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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 25 of 51 |
(In reply to re(4): please read by Ady TZIDON)

"...and the 20th alphabetically sorted comb. is FCDE (not DFEC, BUT THAT IS  BESIDE THE ISSUE)"

20th is FCED (only FDCE, FDEC, FECD and FEDC follow it).

"I challenge  Charlie to write a program and to test it for
10 millions randomly chosen 4-tuples  in search for a failure- reward granted in case of finding a failing set."

See my previous post.  I only had to try 22 randomly chosen 4-tuples to get 7 failures. In all cases, the first available gap was used in decoding.

PLEASE read the objection.  The hoped for result is to identify all the numbers chosen by all the people, in the correct order.  There are too few possible sets of cards A might see to identify which of the possibilities was actually present.  Remember that some of the information conveyed is "wasted" on identifying the 5 possible cards for ordering, and not just on identifying the order, so it's not just 24 possibilities.


  Posted by Charlie on 2007-05-13 03:35:22
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