$2016$ coins are placed on a table with $50$ coins heads up and the remaining coins tails up.

Suppose you are blindfolded, and the only thing you can do is flip some of the coins. Explain how you can separate the $2016$ coins into two groups such that each group has equal number of head.

This is the question one of my friends give me. I don't know if it is true or false. I can't prove it either way.

  • $\begingroup$ See here: quora.com/… $\endgroup$ – Rohan Dec 30 '16 at 6:14
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    $\begingroup$ Run your finger over each coin to determine which side is up, and if it's heads up, flip it. Once all the heads are down, divide the coins randomly into two groups. $\endgroup$ – bof Dec 30 '16 at 7:23
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    $\begingroup$ By the way, next time you tell this problem, make it 1000 coins instead of 2016, so it doesn't sound like a boring contest problem. $\endgroup$ – bof Dec 30 '16 at 7:24
  • $\begingroup$ Are we able to identify with finger the coin side/figure? $\endgroup$ – Sigur Dec 30 '16 at 13:02
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    $\begingroup$ This should be migrated to Puzzling.SE $\endgroup$ – ABcDexter Dec 30 '16 at 13:44

Divide the coins into two groups of 50 and 1966 coins, then flip all the coins in the first group.

After division, the first group will have $50-x$ heads and the second one $x$ heads.

After flipping, both groups will have $x$ heads.


Start with all the coins in a left pile and no coins in the right pile. Clearly there are fifty more heads in the left pile than in the right pile.

Now take (blindfolded) one coin from the left pile, flip it and add the result to the right pile.

If the coin picked at random from the left pile was a head, then you have decreased the number of heads in the left pile by one, and you have kept the number of heads in the right pile unchanged. If the coin picked from the left pile was a tail, you have kept the number of heads in the left pile unchanged while increasing the number of heads in the right pile by one. Either way you narrowed the difference in the number of heads between the two piles by exactly one.

Since we started with a disparity of fifty heads between the two piles, we need only to repeat this procedure fifty times to guarantee both piles have (sight unseen) an equal number of heads.


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