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So My math professor did a card trick with me where he gave me a set of normal 52 cards and let me shuffle them however I like, he even said that i can arrange them however I would like. Next, he asked me to think of any random number from 1-10 inclusive, and to keep it to myself. lets say i chose 7. He then asks me to slowly drop 7 cards, going one after the other, when i reach the 7th card, i observe its value and continue adding cards over the pile with its value. so if the 7th card was 6, I would add 6 more then observe the 6th card and so on. You repeat this process smoothly and continuously so he doesn't know where i am with the counting. Also, jack counts as 4 (j,a,c,k), queen as 5, king as 4, and the joker as 5.

The trick: I kept doing this and then midway while i was counting in my head he started counting out loud the numbers i was in! so lets say i got 7, then 6, then jack, then 9, he started counting with me 1,2,3!

I've tried thinking about this but it just seems crazy! he reassured me that it involves math only and no cheap tricks, but the way i look at it it seems so random, i could've arranged the cards in any way and chosen any number from 1-10!

I don't want full explanations on this, i just want a hint to start thinking about this, I think that the fact there is an extra 8 4s and 6 5s has to do with something but i am not sure how??

AspiringMat
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  • I do not understand the passage "the trick : ..." You count in your mind , and the prof counts loud ? Explain this more exactly using an example. Also, what was the surprise of the trick ? – Peter Jul 09 '16 at 09:40
  • Lets say i chose 7 in the beginning. i start dropping cards...: 1,2,3,4,5,6,7 now i look at the 7th card and it is 6, so i count: 1,2,3,4,5,6 now i look at the 6th card and its jack then i go j,a,c,k i look at the last card and its 3 so i go 1,2,3 and then midway through dropping cards, he starts counting out loud and its exactly the same number I am counting in my head!! – AspiringMat Jul 09 '16 at 09:43
  • So, he finds out the very last number you count out, right ? – Peter Jul 09 '16 at 09:45
  • @AspiringMat Please explain more clearly. I still don't understand what you are trying to say – Roby5 Jul 09 '16 at 09:45
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    Kruskal count.. – Gerry Myerson Jul 09 '16 at 09:46
  • Not the very last number, he finds out a number in the middle. the numbers i am counting in my mind is the cards, so in my example, in the beginning i am thinking in my mind: one, two, three,...... until seven. Then i see the value on the 7th card (6) and start counting: one two, three,.... six. then i see the value on the 6th card (jack) and i count J, A, C, K then i see the 4th card and so on, then suddenly he starts counting out loud exactly in time with me, so he would say J, A right when i am saying J A in my mind – AspiringMat Jul 09 '16 at 09:49
  • Its surprising because lets say i chose 6 instead in the beginning, the 6th card would be different, so i would follow a totally different chain than what i began with, also if i ordered the cards in any way, the chain would always be different – AspiringMat Jul 09 '16 at 09:50

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The magician can determine sets $S\subseteq\{1,2,\ldots,52\}$ with the property that he knows that one of your stopping points must be $\in S$. Initially, he knows that $\{1,2,\ldots,10\}$ is such a set. Let $S=\{x_1,\ldots, x_k\}$ with $x_1<x_2<\ldots <x_k$. Once you display the card at position $x_1$, the magician knows its value $w$ and thus can replace $S$ with $(S\setminus\{x_1\})\cup\{x_1+w\}$. This new set may be as large as the fromer one, but it may also happen to be smaller by one element. For example, the initial $\{1,\ldots,10\}$ may turn into $\{2,\ldots,11\}$ only if the first card is a 10; in all other cases we immediately improve to $\{2,\ldots,10\}$. As soon as the set reaches cardinality $1$, he knows one of the cards you landed at and from then on can count synchronously with you.

Is it guaranteed that this process will in deed produce a singleton before the deck is used up? Not necessarily if the deck is carefully arranged: Assume we have the following cards at certain positions: A 9 at position 9, a 10 at position 10, a 10 at position 18, a 9 at position 20, a 9 at position 28, a 10 at position 29, a 9 at position 37, a 10 at position 39, an 8 at position 45, a 7 at position 46. With such a deck, the counting sequences starting with 9 and 10 will remain different until the end.

Hagen von Eitzen
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