In a 2-3-4 coin game where coins are arranged with 2 coins in the first row, 3 in the second row and 4 in the last row, 2 players are allowed to pick as many coins as they want at each turn and who ever gets the last coin wins. Determine whether the player that goes first can always win. If that is the case, describe the winning strategy and give convincing argument that this strategy always work despite what the opponent does. If there is no overall winning strategy for the player that goes first, explain why. Statements can be phrased as "If my opponent does......., then i will do......". This is a logic game.
If I understand the game correctly, there are additional rules. At each turn, a player must choose at least one coin. Also, a player can only take coins from 1 row at a turn.
Let the coins be:
The first player can always win.
The first move is to take three C's.
(2a) Player 2 takes 1 A. Then player 1 takes 3 Bs leaving an A and C so player 1 wins
(2b) Player 2 takes 2 As. Then player B takes 2 Bs leaving a B and a C. PLayer 1 wins.
(2c) Player 2 takes 1 B. Player 1 takes a C, leaving 2 As and 2 Bs. Player 2 cannot take 2As or 2Bs as player 1 will win, so she takes 1A (or 1B). Player 1 counters by taking 1B (or 1A) leaving an A and a B so player 1 wins.
(2d) Player 2 takes 2 Bs. Player 1 takes 2 As leaving a B and a C so player 1 wins.
(2e) Player 2 takes 3 Bs. Player 1 takes an A leaving an A and a B so player 1 wins.
(2f) Player 2 takes the last C. Player 1 takes a B leaving 2 As and 2 Bs. This is the same situation as in (2c) so player 1 wins.
There are certain combinations that are certain winners or losers. For example, 1,1 is always a loser as is 2,2. Also 1,2,3 is a certain loser as demonstrated.