The Box of Tic-Tacs

Question

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:

Answer 1

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
the following cases can happen. (your take then kidnappers take then your add and the kidnappers add)
$ - 3 - 1 + 1 + 1 = -2$
$ - 3 - 3 + 1 + 1 = -4$
$ - 3 - 4 + 1 + 1 = -5$
So you reduce the number of pieces with anything between 2 and 5. so you end up either in the interval 3-11, if you are in the 3 - 8 interval, you win as described above, if you are in 9 - 11 you do it one more time or twice until you end up in the interval 3 - 8.
if you are between 8 and 13 and the turn is even, you add 1, kidnapper adds 1 and you end up between 10 and 15 with an odd turn. then follow the same logic (take 3 and add 1) and you end up between 5 and 13 on an odd turn, which you can win as described above.
this proves that you can win for every number up to 13. Following the same logic you can shift the interval from 13 to 18 and so on. So it does not matter how many pieces are initially in the box (anything between 55 and 60). You always take 3 and add 1 until you end up in the interval 3-8.

I just hope the OP does not come along and say that my assumptions were wrong because I spent some time trying these cases.