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slowkneenuh Regular user After 5,278+ posts, only credited with 133 Posts |
On occasions when I do impromptu magic, one of the effects I use is Predict or Pick A Number based upon the participant looking at six different number arrays and telling me whether or not their thought of number is there, followed by my reveal. Lately I have been using it to guess someone's age, typically as one of my walk-around effects.
It is simple, direct (no calculations), and visual. However, a restriction I have reached is it's limit to a number in the sixties. I do a lot of volunteer magic with seniors and veterans where ages go well up to 100. Is anyone aware of a method or existing effect based on the same simple principal that would allow me to increase my prediction up to 100 and use the same approach? Thanks, John
John
"A poor workman always blames his tools" |
KarstenMeyerhoff New user Germany 92 Posts |
As these cards/number arrays use the binary representation of numbers, using an additional card should do the trick: It should get you up to 128 - which probably is sufficient.
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S2000magician Inner circle Yorba Linda, CA 3465 Posts |
Quote:
On 2012-11-10 16:55, KarstenMeyerhoff wrote: 127, actually, but you're correct: it should be sufficient. |
slowkneenuh Regular user After 5,278+ posts, only credited with 133 Posts |
Thanks Folks,
If I understand the solution along with the tradeoff, it now means I am going from 6 arrays(cards) with 32 numbers each to 7 arrays(cards) with 64 numbers each. Is that correct? If so, it becomes more cumbersome but still workable. I'll have to tell the folks the order of the numbers to help simplify/speed up their examination of each card for their chosen number. Thanks for your input! John
John
"A poor workman always blames his tools" |
S2000magician Inner circle Yorba Linda, CA 3465 Posts |
Quote:
On 2012-11-10 19:14, slowkneenuh wrote: Yes, that's correct. |
slowkneenuh Regular user After 5,278+ posts, only credited with 133 Posts |
Thanks for the confirmation S2000magician! I have the cards almost done. They will get put to good use on Veterans Day.
John
John
"A poor workman always blames his tools" |
S2000magician Inner circle Yorba Linda, CA 3465 Posts |
My pleasure.
Please let us know how it plays. |
Scott Cram Inner circle 2678 Posts |
You'll need to re-number all the cards to include numbers up to 100.
To simplify things, I've set up a little Wolfram|Alpha calculation to aid you (click the link immediately below to see it in action): Table[(x*BitAnd[x, 2^(y-1)])/(2^(y-1)), ......here y=1 In the above formula, all you have to do is change the variable y at the end. y is simply the number of the card you need, ranging from 1 to 7. Set that number, and hit return, and you'll see a list of numbers to include on that card. For example, setting y=1 in the above formula returns: {1, 0, 3, 0, 5, 0, 7, 0, 9, 0, 11, 0, 13, 0, 15, 0, 17, 0, 19, 0, 21, 0, 23, 0, 25, 0, 27, 0, 29, 0, 31, 0, 33, 0, 35, 0, 37, 0, 39, 0, 41, 0, 43, 0, 45, 0, 47, 0, 49, 0, 51, 0, 53, 0, 55, 0, 57, 0, 59, 0, 61, 0, 63, 0, 65, 0, 67, 0, 69, 0, 71, 0, 73, 0, 75, 0, 77, 0, 79, 0, 81, 0, 83, 0, 85, 0, 87, 0, 89, 0, 91, 0, 93, 0, 95, 0, 97, 0, 99, 0} If you remove the 0s in that list, you get: {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99} ...and those are the numbers that belong on card 1. Change y=2 in the above formula, and it will give you this: {0, 2, 3, 0, 0, 6, 7, 0, 0, 10, 11, 0, 0, 14, 15, 0, 0, 18, 19, 0, 0, 22, 23, 0, 0, 26, 27, 0, 0, 30, 31, 0, 0, 34, 35, 0, 0, 38, 39, 0, 0, 42, 43, 0, 0, 46, 47, 0, 0, 50, 51, 0, 0, 54, 55, 0, 0, 58, 59, 0, 0, 62, 63, 0, 0, 66, 67, 0, 0, 70, 71, 0, 0, 74, 75, 0, 0, 78, 79, 0, 0, 82, 83, 0, 0, 86, 87, 0, 0, 90, 91, 0, 0, 94, 95, 0, 0, 98, 99, 0} Remove the 0s, and you get: {2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99} ...so those are the numbers that belong on card 2. Simply continue this process up to card 7 (which is just the numbers 64-100), and you're set to go! Just recently, I did a blog post on how to play around with var......ge Cards, including a nice game show-themed variation by Brian O'Neill. This might spark some other ideas for you. |
slowkneenuh Regular user After 5,278+ posts, only credited with 133 Posts |
Scott,
Once again, thank you for your support and never ending supply of thought provoking ideas! Wolfram|Alpha and your blog was great info. Best Regards, John
John
"A poor workman always blames his tools" |
WilburrUK Veteran user 389 Posts |
Just to put my oar in, Wouldn't it be possible to use your existing (32 numbers each) cards. but to add a constant (say 30), to each number, assuming you're not going to be doing this for anyone below 30? Then you just have to add 30 to the answer and Bob's your uncle.
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S2000magician Inner circle Yorba Linda, CA 3465 Posts |
Quote:
On 2012-11-14 09:02, WilburrUK wrote: Alas, it's more complicated than that, because all of your index numbers are increased by 30. So, you'd end up subtracting 150, not adding 30. Or, more likely (in practice), subtracting 30 from each index, adding the indices, then adding 30. Not insurmountable, but more complicated than simply adding 30. |
WilburrUK Veteran user 389 Posts |
Quote:
On 2012-11-14 13:04, S2000magician wrote: In that case, I have misunderstood the mechanics of the effect. Ignore everything I have ever uttered on the subject ! |
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