James and Louise are playing a game with a pile of 20 coins, similar to the last game , but with a little difference. Each player alternates turn by removing 1, 2 or 3 coins from the pile. Win the game who remove all the coins. James will start. If both play optimally, who have the winning strategy?
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As before, the second player can always make sure the count decreases by 4 between the starts of the first player's turns.
That is, when James takes k ∈ { 1 , 2 , 3 } from the pile, Louise can respond by taking 4 − k ∈ { 1 , 2 , 3 } from the pile, leaving 1 6 coins at the start of James' second turn. Then continuing, she will leave 1 2 , then 8 , then 4 , at the start of James' turns. At this point, James cannot take all four coins, but he has to take at least 1 , leaving at most 3 for Louise, who can then take however many are left to win.
It follows that Louise has the winning strategy.