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Topic: How to successively reveal more and more cards in a packet?
Message: Posted by: Bob G (Jul 20, 2018 10:13AM)
Dear Magicians,


I've described a trick that I'm trying to create in another thread or two, but I think I can now boil down what I need into a more focused question.


In the course of the trick I'm holding a packet of about 4 cards red-backed cards, along with maybe 3 blue-backed cards. I want to initially show -- and count -- all the red backs while hiding all the blue backs. (The faces are irrelevant to the trick.)


Then I want to count again and show all the red backs again, along with *just one* blue-backed card, which I then openly discard from the packet. I also discard a red card at this point. So now I have three red and two blue. (I'm beginning to feel like Dr. Seuss.)


Finally, I want to count a third time, this time revealing all five cards. (I assume this would just be an ordinary count, but the way I handle the cards should be consistent with what I did during the two false counts.)


The exact numbers of cards aren't that important. I have a story in mind, which I'll be happy to share, and any sequence of false counts that accomplishes something similar to what I just described would be fine -- as long as it fits the story.


I've been reading about false counts but haven't yet learned any (unless you count small-packet DL's and the three-way move in Color Monte). There are so many counts (in Mentzer's book and in Racherbaumer's Counthesaurus) that it's a bit overwhelming. The thing I'm not seeing is how to reveal more cards during each count. For instance, I could hide three blue cards among four red ones using the EC, but I don't see how to then reveal just one of the blue cards, rather than all three.


Since I'm still pretty new to magic, I'd like to stay away from palms and difficult counts (though I imagine none of them are actually *easy*.)


I'll be happy to repost this on Secret Sessions if that seems appropriate.


Thanks for any suggestions! This idea has been keeping me awake at night. :)


Regards,


Bob
Message: Posted by: Wravyn (Jul 20, 2018 10:34AM)
Sounds close to but not exactly like Fred Kaps Homing Card. You can find a video of him doing Homing Card on YouTube. Also you can get the how to from https://trickshop.com/homing-card.html
Message: Posted by: Bob G (Jul 20, 2018 01:08PM)
Thanks , Wravyn. I have the materials you suggested -- they're great! I got them in response to an earlier suggestion that what I wanted to do was similar to the Homing Card Plot (or at least to this one; I guess there are two plots by the same name). At the time I wasn't able to adapt the plot to my story, but it's probably time for a second look.


After I posted, it also occurred to me that there are all kinds of counts that show fewer cards as more and vice versa. That might be a promising avenue to follow.


Bob
Message: Posted by: Wravyn (Jul 20, 2018 01:53PM)
Your welcome.
A Six Card Repeat (throw 3 cards away, still have 6 & repeat...) could have a count for you also. I don't have my books nearby, but I think it may be in one of the Tarbel books, and I'm sure it's in some others.
Message: Posted by: Bob G (Jul 20, 2018 05:05PM)
I've seen that in The Card Magic of Nick Trost -- another possibility! I'll look it up.
Message: Posted by: Claudio (Jul 21, 2018 01:08PM)
Hi Bob

Here’s a simple and consistent handling which relies on the Hamman Count (HC). I'm assuming you're right-handed.

Start with 4R on top and 3B at the bottom. Perform a HC and the packet will back in original order. You have shown 7 red cards.

With left forefinger buckle the bottom card (Blue) and get a left pinkie above it. Start a modified HC: the right hand grabs the packet and picks up the break with the right thumb. The left thumb peels singly three red cards. On the count of four, the left hand transfers its 3 cards to bottom of right-hand packet while the left thumb drags all the cards (RBB) above the break into the left hand. A blue card is visible on top right hand packet, peel it, but sticking out, onto the left hand packet and count five, count 6 and peel another red card, and finally deposit the two remaining red cards as one on top of the packet and count 7.

Table the sticking out blue card and then a red card.

The cards are in RRRBB order. Do a regular HC to show 5 red cards.

I hope this is clear. What I called "Hamman Count Modified", is not in fact a HC. The count has actually a proper name, which escapes at the moment.
Message: Posted by: Bob G (Jul 21, 2018 02:02PM)
1. Note to Wravyn: I took a quick look at Trost's six (or seven?) card repeat; Trost says it's self-working (!). So I'm planning to read it carefully -- it sounds like a great thing to know whether or not I end up using it for my blue/red trick.



2. Many thanks, Claudio, for taking the time to work this out. I'm going to try it and will let you know if I have any questions. I am indeed right-handed. I haven't learned the HC (though I understand the idea and have several sources for the handling), but now that I know the HC works for this I'm excited to learn it.


Great Happiness, as the King said to Macbeth.


By the way, I've been showing friends my variation of the Nick Trost trick for which I needed the H bottom slip cut. If they're seeing anything they shouldn't, they're too polite to mention it! So another thank you for your critique of my video, which was a big help. I'm definitely moving forward in magic, even if slowly (as is my nature).


I hope things are going well with you.


Best Regards,


Bob
Message: Posted by: Claudio (Jul 21, 2018 02:47PM)
Bob, I'm happy to know you've mastered the HL cut. My old teacher was fond of quoting Oscar Wilde: "Nothing worth learning can be taught." So at the end, you got there by yourself.

Keep up the good work!
Message: Posted by: Bob G (Jul 21, 2018 03:26PM)
Thanks for the encouragement! I love Wilde, though I wasn't familiar with that quote. It's a good one. I agree -- but of course getting there oneself with guidance along the way makes a big difference.


See you,


Bob
Message: Posted by: Claudio (Jul 23, 2018 09:45AM)
Bob, rereading your post, I realised I might have misunderstood it. I suggested to count the packet of 7 cards (4R3B) as 7R. But, it looks like you meant to count it as 4R. Is that right?

And on the last count (3rd), you meant to show reds and blues?
Message: Posted by: Bob G (Jul 23, 2018 11:51AM)
Hi Claudio,


Thanks for rereading and checking with me. Yes, I read your post carefully last night and realized that we weren't quite on the same wavelength. I imagine (hope) that your handling can be modified to fit what I had in mind. To compound the difficulties, I realized that I hadn't thought carefully enough about what I needed. Sorry to have wasted your time.


Assuming that you're still interested, here's an example of what would work for me. I've described the story that goes with it below, partly to clarify my own thinking, and partly so that you know the context, which might allow for different counting from what I'm about to describe.



So, here's the counting that I'd like the spectator to see:


1. Magician initially shows three red cards.


2. He openly adds two red cards to his hand. As far as the audience is concerned, he now has five cards in his hand. But when he counts again, he appears to have three red cards and one blue card. The blue card is discarded. From the spectators' viewpoint, the magician now has three red cards and no blue cards.


3. He counts once more, and discovers that he has only one red card left, and two blue cards. The blue cards are discarded.




I *think* this means that the magician actually has to be hiding three physical blue cards at the outset, to allow for the discards. I hope there's still a way to accomplish that! If it helps, one option is to have the magician, when he discards blue cards, also openly discard a red card (this would be as a penalty in the game that I describe below. )


I suspect that other variations would serve the sort of story I want to tell. In case it helps, here' the story. Two devils, one blue and the other red, are playing cards. A player wins if his opponent runs out of cards of opponent's color. During each round, one devil throws a die. If his number comes up then he gets to take a card of his color into this hand from a pile of cards of his color, and the other devil has to discard a card of his color into a pile of cards of his color. If the thrower's number doesn't come up, the opposite occurs. First round, Red D. tosses the die, second round Blue tosses, etc.


Each player starts with three cards of his own color. (The blue devil really does have just three blue cards; the red devil appears to have three red cards, but is also hiding three blue cards. So in Count #1 above, the red devil has to appear to show three red cards, while hiding the three blue.) The red devil wins the first two rounds (by cheating -- the die is loaded.) Each time, as per the rules of the game, the blue devil has to put one blue card in a discard pile, and the red devil gets to take one red card from another pile. So, at the end of the first two rounds, the red devil has five red cards and the blue devil with just one blue card. (That's what I meant when I said that the red devil openly adds two red cards to his hand in #2 above.)


So, the blue devil holds only one card. If the red devil wins one more round, the game will be over.


But now the blue devil, realizing that the game is crooked, asks innocently, "Are you sure you counted right?" The red devil recounts and discovers that he now mysteriously has a *blue* card and only three red cards. (That's Count #2.) Red D. has to give the blue card in his hand to the blue devil.


So now the red devil (apparently) has three red cards, and the blue devil has two blue cards.


The blue devil says, "You know, I'm still not sure you counted right. Where did you learn to count?" So red devil recounts again, and this time finds that he has only one red card, and two blue cards have appeared in his hand! He has to give the latter to the blue devil. So now the red devil has one red card and the blue devil has four blue cards.


Finally, the blue devil gives the die back to the red devil -- but this time the die has changed (using some kind of utility switch, maybe) in such a way that the red devil must lose the last round. He's left with no cards, so the blue devil wins the game.



Ideally I'd ask a spectator to play the role of the blue devil. They would give the red devil (me) back the die, and be surprised to discover that the red devil now held a different die, one with which he couldnl't possibly win.



I hope all that made sense!



Bob
Message: Posted by: Claudio (Jul 24, 2018 05:39AM)
OK, got it. I'll PM you.
Message: Posted by: Jonathan Townsend (Jul 25, 2018 11:22AM)
[quote]On Jul 23, 2018, Bob G wrote:
...So, here's the counting that I'd like the spectator to see:


1. Magician initially shows three red cards.


2. He openly adds two red cards to his hand. As far as the audience is concerned, he now has five cards in his hand. But when he counts again, he appears to have three red cards and one blue card. The blue card is discarded. From the spectators' viewpoint, the magician now has three red cards and no blue cards.
...[/quote]

I'm lost at step 2. Three red cards plus what?
Message: Posted by: Bob G (Jul 25, 2018 04:21PM)
Thanks for your interest, Jon. I don't understand your question, but let me take a shot at clarifying. In Step 1, magi displayed three red cards. (This is supposed to be the red devil's hand at the beginning of a card game.) I don't care whether there are *really* three red cards in his hand, it just has to look that way. Meanwhile he has three blue cards, but the audience doesn't see them.


Next (now we're into step 2), to portray the first round in the game that the devils are playing, magi adds first one red card, then another, to his hand. So it *appears* that he now has five red cards. In reality he also has those three blue cards, which have remained hidden, but which are about to be revealed.


Now the blue devil demands that the red devil count his cards again. This time it appears that the devil has only three red cards, but suddenly he also has one blue card. He's lost 2R and gained 1B. He has to give up the blue card. So it seems that he has 3R and no blue anymore. Actually, he still has 2B that haven't yet been seen.


Blue devil demands another count, and now the red devil has only one red card, and suddenly two *more* blue cards have appeared. So red devil has lost another two reds, and has gained two blues. He has to give up the two blues. So red devil is left holding just one red card.


So basically I want three counts in a row of the red devil's (magician's) hand:


First Count: Magi seems to have 3R. 3B are also present, but hidden.


Then Magi openly adds 2R, for a total of 5R. (3B still hidden.)



Second Count: Now Magi seems to have only three red cards (instead of five), and in addition he's holding a single blue card. (1B has been revealed, 2B remain hidden.)


Then Magi discards the blue card. Presumably he now has three red cards and no blue cards.


Third Count: Magi has only 1 red card, and two blue cards have appeared. So he's somehow lost 2R and gained 2B.


Magi discards both blue cards and holds just one red card.


And then the stuff with the die....


The number of red cards diminishes; that wouldn't have to be all done by sleights. For instance, it could be a rule of the game that the red devil has to openly discard a red card or two as a penalty for holding blue cards, which he isn't allowed to do.


Hope that helps, Jon.



Bob
Message: Posted by: Jonathan Townsend (Jul 26, 2018 10:46AM)
Thanks. Step 1 has you counting three red backed cards. Fine so far. Step 2 has you adding two red backed cards to the packet in your hands. You count five red backed cards. Fine so far. [you mention something about devils and rules] Step 3 has you counting your packet as three red cards plus one blue card. You discard the blue card from you packet. Where do you put that blue backed card? After you dispose of that (apparently single blue backed card) you count the packet again to display three cards, two blue backed cards and one red. You remove the two blue backed cards. (again where do you put them?)

The counts/displays are a minor part of your routine which has story elements. Prop logistics and blocking looks vague so far. Do you have a way to get the cards and die into play to start your routine? In total, how many red backed cards and how many blue backed cards do you want them to see while you get started?

Since you haven't mentioned the faces of the cards - this might help simplify practice and rehearsal. Use red (diamond, heart) cards and black (spade, club) cards. And a coin rather than a die. There are two-headed coins. ;)
Message: Posted by: Bob G (Jul 26, 2018 12:27PM)
Thanks for your careful reading, Jon. I'm still mulling over the form I want the story to take, but here's basically what I have in mind.


The ideal, if practical, would be for the magician to play the role of the red devil and a spectator would play the blue devil. The red devil shows himself to be a bit of a bully, whereas the blue devil is more just mischievous. So our sympathies are with the blue guy. And there will (I hope) be some humor when the audience notices the blatant cheating that the red devil is engaged in.


At the beginning the magician would hand the spectator his three blue cards and then deal himself -- or already have prepared and sitting on the table at his place -- his own red cards, with the 3 blue cards hidden. The spectator (blue devil) counts his cards and finds that he has three blue. The magician (red devil) false counts his cards to show that he has three red cards.


I hadn't thought about how to get the die in play. I think it can just be sitting on the table at the red devil's place right at the beginning. I bought a pair of mis-spotted dice, one of which has a "five" on each face, and the other has twos and threes or something -- and no fives. The die that we start with is the all-fives one. The other die isn't visible at this point.


The red devil says, "Last time we played I trounced you, so I want to be kind to you this time." (He picks up the all-five die.) "I win only if I get a five; otherwise I lose. You win if you get anything *other* than a five. Fair enough?" Blue devil agrees.


Of course, the red devil keeps winning and the blue devil keeps losing. After a while it should be clear to the audience that the die is all fives. In fact, part of magi's job is to make sure that they see that.


Now a little interlude to answer your question about where the blue cards go when they're discarded. There are two decks of cards off to the side (they can just be on the table right from the beginning), one with blue cards and the other with red cards. If you win, you get to add a card of your color to your hand. If you lose, you have to discard a card onto the pile of your color.


As I mentioned, the faces of the cards aren't used.


Back to the story. After a couple of rounds the blue devil realizes that the red one is cheating. He mumbles some words under his breath, turns his finger in circles, whatever -- something to make clear that he's casting a spell. This is the point where the blue devil says, "Are you sure you counted right?" When the red devil recounts, things start going awry for him. He finds that he's mysteriously lost a red card or two, and a blue card or two has appeared in his hand. Under the rules of the game, he has to give the blue card(s) to the blue devil, who adds them to his hand.


Something similar happens, as I described, with the second recount.


At this point the red devil has exactly one red card and the blue devil has exactly one blue card, so the next round is going to determine who wins the game.


I want to set things up so that the blue devil was the last one so far to throw the die. He hands the die back to the red guy -- but actually a switch has occurred: the red devil finds himself holding the other misspotted die, the one with no fives. At this point I'd want to make a big deal of the red devil inspecting the die and noticing that the fives are gone. What can he do?? He throws, and loses.


Of course I could use a double-headed coin as you suggested (which would later be switched out in favor of a double-tailed coin), but I like the idea that the red devil is pretending to give the blue devil a good deal. Coins are fifty-fifty, whereas the die, with the arrangement that the red devil offers, *seems* to give the blue devil a tremendous advantage: with a fair die the red devil would lose five times out of six and the blue devil would win five times out of six.


As you know, Jon, I'm open to suggestions, both about the story and the sleights. Claudio, in a PM, kindly suggested a couple of ways to handle the false counts. But it's always interesting to hear alternative ideas -- it's a great way to learn. The simpler the counts the better. I really like packet tricks, and am working on Color Monte. But I haven't learned the EC or any of the other standard packet sleights. Of course I'll have to eventually! But it would be fun if I could learn a basic count or two with the goal of performing this trick.


Also, any recommendations on how to switch the dice would be welcome! :)


I tend to be overly ambitious -- learning the necessary false counts *and* a die switch (which I imagine would involve some palming and lapping -- things I want to learn, of course, but all these things take time) -- might be too much. I may have to come up with a simpler version if I want to do this trick in the nearish future.


So, there are my thoughts for what they're worth!


See you,


Bob
Message: Posted by: Bob G (Jul 28, 2018 06:30PM)
Hi again Jon,


I think I just figured out what you meant about rehearsing with face-up cards. That way I can distinguish between the different red cards and the different blue cards, right? It's a good idea.



Bob