Timeline for Questions about "The best card trick"
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Oct 17, 2019 at 7:45 | history | edited | YCor |
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| Apr 16, 2019 at 2:33 | vote | accept | David G. Stork | ||
| Apr 15, 2019 at 23:10 | answer | added | Tony Huynh | timeline score: 7 | |
| Apr 15, 2019 at 23:03 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 22:58 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 22:51 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 21:52 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 21:39 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 18:01 | comment | added | David G. Stork | @MichaelJoyce: Interesting point. I know it wasn't your point, but your alternate recoding doesn't change the answer to my questions, does it? | |
| Apr 15, 2019 at 16:59 | comment | added | Michael Joyce | A comment about the trick itself: You can vary where the card identifying the suit appears as follows. After the assistant gives the mystery card to the mark, they add the ranks of the remaining 4 cards modulo 4 and use that number to determine the position of the suit card, which the magician can recover with the same computation. The order of the other three cards is determined by ignoring the suit card. This small addition to the trick makes it much harder for the mark to crack. | |
| Apr 15, 2019 at 8:11 | comment | added | Ilya Bogdanov | Clearly, the 5-tuples with exactly one working sequences are exactly those with exactly one pair of cards of the same suit, and they can be counted easily. | |
| Apr 15, 2019 at 7:13 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 6:07 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 5:37 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:53 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:39 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:37 | comment | added | Alex Kruckman | I didn't claim that every sequence of four cards was possible (this would be $52*51*50*49$), but rather that the method defines an injective function from hands of $5$ cards to sequences of $4$ cards, the image of which is the set you're trying to count. But based on your edit, I see what I was missing - based on your specification of the method, for some hands the assistant has multiple choices. | |
| Apr 15, 2019 at 4:34 | comment | added | David G. Stork | Nope. Can $\spadesuit 3\ \spadesuit 4\ \spadesuit 5\ \spadesuit 6$ ever be an exposed sequence? | |
| Apr 15, 2019 at 4:33 | comment | added | Alex Kruckman | Maybe I don't understand the question, but isn't the answer obviously ${52 \choose 5}$? | |
| Apr 15, 2019 at 4:32 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:20 | history | edited | KConrad | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:14 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:08 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 4:01 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Apr 15, 2019 at 3:48 | history | asked | David G. Stork | CC BY-SA 4.0 |