Consider a twoplayer game Red Dragon There are 17 green drag

Consider a two-player game \"Red Dragon\". There are 17 green dragons and 1 red dragon. All (green or red) dragons are piled together initially. The two players take turns. When her/his turn comes, she/he will take either 1. 2. 3 or 4 dragons. She/he will lose the game if she/he picked the red dragon. Now find the best strategy for the first-mover to win

Solution

For 1 green dragon and 1 red dragon

Player 1 will pick the 1 green dragon, hence the player 2 must take the red dragon, since he has no choice left with him, so player 1 will be the winner

The same case will hold for 2 green dragon and 1 red dragon, 3 green dragon and 1 red dragon and 4 green dragon and 1 red dragon

If there are 5 green dragon, then player 2 will win, what ever number he chooses either 1,2,3 or 4, the player 2 will choose (5-x) number of green dragons, where x is the number of green dragons choosen by the first player

Now for 6 green dragons, if player 1 chooses 1 dragon in the first attempt, then in the second chance player 2 will be going to loose irrespective of number of dragon choosen

For 7,8,9 dragons he will pick 2,3 and 4 in the initial turn

For 10 dragons player 2 will be winner

Hence we deduce that when number of green dragons is divisible by 5, then only player 2 is winning

Hence he must start initially with 1 green dragon because 17 mod(4) = 1