Dathi has the right answer, and for the right reason. Next question, please Dathi!
As Dathi said, the answer is that you split the one pile into two piles, one containing the same number of coins as you've been told there are heads up coins, and the other pile containing the rest. But every coin put into the first pile must be turned over.
In the special case of being told that none of them are heads up, just split the pile into two any way you like, without turning any of the coins over.
In the special case of being told that all of them are heads up, turn all the coins over and then split into two piles any way you like.
Otherwise it works just like Dathi says. I'm don't think I could explain it better.
Quote:
Originally Posted by Daithi
Finally, here is the reason why it works.
Let,
H=total # of heads up coins
x=number of unknown heads up coins we grabbed
So,
heads up coins in stack A = x
heads up coins in stack B = H-x
After flipping stack A we have
heads up coins in stack A = H-x
heads up coins in stack B = H-x
I don't know how good this explanation is, but it does make sense to me. When you flip stack A you get just the opposite of the number of heads ups coins than you grabbed. Also, since stack A contained H number of coins, it can't contain too many or too few coins (this is why flipping stack B won't work -- it doesn't have the right number of coins).
Maybe pdurrant can do a better job of explaining why it works than I have done.
One example with the mathematical explanation.
Suppose you have 100 coins and 15 are heads up. You grab 15 coins at random to form stack A. There are x number of heads up coins in this stack. Let's say that you had grabbed 3 of the heads up coins. Therefore, stack A contains 3 heads up coins, and stack B contains 12 heads up coins (15-3=12). When you flip stack A, which contains 15 coins, then your 3 coins that were heads become tails, and the 12 tails become heads. So, stack A goes from x heads up coins to H-x (15-3=12).
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