I need a little help proving the following question:
A row of fifty coins with integer denominations is given. The sum of all the denominations is odd. Player 1 and player 2 take turns picking a coin from either the left side of the row or the right side. Prove that player 1 will always get more than half the money.
What I have done so far:
I have determined the strategy that player 1 can use, but I don't know how to prove it. This is the strategy: Let's call the left most coin "l". And lets call the right most coin "r". We can also call the second most left coin "l1" and the second most right coin "r1". Player one must subtract the value of l - l1 and r - r1. whichever answer is greater, player 1 would take the "l" or "r" coin. For example, if we had the coins: 1,2,3,4,5,6, we would do "1 - 2" and "6 - 5". Since "6 - 5" is greater, the coin with the value of 6 would be taken. If l - l1 = r - r1, then player 1 would take whichever coin had a greater value.
Thanks in advance
Number the coins from 1 to 50. If the even-numbered coins add up to more than the odd-numbered, then take only even-numbered coins. If the odd-numbered coins add up to more than the even-numbered coins, then take only odd-numbered coins.
This problem is in one of Peter Winkler's excellent books.