From Coins to Magic

Here’s a great problem or trick for a class exploration … or magic for parties.


Grab a small handful of coins (it doesn’t matter how many), randomly flip them onto a flat surface, and count the number of tails.

Randomly pull off from the group into a separate pile the number of coins equal to the number of tails you just counted.  Turn over every coin in this new pile.

Count the number of tails in each pile.

You got the same number both times!



Marilyn Vos Savant posed a similar problem:

Say that a hundred pennies are on a table. Ninety of them are heads. With your eyes closed, can you separate all the coins into two groups so that each group has the same number of tails?

Savant’s solution is to pull any random 10 coins from the 100 and make a second pile.  Turn all the coins in the new pile over, et voila!  Both piles have an equal number of tails.

While Savant’s approach is much more prescriptive than mine, both solutions work.  Every time.  WHY?


You have no idea the state (heads or tails) of any of the coins you pull into the second pile.  It’s counterintuitive that the two piles could ever contain the same number of tails.

Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails.

Enter the power (and for young people, the mystery) of algebra to generalize a problem, supporting an argument that holds for all possibilities simultaneously.


The first clue to this is the misdirection in Savant’s question.  Told that there are 90 heads, you are asked to make the number of tails equivalent.  In both versions, the number of TAILS in the original pile is the number of coins pulled into the second pile.  This isn’t a coincidence; it’s the key to the solution.

In any pile of randomly flipped coins (they needn’t be all or even part pennies), let N be the number tails.  Create your second pile by pulling a random coins from the initial pile.  Because the coins are randomly selected, you don’t know how many tails are in the new pile, so let that unknown number of coins be X .  That means 0 \le X \le N, leaving N-X tails in the first pile, and N-X heads in the new pile.  (Make sure you understand that last bit!)  That means if you flip all the coins in the second pile, those heads will become tails, and you are guaranteed exactly N-X tails in both piles.

Cool facts:

  • You can’t say with certainty how many tails will be in both piles, but you know they will be the same.
  • The total number of coins you start with is completely irrelevant.
  • While the given two versions of the problem make piles with equal numbers of heads, this “trick” can balance heads or tails.  To balance heads instead, pull from the initial coins into a second pile the number of heads.  When you flip all the coins in the second pile, both piles will now contain the same number of heads.


If you work on your showmanship, you can baffle others with this.  For my middle school daughter, I counted off the “leave alone” pile and then flipped the second pile.  I also let her flip the initial set of coins and tell me each time whether she wanted me to get equal numbers of heads or tails.  I looked away as she shuffled the coins and pulled off the requisite number of coins without looking.

She’s figured out HOW I do it, but as she is just starting algebra, she doesn’t have the abstractness yet to fully generalize the big solution.  She’ll get there.

I could see this becoming a fun data-gathering project for an algebra class.  It would be cool to see how someone approaches this with a group of students.



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