Splitting coins in the dark
There are 100 coins on a table, exactly 10 of them heads up. You are blindfolded and cannot feel which side is up. You must split the coins into two piles, flipping any coins you like, so that both piles end up with the same number of heads.
The standard method is to move some number of coins into a second pile and then flip every coin in that pile. Which value of always works?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Suppose the second pile has heads before flipping. How many heads are left in the first pile?
After flipping, the second pile has as many heads as it had tails before.
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Take as many coins as there are heads in total. Whatever the second pile contains, the heads it took from the first pile are exactly matched by its tails, which become heads when flipped.
Step 1. Move any 10 coins into a second pile. If of them are heads, the first pile keeps heads.
Step 2. The second pile has tails. Flip all of its coins: it now has heads.
Step 3. Both piles have heads, whatever is. So .
Why only 10. With coins in the second pile, after flipping it has heads against in the first pile; these are equal for every only when .