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Magic trick (Posted on 2007-05-11) Difficulty: 3 of 5
Two magicians A and B perform the following trick:

A leaves the room and B chooses 4 members from the audience at random. Each member chooses a card numbered from 1 to 100 (each chooses a different card) and after B has seen their cards he chooses a card from the remaining deck of cards. The 5 chosen cards are shuffled by an audience member and handed to A who just returned to the room. Prove that A is able to figure out which cards each member picked. Consider that the chosen members form a row and e.g. the leftmost member picks the first card and the rightmost member (B) picks the last card.

No Solution Yet Submitted by atheron    
Rating: 4.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(5): please read | Comment 31 of 51 |
(In reply to re(4): please read by Ady TZIDON)

Ady,

OK, suppose the 4 audience members chose 20, 40, 80 and 100 (with the same FCDE sequence).  B would pick the first (only) big gap (between 40 and 80) count up to 20 (corresponding to FCDE) and add card 60 to the set.  Now, A would still see 20, 40, 60, 80, 100.  Same as before.  He would think that B added card 40, (as you state in your last post) which is wrong.

Don't you see that any method you propose is doomed?  B can only select up 75 million+ different sets of 5 cards (combinations of 100 cards taken 5 at a time) and give them to A.  But there are 94 million+ different possibilities (permutations of 100 cards taken 4 at a time) to be communicated to A.


  Posted by Ken Haley on 2007-05-14 00:25:20
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