Grabbing coins from the ends of a row

Question

Six coins lie in a row with values (left to right): 3, 9, 1, 2, 7, 5. Players alternate; on your turn you must take either the leftmost or the rightmost remaining coin and keep it. When all coins are gone, each player adds up the values they collected, and the larger total wins.

Can the first player guarantee winning, and by how much?

Accepted Answer

Yes. The first player can guarantee collecting 16 out of the 27 total, leaving the opponent just 11. The trick: color the positions. Number the positions 1–6. Color the odd positions (1, 3, 5) one color and the even positions (2, 4, 6) another. Their sums are: - Odd positions: 3 + 1 + 7 = 11. - Even positions: 9 + 2 + 5 = 16. Here is the crucial fact: the two ends of the row always have opposite colors. With six coins, one end is odd-position and the other is even-position. So whichever color the first player decides to chase, they can always take a coin of that color, and, here is the key, t…

Grabbing coins from the ends of a row

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