Tuesday, January 12, 2010

truk

About the second trick - when one draws 4 different cards from a single deck, there always will be a pair among them whose numerical value differs by not more than 3.

The extreme case would be 1,5,9,12 , but the distance among 12 and 1 is less than 4. Similarly 2 different suits differ by not more than 3 (with appropriate convention).

So if we choose the smaller card of the pair whose numerical difference is less than 4 and use this card as a marker we need 9 different values to describe the mystery card. Let the marker be always the leftmost opened card.

For all 3 open cards we have 2 different possibilities arranging the remaining open cards in either decreasing or incrasing order. The cases where one or 2 cards are closed give us 6 different configurations. So we have 8 different configurations. Since we have only one deck, this is enough , because the “zero” case where the marker is the same as the hidden card will never happen.

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