A Game Of Tokens

Look at a set of 3 rounds, where the players have x+1, x, and x-1 tokens. Each of the players will gain two tokens from the others and give away 3 tokens, so overall, each player will lose 1 token.

Therefore, after 97 sets of 3 rounds, or 291 rounds, the players will have 3, 2, and 1 tokens, respectively. After 1 more round, player A will give away his last 3 tokens and the game will stop. Hence round 292.

In the first round, player $A$ loses three tokens while $B$ and $C$ both gain one each... well, $B$ and $C$ 's number of tokens don't really matter because it's player $A$ who ends up having the lowest count of tokens: $A$ ends up having $97$ tokens left first, then by $B$ and then by $C$ in the succeeding rounds. At the fourth round, $A$ ends up having $96$ left first and so on. Thus, it's only $A$ 's token stock that needs attention in this game as player $A$ 's would be the first to reach $0$ tokens after $n$ rounds.

To find $n$ , we should notice that from the first round when $A$ has $97$ tokens left, after every three rounds, $A$ loses a net of $1$ token. Thus the number of rounds needed to end the game would be

$\begin{aligned} n &= 1 + 97 \times 3 \\ n &= 1 + 291 \\ n &= \boxed{292} \end{aligned}$

>>> stack = [100,99,98];


>>> while (stack[0] or stack[1] or stack[2]):
...     if stack[0] > 0 and stack[1] > 0 and stack[2] > 0:
...             k = stack.index(max(stack))
...             stack[(k+1)%3] += 1
...             stack[(k+2)%3] += 1
...             stack[k] -= 3
...     else: break
...     print stack

Lets start with the terminating condition for this problem

This happens when the players have [3,2,1] then the first player will be bust out of tokens

so that's 1 turn.

before that if we have for example [5,4,3] this means that we would have 3 turns with 5 tokens on one of the players [5,4,3]->[2,5,4]->[3,2,5]->[4,3,2] this goes on for all numbers [N,N-1,N-2] it only doesn't work when N=3 because it is the terminating condition

so this means that the number of turns is ((N-3) * 3) +1

= ((100-3)*3) +1 = 292