• Consider the following $2$ player finite extensive game. Player $1$ moves first. She must choose an integer from $1$ to $10$, and place that number of matchsticks on the table. Next, Player $2$ must add to this by picking an integer from $1$ to $10$, and adding that number of matchsticks to the ones placed by Player $2$. This game goes on until the player places matchsticks such that the total number reaches $30$, hence winning the game.
  • Describe what choices Player $1$ must make to win this game, what number must she start with; then given the choice of Player $2$, what number must she pick and so forth.

For the above question, I found that playing $8$ in the first round would ensure that Player $1$ wins the game, given he plays rationally. I used the logic of backward induction, very vaguely though. In the last stage, for Player $1$ to win, he must ensure that the sum of matchsticks on the table is greater than or equal to $20$, thus ensuring that Player $1$ can choose a number between $1$ to $10$ such that sum becomes $30$ and he wins. For this, it must be the case that in the second last stage of the game, player $2$ chooses a number which can never be less than $19$, so that the only available options for him would be to make the sum greater than or equal to $20$, but not more than $29$. In the previous stage to the above Player $2$ move stage, Player $1$ must ensure that he makes the sum $19$. Thus moving backwards, I came to the conclusion that playing $8$ guarantees a win for Player $1$. If Player $1$ plays $8$ in the first round, Player $2$ can only make the sum greater than $8$ and less than $19$, such that in next stage, Player $1$ can choose such a number so as to make the sum $19$. Now Player $2$ can choose any number and lose since the sum will never be greater than $29$.

This is the only winning strategy I could find for Player $1$. I am not sure if this is correct or whether there are more winning strategies. Please help me in finding the complete winning strategy for Player $1$. Also, how do we formally write down the strategy (winning) set? $8$, $11-s_j^1$, $11-s_j^2$ is the only way I could think of writing the strategy set. Please correct me. Thanks a lot!


You've established that a player getting to 8 or 19 seals a win. You want to understand whether player 1 has any options other than starting with 8, to win.

Consider it from player 2's perspective. What is player 2's best strategy if player 1 picks a number less than 8? What is player 2's best strategy if player 1 picks 9 or 10?

The answer to both, is contained in the first sentence I wrote above.

  • $\begingroup$ Each player would want to reach either 8 or 19 to seal a win, given the other player's strategies. If player 1 does not play 8, player 2's best response would be to choose a number so as to make the sum 8 or 19. Is this correct? $\endgroup$ – S.Rana Feb 20 '19 at 20:56
  • $\begingroup$ @ShinjiniRana exactly! So if player 1 picks anything other than 8, it's a losing strategy. $\endgroup$ – 410 gone Feb 20 '19 at 21:02

The backtracking method you used is a good one. Now that you found a solution, it is nice to discover some patterns and generalizations.

We could generalize this game to reaching any goal G with any choice of integer from n to N. You can find that the winning strategy is making sure that you reach any number G-(n+N)*i, where i is an integer. These are all the winning spots. The lowest winning spot can be found by G modulus (n+N)

Mathematical games are fun, aren't they?

  • $\begingroup$ Thank you for the general solution. What method did you use to derive such a solution? Also, I can't seem to understand the step you used to find the lowest winning spot, G modulus (n+N). Could you please elaborate? And yes, mathematical games are really fun. $\endgroup$ – S.Rana Feb 21 '19 at 13:57
  • $\begingroup$ I just tried a couple of combinations, discovered a reoccurring pattern and tried to understand the pattern. About the modulus operation: that is just subtracting (n+N) from G until the remainder is less than (n+N) You can check the Wikipedia on modulus. en.wikipedia.org/wiki/Modular_arithmetic $\endgroup$ – jos Feb 25 '19 at 11:16
  • $\begingroup$ Oh okay! You seem to have missed the minus sign in the expression G - modulus(n+N). This matches the solution. Thank you! $\endgroup$ – S.Rana Feb 25 '19 at 12:25
  • $\begingroup$ No, you misunderstood me. with G modulus(n+N) I mean the remainder after doing the division G/(n+N). 30/11 = 2, remainder 8. 8 is the lowest winning spot. $\endgroup$ – jos Feb 28 '19 at 12:52
  • $\begingroup$ Sorry for the confusion, Jos. I checked the link you sent after replying to you and seem to have forgotten to delete the comment. I understood it from the Wikipedia Page. Thank You for the reply. $\endgroup$ – S.Rana Feb 28 '19 at 14:36

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