

The principle is simple. In the first down and under deal, you are going to get rid of all the odd cards (1,3,5, etc). The remaining cards in the hand are the original 2,4,6, etc.

When you get to the original 2nd card, if there were an even number of cards, then you will get rid of cards 2, 6, 10, 14, n+4, etc. Leaving you with cards 4, 8, 12, 16, etc

However, if there were an odd number of cards, then when you will keep cards 2, 6, 10, 14, n+4, etc, and you will get rid of cards 4, 8, 12, 16, etc.

If you go the next step, you are just repeating one of the above sequences.

I remember that a formula for figuring out which card survived in a D&U deal was written up in Ibidem. Can't remember which issue, but I know it's in volume 2 of the reprint.

Bye for now

Harold

Thank you, Mr. Cataquet. I have found an explanation at http://www.ams.org/new-in-math/cover/mulcahy5.html, but my mathematical education is not nearly adequate to understand it. It's a good Website, however, and has an interesting trick using the down-under deal and the method for calculating the surviving card.

But if the number is in the form 2 to the nth power such as 8, or 16, or 32 then the calculation above gives you zero which represents the bottom card.

Technically speaking - and isn't this the forum for it? - the instructions as presented by Hushai state that you subtract the largest power of 2 less than the number of cards you have. So if you're starting with a power of 2, say 16, you subtract the power less than that, in this case 8. You then have 8, which you double to 16, in accordance with the instructions. Of course, you could simply remember that with powers of 2, the result will always be the last card, but what's the algebraic fun in that?

Count Elmsley and rgranville,

Yes, I've noticed that you can figure it either of these two ways -- interesting.

For binary geeks I have found a "formula" (ok, almost) for the under/down deal. If the packet has N cards all you have to do is write N in binary then move the leftmost (most significant) 1 to the rightmost (least significant) position. So for example:
N = 5 => N = 101 => card = 011 => card = 3
N = 12 => N = 1100 => card = 1001 => card = 9
Hope I have got this right. It should be the binary equivalent of the above formula but there is a kind of "coolness" to it.

Mr. Scotti, you've got it right. As shown by the article cited by Hushai above, the down-under deal can be described in binary terms as removing the leftmost (most significant) 1 and adding a 0 for the rightmost (least significant) digit (a left bit shift for computer geeks). Since the under-down calculation is the down-under plus 1, your formula works just fine.

I have been studying Phil Goldstein's book "Redivider," and in some of the fine effects there he has you use under-down and down-under deals. He notes that magicians often disdain these deals, but that lay people find them interesting. My new-found understanding of the down-under deal helped me understand better the principles on which those effects in "Redivider" work. Thanks to all of you for your help. -- Hushai

Well, generally I just follow the instructions for the trick without having to go into all this, lol.

Some good stuff with these deals and other principles in the new Howard Adams book mentioned elsewhere. If you liked "Redivider" you will love "Mathcasts Aspellonu"

Paul.

Paul:

I have to admit that the use of down-under deals (or vice-versa) have never appealed to me, mainly because I've never encountered a presentational justification for the procedure (in mentalism, particularly).

I'm curious as to how you and others on this board feel about this matter, and whether or not any of you have come up with ways to justify this procedure in a way that's consistent with the generally accepted ideals of mentalism (e.g., avoiding contrived and unnatural handlings).

I've had a similar aversion to spelling effects in mentalism. Despite the many ingenious formulations, I just can't seem to see the appeal in the context of mentalism.

Any thoughts folks?

Ben Blau

J.K. Hartman has a simple method for putting selections in a position for the down-under deal in a packet of any number of cards stated by the spectator. In other word: Say for example, three spectators select cards, they are lost in the deck, the magician asks each spectator for a number between 5 and 15 (depending on how many selections are involved, any quantity will work, and each selected number can be different), that many cards are given to each spectator. Each does the down-under deal to find their selections being the sole remaining card. It's called 'Free and Easy' in "Card Craft", pg. 314. I use it quite often. It's a very simple mathematical formula and method.
Magically, Jack

Here is another way to figure out what card will appear in a under down deal.

Take the pile of cards that the deal is to be performed with and fan them out, counting them as you go. At the same time, you push forward with your thumb the cards in positions that are powers of two - that means those cards in positions 2, 4, 8, 16 (and if you have that many, the card in position 32). When you get to the end, you count the number of cards beyond the last signpost card. Then double that number and add one; the card at that spot will be dealt last.

From here it's just a matter of placing the first card under the pile, then the next top card on the table and so on.

Magically

Aus

Ben, they have never truly appealed to me for mentalism OR magic either, lol. I don't recall ONE occasion where I have used it in my professional work. (Maybe in the dim and distant past, but I truly don't remember).

But that doesn't mean I would rule out using it. I assume Phil Goldstein and Larry Becker use it, both have published routines. I think it can be justified, but it really boils down to whether you think the effect is worth it or direct enough. Sometimes other card items can be procedural but still have a big pay off, so the procedural part should not necessarily be a bar to doing it.

A good one from Phil Goldstein is "Oct-See, More On" p. 6 of "Fifth". Phil doesn't even try to justify the procedure, the spectators are just instructed to do it (after an initial genuine shuffle of the packet). I'm sure at the conclusion the spectators are too startled to start analyzing the procedure, which would lead them nowhere anyway! I really did consider adding that to my repertoire. As Phil pointed out there, symbols, words, celebrity portraits, scenic photographs etc. could be used.

As it IS procedure, then it is the sort of thing that is downplayed isn't it? Perhaps justifying it is calling more attention to it? If it is regarding card selection perhaps a comment, "To further eliminate random cards..." or some such.

There is a routine that uses it in my book "Mindful Mentalism", a contribution from Peter Kane. Peter's routine is rather whimsical. (It was once marketed.) He does introduce it as the Australian "Down Under Deal" but the routine is themed around a letter from Australia from a psychologist friend. I have not seen Peter use it by I have seen it used by someone else and getting a good reaction.

Maybe what has been done to the cards immediately prior to the down under deal has a bearing. One assumes an apparent shuffle or mix has taken place. If someone just takes 8 cards and starts doing a down under deal it doesn't take a genius to figure they could have been in a particular order.

I AM currently having fun playing around with an effect using the deal from the new Howard Adams book, but the intention is perhaps for testing out on friends and associates. I would consider it more mental magic, as it includes a picture of the tomb of Nostradamus and involves five predictions. Who knows, depending on responses it generates as I continue to work on it, it may end up in professional use one day.

If you don't feel comfortable using it in mentalism, Ben, then don't. There's plenty of other stuff out there.

But you should try something out using if for friends, just go for it. At the end of the day, it is simply another tool at our disposal. It's not something I have personally used often, but on the odd occasions I have, no-one has questioned it.

Having said all this, I realize others are perhaps more in a position to defend the move or have experience of using it more than I!

:) Yeah, Spelling effects in mentalism? With cards? I can understand that. At the end of the day some of them prove VERY effective as card tricks, one I used to use in MAGIC performances was a George Kaplan routine, its in the Sphinx Golden Jubilee book, and the Encyclopedia of Card Tricks. The stack is described wrong in both cases, KH is repeated within the 12 card stack I think, you need to swap one for a KS. I used to follow this up with the original version of Harris's "Whack Your Pack" (Reflex).

I also remember baffling a room full of magicians with another Kaplan spelling effect, "Yogi Spelling Test", p.186, "The Fine Art of Magic".

Spelling tricks are a bit passé these days and seem slow when the average person has a ridiculously short attention span. I have come up with a good hook for one that creates interest which will be in my next book.

From a mentalist viewpoint (got my mentalist head on now) I would say it is extremely difficult for anyone to convince anyone else they are doing something psychic with a card spelling effect!* But, and there is always a but, if you have suspension of belief sometimes you can almost get away with murder.... Every effect should be judged on its own merit and how it would fit into the performance.

Paul
*Though I gave it a good go as a part of another routine "Ultimate Think Stop".

What IS a down/under deal ???