Very good riddle, but you're right -- it could have been told better. The question is: Without looking at the coins, can you divide them into two (not necessarily equal) piles that each contain the same number of heads?

Answer: One easy solution is to consider all 50 pennies to be in one pile, then randomly select 30 pennies for a second pile, turning all those pennies over. Since you started with exactly 30 heads, you are guaranteed to have the same number of heads showing in each pile.

Why does this work?

Consider that you randomly select N heads out of the original 30, leaving 30-N heads in the first pile. Since you selected 30 coins, N of which are heads, this means that you also have selected 30-N tails into the second pile.

Huh. Look at that. You have 30-N heads in the original pile and 30-N tails in the new pile. If only there were some way to reverse the heads and tails in the new pile...

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Guest Message by DevFuse

# cool riddle

Started by bociniki, Dec 13 2007 01:51 AM

13 replies to this topic

### #11

Posted 14 December 2007 - 07:40 PM

### #12

Posted 14 December 2007 - 10:35 PM

the same number of pennies HEADS UP in each pile not the same number of pennies total.

Ah! I think this is where it was unclear. So it's perfectly valid to have one pile of one penny and one pile of 49 pennies, correct?

The way you presented it it seemed you wanted equal numbers in the piles. I see where you are going with this now. Nice one.

### #13

Posted 15 December 2007 - 08:59 PM

That is exactly right!

### #14

Posted 18 February 2008 - 03:32 PM

This problem was actually posed to one of my friends in his interview today. The problem got leaked and he asked me to try to solve it before he went in for his interview. I found it very interesting after I could get to the bottom of it and was thinking to post it on this forum. But looking at the mess in the explanation I thought I would rather drop in the general solution!

So here it goes:

So here it goes:

Spoiler for My general Solution

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