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世界大赛逐场下注复制

World Series

专题
Finance / 金融
难度
L4

题目详情

红袜队与洛矶队打 7 局 4 胜制系列赛。你想做一个“全押系列赛”的双倍或归零下注：

  • 若红袜最终赢系列赛，你净赚 +100；
  • 若红袜最终输系列赛，你净亏 -100。

但你不能直接押系列赛结果，只能每场比赛单独下注。如何选择每场下注金额，使得最终实现上述净收益？

The Boston Red Sox are playing the Colorado Rockies in a best-of-7 World Series (first team to win 4 games). You have $100 on the Red Sox in a double-or-nothing bet (win +100 or lose -100). But you cannot bet on the entire series outcome directly; you can only place bets on each individual game. How do you choose the amount to bet on each game so that, if the Red Sox eventually win the series, you net exactly +100, and if they lose, you net exactly -100?

解析

用动态规划/二叉树回溯。

设状态 (i,j)(i,j) 表示红袜已赢 ii 场、洛矶已赢 jj 场。设 f(i,j)f(i,j) 表示此时你希望“净值”是多少。

在吸收态：

  • 红袜到 4 胜：f(4,k)=+100f(4,k)=+100；
  • 洛矶到 4 胜：f(ℓ,4)=−100f(\ell,4)=-100。

从 (i,j)(i,j) 往下一场下注 yy：

  • 红袜赢则到 (i+1,j)(i+1,j)，净值变为 f(i,j)+yf(i,j)+y；
  • 红袜输则到 (i,j+1)(i,j+1)，净值变为 f(i,j)−yf(i,j)-y。

因此有

{f(i+1,j)=f(i,j)+y,f(i,j+1)=f(i,j)−y,⇒{f(i,j)=f(i+1,j)+f(i,j+1)2,y=f(i+1,j)−f(i,j+1)2.\begin{cases} f(i+1,j)=f(i,j)+y,\\ f(i,j+1)=f(i,j)-y, \end{cases} \Rightarrow \begin{cases} f(i,j)=\frac{f(i+1,j)+f(i,j+1)}{2},\\ y=\frac{f(i+1,j)-f(i,j+1)}{2}. \end{cases}

从终局向前回溯即可得到每个状态下的 f(i,j)f(i,j) 与下注额 yy。例如 (3,3)(3,3)：

f(3,3)=100+(−100)2=0,y3,3=100−(−100)2=100.f(3,3)=\frac{100+(-100)}{2}=0,\quad y_{3,3}=\frac{100-(-100)}{2}=100.

Original Explanation

Let (i,j)(i,j) denote the state where the Red Sox have ii wins and the Rockies have jj wins. Let f(i,j)f(i,j) be your current net amount of money. At absorption:

  • If Red Sox have 4 wins (4,k)(4,k), you want f(4,k)=+100f(4,k)=+100.
  • If Rockies have 4 wins (ℓ,4)(\ell,4), you want f(ℓ,4)=−100f(\ell,4)=-100.

From state (i,j)(i,j), you bet yy on the next game for the Red Sox. If they win, new state (i+1,j)(i+1,j) and your net changes by +y+y. If they lose, new state (i,j+1)(i,j+1) and your net changes by −y-y. Thus, {f(i+1,j)=f(i,j)+y,f(i,j+1)=f(i,j)−y,⟹{f(i,j)=f(i+1,j)+f(i,j+1)2,y=f(i+1,j)−f(i,j+1)2.\begin{cases} f(i+1,j) = f(i,j) + y,\\ f(i,j+1) = f(i,j) - y, \end{cases} \quad \Longrightarrow \quad \begin{cases} f(i,j) = \frac{f(i+1,j)+f(i,j+1)}{2},\\ y = \frac{f(i+1,j)-f(i,j+1)}{2}. \end{cases}

By working backward from the final states, you find the unique f(i,j)f(i,j) and bets yy. Example:

  • f(3,3)= 100+(−100) 2=0f(3,3) = \frac{\,100 + (-100)\,}{2} = 0.
  • The bet at (3,3)(3,3) is y3,3=100−(−100)2=100y_{3,3} = \frac{100 - (-100)}{2}=100.

You can fill out the entire table or a binomial tree to get the bets and net amounts.