Actually B claims that A should have 1 coin and B 7 coins because A has also eaten 5 pieces ( out of 15 pieces) of B . B has also eaten 3 pieces ( out of 9 pieces) of A. Thus A has to give coins to B for that extra 2 pieces ( 5-2) eaten. Thus from 3 coins of A , he has to give 2 .
Thus at the end A would have 1 coin and B is left with 7 coins. ( If the pieces eaten from A and B are distributed equally first from A and then from B). It all depends on probability