This is a Bofore After question.

We have 5 Head and 7 Tails.

Just repeat until all Tails.

This is interesting and I can try this “magic trick” too.

To think about it, it is only logical so. But I don’t know how to put these in Mathematical terms…

There are only 5 heads in the original group of 5 (“High-Five” group). So if 1 head left High-Five group to join the bigger group (“Lucky-Seven” group), 1 tail must come to fill the gap in High-Five group and get flipped (into 1 head) with the other 4 heads (flipped into 4 tails) in the High-Five group. Naturally as a result, there is only 1 head in Lucky-Seven and 1 head in High-Five group.

Likewise if N head(s) left High-Five group to join the bigger group (“Lucky-Seven” group), N tail(s) must come to fill the gap in High-Five group and get flipped into N head(s), with the other 5-N head(s) (flipped into 5-N tail(s)) in the High-Five group. Naturally as a result, there is only N head(s) in Lucky-Seven group and N head(s) in High-Five group.  

Let the number of coins with heads facing upwards in the smaller pile be x.

The number of coins with heads facing upwards in the larger pile will be 5-x.

The number of coins with tails facing upwards in the smaller pile will also be 5-x. Once the coins in the smaller pile are flipped, the number of coins with heads facing upwards in the smaller pile will be 5-x, which is equal to the number of coins with heads facing upwards in the larger pile.

let the number of heads in Small pile (5) be H

then, there will be

(5-H) heads in Big pile  (since there are total of 5 heads) and

(5-H) tails in Small pile (since total coins in small pile is 5)

So, when you flip the coins in the small pile, all the tails will become heads -> (5-H) heads in small pile after flipping.

cheers.

Thank you all for your reply.  Greatly appreciate it.  Will get ds to try.