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| ||bass Posts: 12/594 |
Ok. That was better then last time.
drizzt do'urden's answer is WRONG for making assumptions about things I didn't mention in the question. Cellar Dweller's answer is CORRECT. The proper strategy is to, on your turn, always return the coin count to a multiple of 3 by drawing however many coins are required to do so. |
| Cellar Dweller + Posts: 13/138 |
For each turn, draw however many coins is needed to force the number of coins in the pile to a multiple of three (zero inclusive, of course).
I solved this by making a list of numbers and assigning an advantage to a player. 1, 2, and 4, were assigned to the current player, and 3 was assigned to the next player. By manualy going through the possible moves, I assigned advantages up to about 6, when I saw that the advantage would always go the next player if there was a multiple of 3 coins in the pile. Shortly after that, I realized that the choices of numbers of coins to remove made it possible for a player to force the number of coins on the pile to a multiple of 3 as long as there is not a multiple of 3 coins in the pile. |
| drizzt do'urden Posts: 10/57 |
P1: 2 coins
P2: 4 coins P1: 2 coins P2: 4 coins P1: 2 coins P2: 4 coins P1: 2 coins P2: 4 coins P1: 2 coins P2: 4 coins P1: 2 coins P2: 4 coins P1: 2 coins P2: 1 coin P1: 1 coin Player 1 takes out 2 coins every turn while player 2 takes out 4 coins until there is 2 coins left in which they will both take just 1 coin out each. Player 1 will take the final coin. |
| ||bass Posts: 11/594 |
Ok. Since you all did so poorly on the last question. I've decided to have a much easier question this time.
The Coin Game: The rules are as follows The game has 2 players, alternating turns. You begin with a pile of 40 coins. On each turn, a player may remove 1,2 or 4 coins. The player to remove the LAST coin wins. Find a strategy that is guarinteed to win for the FIRST player. |
