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  1. suppose instead, take 1 from each chest (COUNT is 3) you will have 2 coins the same & the odd box would either the gold 50 (now 49) or the silver 40 (now 39). the matching, let's say "silver" coins one is the 39 the other 19 - sadly you may not be lucky enough to draw gold from the 20 er 19 chest until the 10th coin from that chest is drawn- when a gold coin is found (or not) we know each of those 2. So without a merciful reveal the total will have been 3 + 10. But maybe , if y're fortunate... the 4th draw will be the reveal. My previous consideration would have come to 11 plus 1 = 12
  2. are they simply "mystery trunks" - not revealing count? & not the Letter?! only that there are 50 G and 40 S Eleven would determine 10-10 ----> ran through all the S or all the G if it is different - then it is the "10-10" one would then need to move to a different box ------> & there you have it!! where the next draw reveals itself as either all G or all S [if G then it is A]; [if S, then it is B] We know that chest A have more coins than chest B. one chest contains 50 gold coins and next one contains 40 silver coins and the last one contains 10 gold coins and 10 silver coins. BUT WHAT IF the 11th draw was the same?? i one would then need to move to a different box ------> & either the mix or the other "all" what then. lift the box 50 or 20? i know that's lame.... gonna go think some more...
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