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 Posted: Thu Sep 02, 2010 6:01 pm Post subject: |
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Marvo Marky
Senior Member
Joined: 08 Mar 2007
Posts: 650
Location: Newcastle, UK (30:AH)
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 Posted: Fri Jan 18, 2008 12:54 pm Post subject: Have you come across this procedure? |
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Being known at work as 'the magic guy' I get accosted with allsorts showing me 'card tricks'.
If I see that 21 card trick again or that infuriating bank robbers one I may actually give up magic.
Anyway, someone showed me something today that I had not come across before. And it was very intriguing.
I immediately saw a use for it, perhaps as a multiple force. I'll describe what happened. It sounds a little technical, though I have tried to keep the description simple. But it certainly isn't complicated when it's performed.
A shuffled deck is laid out in rows of eight until the deck is finished. A card off the top row is chosen by a spectator and this is where they start. The spectator then moves along the cards, starting at their chosen card, and they move according to the value of his/her card. So if they chose an eight, they move eight spaces along to the right. If they get to the end of the row they drop down a level and then start moving left (Snakes and Ladders style) until their move is finished. Now they note the value of the card they land on.
Now, the process is repeated. They then move along according to the value of the new card, exactly the same as before.
Picture cards and aces count as a 'one'.
The game ends when they land on a card that gives a move that cannot be finished - ie landing on a ten when there are only four moves left.
You will find that the outcome of the game is always the same card, regardless of the starting point chosen by the spectator.
Right then. The effect here and secret is a mathematical one, and pretty much guarantees that the spectator will land on the same card no matter which top card they start from. I have tried it a few times and it does seem to work, at it's worst I am told there were no more than two endings to the game.
Has anyone come across this before? Does it have a name?!?
I have had it explained to me by a mathematician and apparently the solution is slightly more complicated than I first thought.
If you have an idle moment today give this a try! I think it could be used as a multiple force, and although I know there are several better, simpler and more convincing methods to force multiples it's always nice to try something new, especially if it's not widely known.
Regards,
Mark |
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Adrian Morgan
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Joined: 06 Dec 2007
Posts: 209
Location: Adelaide, Australia (30:EN)
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| Posted: Fri Jan 18, 2008 3:18 pm Post subject: Re: Have you come across this procedure? |
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| Marvo Marky wrote: | | I have had it explained to me by a mathematician and apparently the solution is slightly more complicated than I first thought. |
The best way to understand things like this is not to scribble numbers all over the place but to find an explanation in terms of metaphors and visual images that your intuition can easily grasp.
Think of each of the eight cards on the first row as the starting points of eight different paths. Allocate to each of these eight paths a colour, and mark the starting card of each path by placing upon it a poker chip of the corresponding colour.
Now for each path, work out which is the second card in that path. Place a poker chip of the colour for that path upon that card.
Some of the cards will now have two poker chips on them, of two different colours. This means that the paths corresponding to those two colours merge together at that point. Think of it as a Y-junction. Once two paths have merged, they can never seperate again, so you only need one of the two colours to represent it from then on.
You can instantly dismiss any path that doesn't make it to the second row by its second card, because in that case it has already merged with whichever path its second card is the start of (since each card in the first row is the beginning of a path). The path beginning at the first card only makes it to the second row without merging if it's at least an eight, the path beginning at the second card only makes it to the second row without merging if its at least a seven, and so on. So you'll probably find there are very few paths which make it as far as the second row without merging with another. Forget all the paths except for the few survivors.
For the paths that make it, continue placing poker chips of the right colour on every card in that path, until eventually you'll get to a card where the only paths that still haven't merged with each other finally do so. From then on, well, there's only one path and that's why the force works.
You should be able to see intuitively that the chances of two or more paths remaining independent all the way to the end are very low. Proving it's actually impossible is not relevant to the intuitive exercise; it is sufficient to satisfy yourself that it's very, very unlikely. One thing to note is that the longer the paths go on the more opportunity there is for the paths to merge, and another thing is that the final merger is quite likely to happen where there's a bunch of ace/king/queen/jack cards all in a row. It's easy to see why - it's because a row of cards like that is a bit like a swamp. You tend to get stuck there.
Look, I'll show you.
In this example, the colours of the paths starting at each card, from left to right, are blue, red, dark purple, orange, green, black, yellow and light blue. The blue path merges with the red path at the six of spades (represented from now on by red), and the orange path merges with the purple path at the two of hearts (represented from now on by orange). Orange merges with black at the jack of diamonds, and that merges with yellow at the king of hearts (this merged path is from now on represented by yellow). At the four of spades, the red, yellow and light blue paths all merge (represented from now on by light blue). By the time we get to the second row, every colour except green has merged into the path represented by light blue.
The green and light blue paths eventually merge together at the ace of diamonds, in row four, showing that whatever path you started on, you are forced to eventually wind up there.
If you're not absolutely sure you follow this, try it yourself with cards and poker chips, and you'll probably get it. |
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bronz
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Joined: 28 Apr 2006
Posts: 1204
Location: Ashford, Kent, UK (28:AH)
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| Posted: Fri Jan 18, 2008 7:03 pm Post subject: |
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Nice one Adrian, that's a pretty clear way to explain it.
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The artist who does not rise, descends. |
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moodini
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Joined: 22 Feb 2005
Posts: 1376
Location: Canada (34-WP)
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| Posted: Fri Jan 18, 2008 9:08 pm Post subject: |
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| bronz wrote: | | Nice one Adrian, that's a pretty clear way to explain it. |
I was thinking the same thing myself....and this is silly how simple your exlanation makes this make sense! Are you one of those guys that does Mensa puzzles?
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Moodini
www.magicoftrevormoore.com |
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Adrian Morgan
Preferred Member
Joined: 06 Dec 2007
Posts: 209
Location: Adelaide, Australia (30:EN)
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| Posted: Sat Jan 19, 2008 8:44 am Post subject: |
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| moodini wrote: | | Are you one of those guys that does Mensa puzzles? |
Nah, but I've been known to read Ian Stewart books . . .
Elsewhere on the Internet, there's a discussion of the trick (with slightly different rules), and also a simulation.
Edit: The thought just occured to me that, if you were doing tricks on a theme of equipment from games but wanted more variety than just playing cards, then this effect could easily be adapted to a set of dominoes. Remove the blanks. Some work would be needed to check and adjust the probabilities involved.
In other news, a few days ago I had an idea for a presentation of the trick that Marvin Kaye describes under the title Find The Ball. I mention this here because it too is mathematical (and even Marvin says it is best presented as a puzzle). The basic change I've made seems really obvious to me (so obvious that I'm sure some other people have thought of it too), and makes life much simpler for magician and spectator alike. I'd be happy to explain my presentation idea to anyone who knows the original. |
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Marvo Marky
Senior Member
Joined: 08 Mar 2007
Posts: 650
Location: Newcastle, UK (30:AH)
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| Posted: Sat Jan 19, 2008 2:57 pm Post subject: |
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Cheers Adrian that's a thorough and a far more visual way to understand the problem. Thanks for taking the time to write it.
The heuristic argument by Tony Philips in the link you gave seems pretty spot on actually, though I suspect the proof'll probably be in a converging series of some kind. Mind you I'll be dammed if I can remember any of the tests for such things.
Going to have a play around with it this afternoon.
Tra!
EDIT: just got Adrian's EDIT. That's a good idea adrian. Yes I wonder what rule changes would have to be made - I shouldn't think it would be too difficult. |
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