What’s My Card?

A student is given five random cards from a standard deck of 52 cards. He must choose four cards to reveal to his professor and the order in which to reveal them; his goal is to reveal four cards in such a way that the professor knows what the fifth hidden card is. How does he do this?

Communicated by Jeff Miller.
cards

70 Comments

  1. ANDY PATEL on August 12, 2017 at 1:34 am

    can i use binary no pattern
    if yes then can i put all those 4 card face up or down



  2. Steven Miller on August 11, 2017 at 10:49 pm

    sadly cannot put in different spots on the table. imagine there are four
    rectangles and must place cards in that…. it can be done –s

    NOTE: the email address you gave me doesn’t work, email me at [email protected]



  3. ANDY PATEL on August 11, 2017 at 7:22 pm

    use binary pattern and arrage all four card to represent corespondant the 5th card
    that is from 0001 (1) to 1101 (13) face up=1 and face doun represent 0
    for color of card top edge of the table= red
    little lower=spade
    center of table=dimond
    bottom edge of table =club



  4. Steven Miller on May 31, 2015 at 1:46 am

    all face up
    no timme delay
    no placing at angles…



  5. kf on May 30, 2015 at 9:31 pm

    Do the cards all have to be face up? Can they be put down in different positions in the row of cards in a different order? I think I have something that works if you’re allowed to convey information by putting the second card into the third slot instead of the second slot, etc.



  6. Steven Miller on August 1, 2014 at 2:23 am

    Glad to hear!



  7. Steven Miller on June 4, 2014 at 1:00 am

    Glad to hear. //s



  8. kyilee palade on June 3, 2014 at 11:20 pm

    i love this website!!!!!!!!!



  9. Steven Miller on May 13, 2013 at 1:09 pm

    Email to you bounced. Please email me directly at sjm1 AT williams.edu



  10. Mark on May 13, 2013 at 1:07 pm

    Solution please?



  11. Steven Miller on December 9, 2012 at 3:04 am

    sure — sjm1 AT williams.edu



  12. Patrick Boyhan on December 8, 2012 at 6:44 pm

    Would you please send solution to this puzzle? Thank You, Patrick



  13. Steven Miller on October 18, 2012 at 1:03 am

    email me: sjm1 AT williams.edu



  14. sam on October 17, 2012 at 10:23 am

    I’m thinking that a card would be 1 to 13 (J=11, Q=12, K=13). Having four binary places can cover up to 15. But how can I make a card represent 0 or 1? Am I on the right track?



  15. Steven Miller on September 7, 2012 at 12:58 am

    ok



  16. Gayle.L on September 6, 2012 at 4:46 pm

    solution please 🙁



  17. Steven Miller on September 4, 2012 at 1:44 am

    sergio: email me at sjm1 AT williams.edu to discuss



  18. Steven Miller on July 27, 2012 at 4:43 am

    not sure what 1 means — you can email me at sjm1 AT williams.edu



  19. gerardo on July 26, 2012 at 1:16 pm

    1



  20. Steven Miller on July 25, 2012 at 1:59 am

    I wanted to use 4! + 4 + 4! = 52 but couldn’t get that to work — emailed you a hint (sjm1 AT williams.edu)



  21. Steven Miller on July 25, 2012 at 1:58 am

    yes, every situation (sjm1 AT williams.edu)



  22. Stephanie on July 24, 2012 at 7:38 pm

    Assuming the hidden card is the/a repeated suit, is it possible to do the “trick” with either card hidden, or might there be a way to do it with just one of the cards hidden? In other words, will the trick work for EVERY situation?



  23. Noah on July 24, 2012 at 3:11 pm

    So if we number all cards from 1-52, and if we always are dealt at least one card from 1-24, we can set that card as the fifth and use the other four cards to represent its number (4! arrangements of their numbers). But can you give me a hint as to how you would deal with getting all five cards >24? Or am I going about this wrong? I know this method doesn’t use all the possible information, but I’m not coming up with another way after about 30 minutes thinking about it



  24. Steven Miller on July 12, 2012 at 12:42 am

    no — you can do it wi’out orientation



  25. hanan on July 10, 2012 at 12:26 am

    using only the order of the four cards gives half the required information.
    can I use the orientation of the fifth card (e.g holding it horizontally or vertically) to convey the missing “bit” of information?



  26. Steven Miller on June 25, 2012 at 4:34 am

    Mike L: you’re on the right path. The problem is that you can’t control how the cards are put on the table, only their order. your friend walks in after the cards are all laid down, face up, same orientation. you only get to choose the order. email me at sjm1 AT williams.edu to chat more //s



  27. Steven Miller on June 24, 2012 at 3:21 am

    Mike L: not posting as this gives away a lot. Email me at sjm1 AT williams.edu — you’re very close, but you are using more info than you are allowed. The person coming in doesn’t know the order in which the cards were placed, and the four cards must go in four specific slots with specific orientations, so no way to use that to encode info.



  28. Steven Miller on June 13, 2012 at 1:06 am

    you can do it without doing a trick like this. in other words, you tell ME what order your want the card to be placed and I’ll place them, so no signals through delays, spacing….



  29. Irving on June 12, 2012 at 2:34 pm

    You lay them out with the correct spacing in between… if you have a 2,5,7,8,J…. You leave 2 card slots between 2 and 5, 2 slots between 2 and 8, and 2 between 8 and J. Then you point to the slot before the 8 and they figure it out.



  30. Steven Miller on June 1, 2012 at 7:16 pm

    I’ll send a hint first //s



  31. Bryan on June 1, 2012 at 2:55 pm

    Can you please send me the solution to this 5 card trick? I have tried about everything I can think of.



  32. Steven Miller on May 31, 2012 at 1:16 am

    possible! email sjm1 AT williams.edu for a hint ..s



  33. Anonymous on May 30, 2012 at 12:47 pm

    not possible



  34. Anonymous on May 29, 2012 at 2:30 am

    is he allowed to discuss a strategy before hand?



  35. Steven Miller on May 16, 2012 at 7:49 pm

    sure /s



  36. Anonymous on May 16, 2012 at 4:27 pm

    Can I have the answer please?



  37. Steven Miller on April 17, 2012 at 2:09 pm

    if you need to do that, sure — however, it’s possible to do it by saying there are 4 boxes: [] [] [] []. you first put a card in box 1, then box 2, ….



  38. Steve on April 17, 2012 at 4:46 am

    Do you get to pick the order in which you lay them down? In other words, can I lay card number 1, then card 2 to its right, then card 3 to the left of those two, etc.? Or do I have to convey the final card simply with the order of the first four?



  39. Steven Miller on April 12, 2012 at 6:19 am

    absolutely!



  40. anonymous on April 12, 2012 at 6:03 am

    can you work out different things with the teacher before you do it



  41. Steven Miller on April 1, 2012 at 5:04 am

    Email me at sjm1 AT williams.edu for a hint. //s



  42. Mario Loepz on April 1, 2012 at 4:04 am

    Confused



  43. Steven Miller on March 31, 2012 at 6:18 am

    I don’t see how this allows the other person to get the hand as he can’t say what the hand is. /s



  44. Kelly on March 31, 2012 at 3:16 am

    Could he organize them by the poker hand he holds and reveal what his hand is, eg “full house” while also organizing by increasing or decreasing numbers? If I’m wrong, please email me the solution. Thanks.



  45. Steven Miller on March 18, 2012 at 4:30 am

    sure, sent ..s



  46. Will K on March 17, 2012 at 9:21 pm

    i’d love the solution



  47. Steven Miller on January 13, 2012 at 4:24 pm

    You’re on the right track. The problem is you have to put down the other four cleanly. No rotation, all facing the same way. You control the order of the four cards, but that’s it.



  48. Andrew on January 13, 2012 at 10:44 am

    You find two cards that share the same suit, since there is 5 cards atleast 2 will have to have the same suit. You will use one of these cards as the card you want the professor to guess for.. You take the other 4 cards, put them face down on a table and spell out the number that is the card the professor is trying to guess. 2 for 2, 9 for 9, K for king, etc. Leave the card with the same suit face up while the other 3 stay down and he should be able to guess the card. Am I right?



  49. Steven Miller on December 30, 2011 at 9:31 pm

    Let me try explaining the problem again. You are given 5 cards, say 2hearts, 5spades, 7diamonds, 8diamonds, Qclubs. You get to choose one of the five to keep hidden, and then place the other four down in such a way that your partner can glance down and deduce what the missing card is. You can’t ‘clue’ him in by things such as the angle the card makes with the table…..



  50. MooMoo on December 30, 2011 at 4:43 pm

    2, 4, 6, 8? the next card would be ten…..so he would go by every 2…..unless i dont get the question…



  51. Steven Miller on December 13, 2011 at 9:15 pm

    but then he’s always showing the same thing…. email me at [email protected] to chat more.



  52. Anonymous on December 13, 2011 at 9:11 pm

    he should show them from lowest to highest, in order to show some kind of pattern



  53. Steven Miller on November 28, 2011 at 2:26 am

    You’re on a possibly right track, but the question is can you make it rigorous and work? There is a very simple way to do it where you ALWAYS put the cards down in a line, and thus don’t use 2-dimensions….



  54. Ben on November 28, 2011 at 12:02 am

    Isn’t that the soln? I mean, you are saying that you can arrange with your professor positions on the table to represent operators (next to means add, on top of means multiply, etc). So he arranges these four operators and a way to show suit (which also shouldn’t be that hard) and also a way to say “this card isn’t used” (just put them off to the side) and the question becomes “given four random numbers between 1 and 14, is it possible to arrange some of them and put operators between them so that it totals a different random number between 1 and 14?” That, I believe, is yes. I have not given exact arrangements because that is for the student and the professor to decide what is easiest for them.



  55. Steven Miller on November 27, 2011 at 4:46 am

    email me at [email protected] for hints / solns



  56. D on November 27, 2011 at 3:50 am

    Nooooo its toooooooo hard



  57. Steven Miller on November 11, 2011 at 10:50 pm

    yes



  58. Soroush on November 11, 2011 at 7:20 pm

    Can the student see the five cards?



  59. Steven Miller on November 10, 2011 at 8:27 pm

    sure ..s



  60. varun on November 10, 2011 at 1:17 pm

    sir please send me the solution



  61. Steven Miller on November 6, 2011 at 3:23 pm

    it’s possible — email me if you want a hint ([email protected])



  62. jess on November 6, 2011 at 6:46 am

    Impossible



  63. Steven Miller on August 26, 2011 at 2:03 am

    Sure: the way you place the cards on the table can be used to represent numbers. You can also assign each card a number, and thus just need to ‘add’ or ‘subtract’ from a certain card. If you want another hint, let me know. /s



  64. K on August 25, 2011 at 4:11 am

    Could you give a hint?
    And from Shane’s question, can I assume from the way you answered him that the student and professor did work out a strategy beforehand?
    Thanks.



  65. Steven Miller on July 28, 2011 at 1:57 pm

    I’m at a liberal arts college. We’re all about encouraging student / faculty interaction. We actually meet for dinner and a Red Sox game and discuss the details in Fenway Park.



  66. Shane on July 28, 2011 at 1:55 pm

    Is it assumed that the professor and student can not work out a strategy in advance? Or can they talk first?



  67. Steven Miller on July 7, 2011 at 5:59 pm

    Did you get my email?



  68. Steven Miller on July 1, 2011 at 1:02 pm

    Not sure what ‘how come’ means. Please email me at sjm1 AT williams.edu.



  69. Anonymous on July 1, 2011 at 11:10 am

    how come??



  70. Steven Miller on June 30, 2011 at 1:37 am

    Done. If you didn’t get it let me know. //s



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