I have a question that may really be about mathematical modelling as much as math itself, but I will try to give it a formulation suited for this site.
Suppose that I am going to play some game, $G$, (for example some card game) repeatedly against an opponent. The way we do it is that we have agreed to play a match of $N$ games and whoever has the most points at the end will will the match. If we are tied on points, we flip a coin to decide the winner. We do not care about the result in an individual game, except in so far it helps us win the match.
The game $G$ can not end in a tie and only one the winner receives any points, but a win can be worth a different amount of points. You can think of backgammon, where the winner may receive $1$,$2$ or $3$ points.
Lets assume that the game $G$ is a symmetric game, by which I mean that the players have access to the same set of strategies and that the distribution of pay-offs is a symmetric function of the choice of strategies. This also implies that each player can guarantee himself a positive probability of winning. (The symmetry assumption is rather strong but you can imagine we flip a coin to decide who gets to be player I before each game.)
It is clear that my strategy towards the end of the match should be impacted by the current score. If I am up already, when we start the last game, it is unimportant whether I win $1$ point or $2$ point, so I should not risk anything in order to get a better score. But intuition suggests that at the beginning of a long match I should play a strategy that maximizes the expected value of my score, or at least something close to such a strategy.
Can someone prove this?
Let me try to be more precise: Let $p(t,n)$ be my probability of winning when I am up by $t$ points (could be a negative number) and there are $n$ games to go in the match, assuming optimal strategy. So, for example, for any positive $n$ it is true that $p(0,n)=\frac{1}{2}$ and we always have $p(-i,n) = 1-p(i,n)$.
I am thinking that it should be enough to prove that for any integer $i$ it is true that $$ \lim_{n\to \infty} \frac{p(i,n)-1/2}{p(1,n)-1/2} = i.$$
This would mean that winning by $i$ points is worth $i$ times as much as winning by $1$ points, just like it would be in the game $G$ considered in isolation.