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You have $n$ coins in front of you, and you are blindfolded. You know that exactly $10$ of the coins are showing heads, the rest are showing tails. How can you sort all the coins into two piles such that the number of coins showing heads are equal in each pile? You are allowed to turn coins, but you have of course no way of distinguishing which side a coin is showing.

Auclair
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  • A question : You must know when you created such a partition or you simply have to pass through such a partition at some point with the strategy ?? –  Dec 15 '15 at 18:08
  • I think you have to stop at some point and declare "now it's correct" – Auclair Dec 15 '15 at 18:10
  • Related: http://math.stackexchange.com/questions/659481/the-blind-man-and-coins-puzzle – JMoravitz Dec 15 '15 at 18:13
  • Is the flipping of a coin random (you can get to either tails or heads on each flip) or you just switch its side (from tails to heads and from heads to tails) ?? –  Dec 15 '15 at 18:16
  • Ah, that was a ambiguous of me. You can turn the coins upside down, not flip it in the air – Auclair Dec 15 '15 at 18:50

1 Answers1

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Split the pile into a set of ten on the left and a set of $n-10$ on the right. Flip all coins in the left pile.

Let $h_l$ and $h_r$ denote the number of coins showing heads in the left pile and right pile respectively.

We know that $h_l+h_r=10$ and so $10-h_l = h_r$.

After having flipped the left pile, all the tails there will become heads and all the heads there will become tails. I.e. there will be $10-h_l$ heads in the left pile after flipping.

The two piles will have the same number of heads.

JMoravitz
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