美国大学的原题参考：
On a table there is a row of fifty coins, of various denominations (the
denominations could be of any values). Alice picks a coin from one of the
ends and puts it in her pocket, then Bob chooses a coin from one of the ends
and puts it in his pocket, and the alternation continues until Bob pockets
the last coin. Prove that Alice can play so that she
guarantees at least as much money as Bob.
-Answer:
Alice adds the values of the coins in odd positions 1st, 3rd, 5th, etc.,
getting a sum Sodd. Then she does the same with the coins placed in even
positions 2nd, 4th, 6th,etc., and gets a sum Seven. Assume that
Sodd ≥ Seven. Then she will pick all the coins in odd positions, forcing
Bob to pick only coins in the even positions. To do so she stars by picking
the coin in position 1, so Bob can pick only the coins in position 2 or 50.
If he picks the coin in position 2, Alice will the pick coin in position 3,
if he picks the coin in position 50 she picks the coin in position 49, and
so on, with Alice always picking the coin at the same side as the coin
picked by Bob. If Sodd ≤ Seven, then Alice will use a similar strategy
ensuring that she will end up picking all the coins in the even positions,
and forcing Bob to pick the coins in the odd positions—this time she will
pick first the 50th coin, and then at each step she will pick a coin at the
same side as the coin picked by Bob
