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RiderBacks
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Again, I preface this with the disclaimer that I am new to memdeck work. But I was thinking about one problem today, and a solution (using a memdeck) occurred to me. I thought I'd share and see, yet again, what folks think. Once I have my stack down perfectly, I'll start looking into purchasing the memdeck literature (starting with _Bound to Please_ and _Try the Impossible_.) But in the meantime, I'm just toying with ideas, so again, forgive me if this is elsewhere in print.

Consider a situation in which you want to force an item out of a group of several. Suppose you have four items, and you want to force one of them. Let's make those items be sealed envelopes, say. (They could be pieces of fruit or anything else, obviously.) With a memdeck, I think you could pull off a neat version of a magician's force...

Let's call the envelopes "1", "2", "3", and "4." You want to force number three.

Ask the spectator to take an envelope. Suppose they pick 1. Now ask them to select a number in the range (inclusive) of 1-52. The spectator selects the number five, suppose. That's the 9S in the Aronson Stack. Here you might say, "If the fifth card is odd, we toss your envelope. Otherwise we keep it." Of course, the spectator will have to toss the envelope.

Ask the spectator to take another envelope and name a (different?) number in the range (inclusive) of 1-52. Suppose they select the number fifteen. That's the 7D. You want to force a toss, since we're interested in forcing the third envelope, so you say: "If the fifteenth card is black, keep the envelope. Otherwise, toss it!"

Next, the spectator takes another envelope. Suppose this time it's the third one, which you want them to keep! Ask them to pick a number... Well, you get the idea.

You have face/non-face, one-way/two-way (not recommended), male/female, and many other (obvious) options available here to mix it up. You could move through a fair number of items and force the exact one you want using this memdeck technique. Give the deck regular false shuffles, and you have a completing convincing magician's force that you cycle through to force a single item out of a fairly large number of items!

I think it's a great idea. It's so great, and simple, that it's probably in print somewhere. But I welcome thoughts (or references to folks who have already come up with this!)
BalloonThief
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I’ve never heard of anything like that before. I like the idea but in practice it sounds a little iffy. Why are you qualifying the cards after the number is named. Shouldn't you say it before the number is named?

For example: "Name a number and if the card at the number is odd we toss your envelope."

Also, wouldn't this process look a bit weird if they chose the force envelope last.

Just trying the give you some food for thought. I'm sure some of these problems could be solved with the right motivation and routining. Keep working on it.
Tim Cavendish
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Magicians in your audience might enjoy it.

For regular people, that's a massively convoluted process to choose an envelope.

It also makes extensive use of an object -- a deck of cards -- that magicians are known for manipulating with great skill.

It's kind of like Mozart pulling out 4 envelopes and saying, "Let's choose an envelope and then we'll each mash down some keys together on this here piano. If our notes sound nice together, you keep the envelope!"
RiderBacks
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On Nov 4, 2015, Tim Cavendish wrote:
Magicians in your audience might enjoy it.

For regular people, that's a massively convoluted process to choose an envelope.

It also makes extensive use of an object -- a deck of cards -- that magicians are known for manipulating with great skill.

It's kind of like Mozart pulling out 4 envelopes and saying, "Let's choose an envelope and then we'll each mash down some keys together on this here piano. If our notes sound nice together, you keep the envelope!"


It only need be massively convoluted on the rare occasions that they pick all the wrong items first. The deck could even be in their hands during the process. They can count the cards (hopefully!) themselves. (Though they cannot shuffle themselves.) As a bonus, depending on the number they pick and what force you have available, you could ask them to choose whether to deal off the top or the bottom! I think that this sort of concern can be easily worked around in most situations. The more I think about this, the more I think I'm going to try it! But I'm just spitballing at this point.
RiderBacks
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On Nov 3, 2015, BalloonThief wrote: I’ve never heard of anything like that before. I like the idea but in practice it sounds a little iffy. Why are you qualifying the cards after the number is named. Shouldn't you say it before the number is named?


While that can be done, I think it's fair to say it requires more effort. I was just throwing out the basic idea which could be pulled off in a fairly effortless manner.

Quote:
On Nov 3, 2015, BalloonThief wrote: Also, wouldn't this process look a bit weird if they chose the force envelope last.


You can always add other items back to give them an option. If we have four items, and we've tossed the first three, then add a second back in to give a choice. That's just one possible fix to your concern. There are others, but I'm not sure this is a real problem provided you start out with the right phrasing/routining.

Quote:
On Nov 3, 2015, BalloonThief wrote: Just trying the give you some food for thought. I'm sure some of these problems could be solved with the right motivation and routining. Keep working on it.


Totally appreciate it.
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Interesting idea.

I think the biggest challenge would be coming up with a presentation that would justify this selection procedure - rather than just having them pick an envelope.

But it's fun (and worthwhile) to brainstorm an idea which is original to yourself - and who knows what it could morph into?

Some quick thoughts:

Rather than always naming a card from 1 - 52 which could be very cumbersome, it might be better just to say "Name a number from 1 - 10", and deal to that point. The next iteration you would repeat this, continuing to deal from the last stopping point. This would give it a feeling of randomness (not continually putting the deck back together, but just messily continuing the deal), and potentially be a lot less counting (boring). At least to me this seems like it would feel like you care less about the outcome of the count than if you continually reassembled the deck.

Also, if the force is the last item chosen, you wouldn't go through the count again, but simply say "... so that leaves the ____" after the third selection is tossed.

Finally, Balloon Thief is obviously correct that qualifying the cards after the number is named is much weaker. Sleight of hand also would weaken the procedure, but maybe you could combine sleight of hand with subtlety and equivoque to get something workable. For example, maybe you tell them the qualifying statement, then they name a number for the first question. If they hit, great, if they miss, you have them turn over the "next card" (if that card would be a hit). Once you deal the "next" card, you have to do the remaining ones that way as well. To speed up the process, grab the packet and you deal for a question and do a double lift. You would take advantage of segments of your stack if there's a high amount of black cards in the next 10, or odd cards, etc.

You'd want to play the whole thing off as a casual way to quickly make a random choice, like you're making it up as you go, so that this procedure is almost forgotten later.

Have fun playing with it!
RiderBacks
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Quote:
On Nov 5, 2015, Cohiba wrote: Interesting idea.

I think the biggest challenge would be coming up with a presentation that would justify this selection procedure - rather than just having them pick an envelope.

But it's fun (and worthwhile) to brainstorm an idea which is original to yourself - and who knows what it could morph into?

Some quick thoughts:

Rather than always naming a card from 1 - 52 which could be very cumbersome, it might be better just to say "Name a number from 1 - 10", and deal to that point. The next iteration you would repeat this, continuing to deal from the last stopping point. This would give it a feeling of randomness (not continually putting the deck back together, but just messily continuing the deal), and potentially be a lot less counting (boring). At least to me this seems like it would feel like you care less about the outcome of the count than if you continually reassembled the deck.

Also, if the force is the last item chosen, you wouldn't go through the count again, but simply say "... so that leaves the ____" after the third selection is tossed.

Finally, Balloon Thief is obviously correct that qualifying the cards after the number is named is much weaker. Sleight of hand also would weaken the procedure, but maybe you could combine sleight of hand with subtlety and equivoque to get something workable. For example, maybe you tell them the qualifying statement, then they name a number for the first question. If they hit, great, if they miss, you have them turn over the "next card" (if that card would be a hit). Once you deal the "next" card, you have to do the remaining ones that way as well. To speed up the process, grab the packet and you deal for a question and do a double lift. You would take advantage of segments of your stack if there's a high amount of black cards in the next 10, or odd cards, etc.

You'd want to play the whole thing off as a casual way to quickly make a random choice, like you're making it up as you go, so that this procedure is almost forgotten later.

Have fun playing with it!


All good points. As I said, I'm just spitballing. =) Some refinement is definitely required. This is to be considered a bare-bones idea. I'll definitely think about some of your great advice! I think there's some potential here, and I'd like to work out some of the details more!
Tim Cavendish
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I realize that this thread isn't actually about choosing 1 of 4 envelopes, but since it's the example at hand I feel compelled to mention that L&L is currently selling Multiplicity -- Max Maven's superb DVD devoted to equivoque and multiple outs -- for just $10.

Just sign up for their email list and they'll send you an email with the special coupon code.

http://www.llpub.com/

More apropos to this forum, I'll also note that L&L's 3 DVD set of Simon Aronson's work is also on sale for $10/DVD (with coupon code).
alecStephenson
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You might want to look at number 147 from Annemanns 202 methods of forcing (which is public domain so you can just download it) for a similar way of selecting one of four things (envelopes, piles, books) using a stack.

Mnemonica also has a short paragraph about using a stack to force (practically) any number.
Nick Pudar
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A related approach to force one of three items (that might be able to be turned into an entertaining process) is the following:

On one item, place two red dot stickers. On the second item, place two black dot stickers. On the third and the one you want to force, place one red and one black dot sticker. Start with a deck that alternates colors (red, black, red, black,...)

As you false shuffle the deck, say that the spectator will select one of the objects using the colors of the cards and some real randomness. Ask the spectator to riffle shuffle the cards (just once). Or it could be a rosette shuffle on the table. The important thing is that the spectator knows they did the shuffle, and that they were the source of randomness. You need to glimpse the top two cards, and make sure that they contain one red and one black. If they are the same color, then secretly shift the top card to the bottom. Due to the Gilbreath principle, each pair of cards from the top down contains one red and one black card.

Tell the spectator to name any number between 1 and 26, and that you will count down to that pair. The spectator shuffled and they named the random pair, so it appears that you have no control of the outcome. Count down to the correct pair. For example, if the number called out was 4, then take the first two cards, and drop them face down on the table while you call out "this is the first pair". Take the next pair of cards, and drop them on the table, as you count "2". Then drop successive pairs of cards as you continue to count to their number. When you turn over the pair at their named number, it will be one red and one black card, which will force the appropriate object.

Fairly simple, and it's most likely already be an existing idea -- I just don't have the time to investigate the literature.

Now, if you are a memdeck purist, you can add a wrinkle. You will be using your memdeck in its correct starting order, and give it a few false shuffles before placing the deck face down on the table. Start by giving three red dot and three black dot stickers to the spectator. Instruct them to put two stickers on each object in any combination they want. You just need to know what the resulting color sequence is on the force object. (It might be two blacks, two reds, or one of each -- there is no other combination possible.) Tell the spectator to name any number, and you will go to that pair in the deck (which only you have false shuffled). Since you know what the color pairs are throughout the entire deck, and you know what color pair the force object has, then it is a simple matter to secretly shift one or several cards from top to bottom (or vice versa) to ensure that the right color combination is at the right depth in the deck.

For example, let's suppose you're using the Aronson Stack, and the number called out is 5. If your target object has a red and black dot sticker, then you are all set since cards at positions 9 and 10 are already red and black (8D and AC). If your target colors are two blacks, then you have to shift one card form top to bottom, so that the target cards are two blacks (AC and 10S). If your target colors are two reds, then you would need to shift three cards to get two reds (5H and 2D). For this approach, you just need to know your stack well, and think quickly. (Just in case it is not immediately obvious, whatever number the spectator names, you multiply it by 2 then subtract one to know the starting card you need to think about for the colors. For example, if the number called is 15, then you calculate 2*15 - 1 to get 29.)

So, in the first approach, you have pre-assigned the colors to the objects. The spectator shuffles the cards and names a random pair position. In the second approach the spectator assigns the object colors and selects the pair number position, but only you (false) shuffle the deck (which you should do before the spectator does any of his/her tasks). Either approach can be made to appear extremely fair and without your influence.

Granted, all of this is a contrived selection process, but I think it can be framed up as an "unique randomized selection approach". I'll leave it to you to justify the approach. Smile

Nick
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RiderBacks
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Quote:
On Nov 9, 2015, Nick Pudar wrote: You need to glimpse the top two cards, and make sure that they contain one red and one black. If they are the same color, then secretly shift the top card to the bottom. Due to the Gilbreath principle, each pair of cards from the top down contains one red and one black card.


Not sure why the glimpse is necessary if you're working with a deck of fifty-two cards set in alternating red/black order. Other than that, I like your thinking here!
alecStephenson
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Nice ideas Nick.

And on the R/B riffle you may be able get away without doing the two card peek:

(i) If the two cards on top following the riffle are from the same half (i.e. are not interlaced), then you are done.
(ii) If you can peek at the bottom card of the top half prior to the riffle: if it is R you are done, if it is B and (i) does not happen then you know there are two R's on top and you need to shift.
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Not sure why the glimpse is necessary if you're working with a deck of fifty-two cards set in alternating red/black order. Other than that, I like your thinking here!


You might end up with RR as the top two cards after the riffle (not very likely, but possible)
alecStephenson
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Actually, I've decided (rightly or wrongly) that this will not necessarily work.
Obviously if the top card of the bottom half is B then you have a Gilbreath shuffle and you are fine.
But if it is R then this ensures that at most two cards of the same color can appear in sequence, but this is not enough for the force because you can't identify when to shift the top card.

For example with eight cards RBRBRBRB then halved to RBRB RBRB and riffled to RBRRBRBB and you are screwed, because you would need to move the top card to the bottom but you wouldn't know it from glimpsing the top two cards.
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On Nov 10, 2015, alecStephenson wrote:
Actually, I've decided (rightly or wrongly) that this will not necessarily work.


Alec,

Upon further reflection (and some simulations in StackView), you are absolutely correct -- there is an error. My initial recollection of how the Gilbreath Principle works was not complete. It turns out that (after the riffle shuffle) you have to cut the deck at the first matching pair of colors for this to work. So, when the first two cards are of the same color, shifting the top car to the bottom will make each successive pair have one of each color. But if the top two cards are different colors, then you have to look deeper to find the first pair of matching colors, and cut the deck between them.

So, to fix the routine, after the spectator riffle shuffles the deck, take the deck and spread it face up across the table to show how well mixed up the colors are, and reinforce the randomness. Use this opportunity to see where the the first pair of matching color is. When you pick up the deck, either separate it at the right point and cut while you pick up. Or remember how many cards you need to shift, and do a pass after you assemble the cards.

Thanks for finding the error!

Nick
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On Nov 10, 2015, RiderBacks wrote:
Not sure why the glimpse is necessary if you're working with a deck of fifty-two cards set in alternating red/black order.


RiderBacks,

The glimpse is necessary because the spectator has just riffle shuffled the deck, and there needs to be one separation of colors for the Gilbreath Principle to work. Be sure to see my one post just before this one -- Alec found an error in my original description. I've corrected the handling.

Nick
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Nick Pudar
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Hmm... one final observation.

For the Gilbreath Principle application described earlier, when the color pair separation cut results in a cut of an even number of cards, then you do not need to do the cut. But if the color separation results in a cut of an odd number of cards, then you need to do the cut. This is why when the top two cards are of the same color, you need to cut one card (an odd number) to the bottom. Alec, in your example, the two red cards appear at positions 3 and 4, so the separation cut would be odd, and it needs to be done.

And after further thinking about the Gilbreath Principle workings, in my description of an odd number of cards needing to be cut, it turns out that you only need to cut one card to get the sequence right. So the final description would be see if the separation cut would be an odd number of cards -- then just cut one card. If the separation cut is an even number of cards, then you do not need to do one.

This just goes to show that you really have to think about things before writing about them!

Nick
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alecStephenson
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Agreed on all of this Nick. Perhaps the most direct way then, would be to identify first occurence of two colours together (RR or BB) and cut the deck between them, as this would work irrespective of their position (even or odd).

A more indirect way would be to identify the Xth position of first card (first R in RR or first B in BB), then if X is odd (which can only happen in the RR case) you cut top card (or any odd number of cards) to bottom. Or bottom to top. Alternatively, you can run the first X cards, or any odd number less than or equal to X, because the top stack is reversible.
RiderBacks
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Quote:
On Nov 10, 2015, Nick Pudar wrote: The glimpse is necessary because the spectator has just riffle shuffled the deck, and there needs to be one separation of colors for the Gilbreath Principle to work. Be sure to see my one post just before this one -- Alec found an error in my original description. I've corrected the handling. [...] This just goes to show that you really have to think about things before writing about them!


I know, right? I was a wee bit puzzled by your initial description, and I didn't want to have to think too hard to sort it out. Thus my question. Thanks for working with Alec on fixing the handling. =) Definitely some good food for though here. I like it when my brain is hurt, but when I'm too many whiskeys in I sometimes opt for the easy out and just ask a question. =)
alecStephenson
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By an odd coincidence, I've just been reading the card magic of Nick Trost (which is brilliant) and in an appendix his discussion of gilbreath has a somewhat misleading sentence in it based on exacly this issue. The sentence (for no-reversing rotating colors) is:

"After the shuffle, if there is a red and a black card on top (or on the face), the rest of the cards are in red-black, or black-red, pairs throughout. If there are two cards of the same color on top (or on the face)..."

which is not correct e.g. eight cards RBRBRBRB then halved to RBRB RBRB and riffled to RBRRBBRB whereupon the top is RB and the face is RB but they are not paired throughout.

If this appears here then it probably appears in other places as well, so reader beware.
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