How would u explain your strategy to win this math game/problem?

Question

Rules:
2 teams of 1 or more players (stick w/ 1 on 1),
Begin by putting 10 items (coins, cubes ect...) on table betweenthe two of you.
Designate one of the items as "poison"
The 1st time u play, toss a coin to decide who goes 1st, after that alternate who begins.
For your turn you must either grab 1 or 2 of the items
The team forced to take the last "poisoned" item loses.

Instructions:
1. Play a few times, as u play, consider these ?'s-
a:how many u picked up
b:why you picked up that many and who won.

2. Reflect on the game records and see if u can devise a winning strategy.
a: Test it out once u have come up with one against other player.

3. Once u have a winning strategy, try starting with a diff. # of items than 10.
a: How does the winning strategy change?

4. If you had to play, starting with 431 items, would u care whether u went 1st or 2nd?

PLEASE EXPLAIN YOUR STRATEGY AND HOW IT WORKS OR WON'T WORK WITH A DIFF. # OF STARTING ITEMS.

Answer 1

2. The second player can always win. Reason: if they pick a different number to what the first person picks, the total removed by both players will always be 3. After three turns there will only be the poisoned item left and the first player must pick it.

3. If the number of items is one more than a multiple of 3, the second player always wins using the above strategy.
If the number of items is two more than a multiple of 3, the first player always wins by taking one item on the first turn and then using the same strategy (he effectively becomes the second player in a game with one more than a multiple of 3 items).
If the number of items is a multiple of 3, the first player always wins by taking two items on the first turn and then using the same strategy - again reducing the problem to one where he plays second and the number of items is one more than a multiple of 3.

4. Given the previous answer, you always care! In this case 431 = 3*143 + 2, so you would want to be the first player.

We can generalise this to the case where you can take anywhere between 1 and k items at a time. Your opponent can always make the total up to k+1, so the second player wins if the number of items is one more than a multiple of (k+1), and the first player wins in all other cases by taking away enough items on the first turn to leave one more than a multiple of (k+1).

Answer 2

The winning strategy is obvious. Consider the following:

1 is a losing number. If your turn starts with only one item left, you've no choice but to take it, and so you will lose.

If n is a losing number, n+1 and n+2 are both winning numbers, since you can remove enough items to leave your opponent with exactly n items, which (assuming optimal play from you), will guarantee his loss.

If n is a losing number, then n+3 is a losing number, because regardless of whether you play 1 or 2, your opponent can play a move that will leave you with n, guaranteeing your loss.

From this we conclude that any number of the form 3k+1 for some integer k is a losing number, and all other numbers are winning numbers. The optimal strategy is to always play so as to leave your opponent with 3k+1 items for some integer k, which is always possible, unless you start on a losing number, in which case your opponent can force a win. You can also force a win if at any time your opponent makes a mistake.

If the game starts with a winning number of items, then you should attempt to play first. If the game starts with a losing number of items, you should attempt to play second. In particular, you should always attempt to play second in the 10-item game (since 10=3*3+1), and attempt to play first in the 431-item game (since 431 = 3*143 + 2, you can leave your opponent with a losing number by playing 1 on your first move).