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Originally Posted by milo
Hes dutch
I get the principle ie take 11 random coins into one pile and 40 random coins in another. If by chance you picked completely correctly theres 11 heads in one pile and no heads in the other, by flipping the pile with 11 heads both piles have no heads. But lets say you get 1 head in the '40 pile' and 1 tail in the '11 pile' by flipping the 11 pile you've now got one head and the others are tails.
I 'get' the solution but what would a generalised formula for the riddle be*?? the flipping over bit is confusing me, if you have N coins
N=n(T)+n(H)
Pile 1 = N-n(H)
Pile 2 = N-n(T)
Flipping over pile 2 would be 'what' in maths terms?
Nod?
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I'm not sure what youre asking.
If you just want a proof, then assume you have N coins of which H are heads. Split the coins into 2 piles by selecting H random coins (call this pile 1), and call the coins left over pile 2.
Let H1 be the number of heads in pile 1, and H2 be the number of heads in pile 2. Since H = H1 + H2, the number of heads left in pile 2 is H2 = H - H1. So we need to show there are H - H1 coins in pile 1.
pile 1 contains H coins, of which H1 are heads (by definition). Affter you flip them all over, you obviously have H - H1 heads. So both piles now have an equal number of heads.