plainglazed

answer

yes, but...

for a set of five...

another seven and I think an example of six? still working on five...

yeah, think I agree with that statement, dealer.

better than 50%?

I think I see the pattern here and am encouraged by the beauty of binary symmetry so here goes:

@aiemdao - Yes, nice solve indeed Aiem. Well done. @bonanova - thanks.

hey bn - Thanks for getting this one started. You are correct in all you say above including your initial summary of the problem. Have edited the OP to include that explanation. I assure you there is a scheme in which one can guarantee finding seven heavy coins. Perhaps start off by finding one guaranteed heavy coin in two uses of a balance scale. Or not. I have no doubt you or others here will discover the method for finding seven heavy dimes.

You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification

okay, not too proud to start this off with some probably way simplistic maths:

Not Rats

had thought perhaps then perhaps rats, still thinking

ooh, nice one Cyget. Spoiler for some additional thoughts:

1. EDIT: yeah, as I continue to wake up here, am less convinced of my logic above.

Not a generalized formula but better than strategy for three?

dang, like that answer MM - nice

Silently be in doubt and subtle...

legalmente?

perhaps or