# The Box of Tic-Tacs

You were kidnapped and had no way of escape. The kidnapper opens the lights, and in front of you is an opened box of Tic-Tacs. He claims that he ate from 0-5 Tic-Tacs, (with 60 from the newly-bought box). Now, here is the game.

The game is quite simple. You and the kidnapper may take 1, 3, or 4 Tic-Tacs out of the box. Then, at every even-numbered round, you, and the kidnapper, must return a certain number of Tic-Tacs that you decide (you must return at least 1). The person who gets the last Tic-Tac wins. This means that if there is one in the box and it is your turn, you win. BUT if there are two, for instance, you lose.

Finally, the kidnapper tells you to begin in 10 minutes. Use this time to formulate a game plan to have the highest chances of winning. It is a life or death situation, really.

Good thing this is a hypothetical situation, here is a Tic-Tac to help you solve it:

• "You" is also plural in English, so it's not clear whether the kidnapper also returns any tic-tacs. – Bass May 4 at 4:01
• @Bass fixed that! – tdserapio May 4 at 5:02
• The ordering of events on even-numbered rounds is still ambiguous (takes-returns-takes-returns or takes-takes-returns-returns), and it's not clear if both have to return the same amount. Also, it's not clear if the kidnapper has to return at least one, again because the "you" can refer to "me" or "the player whose turn it is". – Bass May 4 at 6:13
• Based on @Bass' comment, I think it would be helpful to give a simple illustrated example with a box of , say, 10 tic-tacs. – Earlien May 4 at 6:44

Partial (assuming I understood the requirements correctly):

My assumptions: You (captive) start.
You and the kidnapper always return on the even turn the same number of tic-tacs and you decide the number.

Here goes:

Case 1.

If you end up at one point with 1 piece in the box.
if it's an odd turn, you take it and win.
if it's an even turn, you add 1 the kidnapper adds 1 so there are 3. Next turn you take 3 and win.
conclusion is that if there is 1 piece left when it's your turn you always can win.

Case 2.

If you end up at one point with 2 pieces in the box.
if it's an odd turn, you take 1 and lose (kidnapper takes the other one).
if it's an even turn, you add 1 the kidnapper adds 1 so there are 4. Next turn you take 4 and win. Conclusion. Try to not end up with 2 left on your odd turn.

Case 3.

If you end up at one point with 3 pieces in the box.
if it's an odd turn, you take 3 and win
if it's an even turn, you add 1 the kidnapper adds 1 so there are 5. Next turn you take 3, kidnapper has to take 1 from the remaining 2 and there is 1 left on your turn. You win, as shown in case 1.
Conclusion. You can always win if there are 3 pieces left.

Case 4.

If you end up at one point with 4 pieces in the box.
if it's an odd turn, you take 4 and win
if it's an even turn, you add 1 the kidnapper adds 1 so there are 6. Next turn you take 4, kidnapper has to take 1 from the remaining 2 and there is 1 left on your turn. You win, as shown in case 1. Conclusion. You can always win if there are 4 pieces left.

Case 5.

If you end up at one point with 5 pieces in the box.
if it's an odd turn, you take 3, kidnapper has to take 1 from the remaining 2, leaving 1 and you win (case 1).
if it's an even turn, you add 2 the kidnapper adds 2 so there are 9. Next turn (odd) you take 4, kidnapper takes 1, 3 or 4 leaving 4, 2 or 1 and the next turn is even. So you win (cases, 4, 2 on even and 1)
Conclusion. You can always win if there are 5 pieces left.

Case 6.

If you end up at one point with 6 pieces in the box.
if it's an odd turn, you take 4, kidnapper has to take 1 from the remaining 2, leaving 1 and you win (case 1).
if it's an even turn, you add 1 the kidnapper adds 1 so there are 8. Next turn (odd) you take 1, kidnapper takes 1, 3 or 4 leaving 7, 5 or 4 and the next turn is even. You win in cases 5 and 4. And you "stalemate" in case 7. Conclusion. You can never lose if there are 6 pieces left.

Case 7.

If you end up at one point with 7 pieces in the box.
if it's an odd turn, you take 1, kidnapper can take 1, 3, 4 leaving 5, 3, 2. You win (cases 5, 3 and 2 even).
if it's an even turn, you add 2 the kidnapper adds 2 so there are 11. Next turn (odd) you take 4 and kidnapper takes 1, 3 or 4 leaving you in cases 6, 4, 3 so you win or stalemate. Conclusion. You can never lose if there are 7 pieces left. You can always keep it steady at 6 or 7.

Case 8.

If you end up at one point with 8 pieces in the box.
if it's an odd turn, you take 3, kidnapper can take 1, 3, 4 leaving 4, 2 even or 1. You win.
if it's an even turn, you add 1 the kidnapper adds 1 so there are 10. Next turn (odd) you take 3 and kidnapper takes 1, 3 or 4 leaving you in cases 6, 4, 3 so you do not lose
Conclusion. You can never lose if there are 8 pieces left.

now the general part.

We notice that for every number between 3 and 8 you win.
Let's say there are X pieces where $$8 < x <= 13$$. and the turn is odd. So you have to take.
you always take 3 and add 1
$$- 3 - 1 + 1 + 1 = -2$$
$$- 3 - 3 + 1 + 1 = -4$$
$$- 3 - 4 + 1 + 1 = -5$$