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In a counting game to a specific number, if the rules are

  • only two players.
  • I am the first player always.
  • one can say at least one number and at most 2 numbers.
  • the number we are counting to is bigger than 3.

for example

Player 1: 1,2

Player 2: 3,4

Player 1: 5,6

Player 2: 7

Player 1: 8

Player 2: 9

Player 1: 10

I can always win if i followed the right pattern depending on the number. Is this theory right\applicable to all numbers?

browngreen
  • 1,898
Amr
  • 103
  • It is true that there is a winning strategy for one of the players. But this need not be the first player. For example, if we were going up to six, instead of ten, then the second player has a winning strategy. – Sarvesh Ravichandran Iyer Apr 09 '17 at 07:04

1 Answers1

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You can guarantee to win by following the right pattern as long as the number you are counting to is not divisible by three. If it is divisible by three however, then player 2 can guarantee to win. Whenever you say one number, he can say two, and when you say two, he can say one. This way, he will always reach every number divisible by three, including the winning number.

browngreen
  • 1,898
  • ok, on the other hand i can push him to reach that number (other than numbers % 3==0) in the game that i guarantee wining, right? – Amr Apr 09 '17 at 07:58
  • If the winning number is not divisible by 3, then you can guarantee winning by picking a number such that after your turn, the remainder until the winning number is divisible by 3. If you do this every turn, you will reach the winning number. – browngreen Apr 09 '17 at 18:06
  • Aha thanks a lot it helped me much. – Amr Apr 09 '17 at 18:08