Question

Two players take turns at removing 1 to 4 coins from an original pile of 16 coins. Is there a winning strategy for either player?

Answer 1

The answer to the above question is given below:

Yes, there is a winning strategy for the the player who takes the first turn of the game.

Let's work backword in the game.

• If number of coins in the pile is 1 or 2 or 3 or 4, then 1st player (current player with the turn) wins.
• If number of coins in the pile is 5, then 1st player loses the game.
  • If 1st player removes 1 coin, then 4 coins will be left which will be removed by 2nd player. So, 1st player loses
  • similarly if 1st player removes 2 or 3 or 4 coins, then 3, 2, 1 coins respectively will be left which will be removed by 2nd player. So, 1st player loses.
• Similarly if number of coins is 6 or 7 or 8 or 9, then 1st player wins.
• If number of coins in the pile is 10, then 1st player loses the game.
  • If 1st player removes 1 coin, then 9 coins will be left which will be removed by 2nd player. So, 1st player loses
  • similarly if 1st player removes 2 or 3 or 4 coins, then 8, 7, 6 coins respectively will be left which will be removed by 2nd player. So, 1st player loses.
• Similarly if number of coins is 11 or 12 or 13 or 14, then 1st player wins.
• If number of coins in the pile is 15, then 1st player loses the game.
  • If 1st player removes 1 coin, then 14 coins will be left which will be removed by 2nd player. So, 1st player loses
  • similarly if 1st player removes 2 or 3 or 4 coins, then 13, 12, 11 coins respectively will be left which will be removed by 2nd player. So, 1st player loses.
• Similary if number of coins is 16 or 17 or 18 or 19, then 1st player wins.

Conclusion: We can conclude that :

• The first player loses if the number of coins in the pile is multiple of 5.
• The first player always wins if the number of coins in the pile is not multiple of 5.

Since, the number of coins in the pile is given as 16. Thus, we can say that there is a winning strategy of the first player.

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