Blind Man and Cards


A blind man is handed a deck of 52 cards and told that exactly 10 of these cards are facing up. How can he divide the cards into two piles, not necessarily of equal size, with each pile having the same number of cards facing up?


Heard from a Googler in 2006.


If the original pile has c cards with f cards facing up, then the blind man divides them into piles of size f and c - f. Then he flips all cards in the pile with f cards. Let's see why it works for c = 52 and f = 10. The blind man would divide the cards into two piles with 10 and 42 cards each. If there are k face-up cards in the 10-card pile, then there must be 10 - k face-up cards in the 42-card pile (because the total number of face-up cards is 10). So by flipping all cards in the 10-card pile, the number of face-up cards in both piles would become equal to 10 - k.

Previous Puzzle: Non-Transitive Dice

You and your opponent shall play a game with three dice: First, your opponent chooses one of the three dice. Next, you choose one of the remaining two dice. The player who throws the higher number with their chosen dice wins. Now, each dice has three distinct numbers between 1 and 9, with pairs of opposite faces being identical. Design the three dice such that you always win! In other words, no matter which dice your opponent chooses, one of the two remaining dice throws a number larger than your opponent, on average.

How many steps are required to break an m x n bar of chocolate into 1 x 1 pieces? We may break an existing piece of chocolate horizontally or vertically. Stacking of two or more pieces is not allowed.

12 Sep 2008
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