The following problem has been bugging me for several days now: You have 100 coins scattered on the floor. 90 are heads, and 10 are tails. Devise a method to separate the coins into piles with the same number of tails. You cannot see the coins, only flip them.
How certain do you have to be that the two piles are exactly equal in tails ? If approximate will do, just flip half the coins into one pile, half into another.
I have this inkling that probability would, at some point, equalize the proportion of heads vs tails from random flipping. The more flipping, the more equalized. Of course, I've also heard that if you overshuffle a deck of cards, you can unshuffle it. So, probability may not quite work here.
Invert that is nearly impossable (i say nearly) the chance of that happering would be 1:52*51*50*49 ect and no there is no way in hell i am going to work out what that would be
Ashura answered the OP. (My suggestion depends on a different reading of the word "flip".) As far as unshuffling - randomization increases up to IIRC 7 perfect riffle shuffles. After that, it fluctuates - it doesn't unshuffle, but it doesn't get any more random and occasionally gets a little less in human time. Given enough time - age of universe type time - it would eventually form any pattern it can given your shuffling technique. That has no practical effect on card games.
I wonder if this riddle wants you to create 50/50 piles with heads in one, and tails in the other--- or is this a play on words (piles with the same number of tails....IE - 10 piles with 1 tail in each; 2 piles with 5 tails in each)? edit -- same thing ashura was saying
Stupid question. Ambiguous at best. Simple answer, don't flip the fuckin' things just arrange them in two piles, each with 45 heads and 5 tails. Derr! Too simple.
That's true, just because of the way humans shuffle. If you could get a *perfect* shuffle each time (interleaving the cards one by one from each pile) you would end up with the exact same deck on your second shuffle. If you devised an algorithm to shuffle the cards in a non mechanical fashion it wouldn't tend toward "unshuffled" except by random chance. Maybe the riddle is a trick. After all, every coin has a head and a tail. Thus, no matter what piles you made, they would be equal. That's a valid point.
This one's got me puzzled too LOL. I keep wanting to use the same solve method as the riddle with the 8 pennies (and one is counterfeit that's heavier than the others, and you have to weigh them with a balance scale in two tries)......but I can't seem to find a neat little trick for telling which one it's NOT like that riddle does. I also tried looking for a uniform group where flipping the same number would give the same number regardless of the permutation...... but that doesn't seem to work either Ex- 25 sets of 4 (first 5) 1111 1110 1100 1000 0000 Flip first three 0001 0000 0010 0110 1110 ........no dice here :scratchin: I'm tempted to dump out a jar of pennies, and start brute forcing LOL
As far as I can tell, this is impossible. With the counterfeit coin puzzle, the coins have a difference in the weight, and you can exploit that difference to get more information. In this puzzle, there is absolutely no difference (except maybe if you could feel which ones were heads just by touching them). With that in mind, the only possible way of doing it would be to rely on probability, especially if by "flip" it means "toss it up and expect a random result". I'm convinced this is a trick question in some way, or there is some sort of other way of interpreting it. But the problem still remains as to what the trick is.
Yessss! I just got it. Please Register or Log in to view the hidden image! The key is that it never says the piles have to be equal in size. The other key to the puzzle is that you know exactly how many heads there are, and that a head is the opposite of a tail. Here's how it works. You separate a pile of 10 coins and a pile of 90 coins. Choose any coins you want as long as the piles are 10 and 90. Then take the pile with 10 coins. Let's say, for the sake of example, that none of them are tails. That means there are 10 heads in this pile, and 10 tails in the bigger pile. Well that's simple because you can just flip over all 10 in this pile, and have 10 tails in both piles. So then let's assume there was 1 tail in this pile. That means 9 tails in the other pile. Flip all 10 coins and get 1 tail in this pile. Once you realize that the number of heads in one pile ALWAYS equals the number of tails in the other, it becomes intuitive.
Oh - that was neat. That was a really cool solution. You're right -- the 10/90 arrangement gives the same number in each no matter what the arrangement of the 10 pile looks like.
