Problem 661

A Long Chess Match

Two friends $A$ and $B$ are great fans of Chess. They both enjoy playing the game, but after each game the player who lost the game would like to continue (to get back at the other player) and the player who won would prefer to stop (to finish on a high).

So they come up with a plan. After every game, they would toss a (biased) coin with probability $p$ of Heads (and hence probability $1 - p$ of Tails). If they get Tails, they will continue with the next game. Otherwise they end the match. Also, after every game the players make a note of who is leading in the match.

Let $p_{A}$ denote the probability of $A$ winning a game and $p_{B}$ the probability of $B$ winning a game. Accordingly $1 - p_{A} - p_{B}$ is the probability that a game ends in a draw. Let $E_{A} (p_{A}, p_{B}, p)$ denote the expected number of times $A$ was leading in the match.
For example, $E_{A} (0.25, 0.25, 0.5) \approx 0.585786$ and $E_{A} (0.47, 0.48, 0.001) \approx 377.471736$ , both rounded to six places after the decimal point.

Let $H (n) = \sum_{k = 3}^{n} E_{A} (\frac{1}{\sqrt{k + 3}}, \frac{1}{\sqrt{k + 3}} + \frac{1}{k^{2}}, \frac{1}{k^{3}})$
For example $H (3) \approx 6.8345$ , rounded to 4 digits after the decimal point.

Find $H (50)$ , rounded to 4 digits after the decimal point.

漫长的棋局

两位好友 $A$ 和 $B$ 都很喜欢下棋，但是每下完一局，输的一方都想要再下一局（想要赢回来），而赢的一方则不想再下（见好就收）。

他俩想到了一个办法：每下完一局，他们会抛掷一枚（不公平的）硬币，硬币有 $p$ 的概率正面朝上（因此反面朝上的概率是 $1 - p$ ）。如果硬币是反面朝上，他们就接着下一局，否则就结束。此外，每下完一局，他们都会记录目前在总获胜局数上谁暂且领先。

记 $A$ 赢得单局的概率为 $p_{A}$ ， $B$ 赢得单局的概率为 $p_{B}$ ，进而两人平局的概率为 $1 - p_{A} - p_{B}$ 。记 $E_{A} (p_{A}, p_{B}, p)$ 为整个棋局过程中 $A$ 领先的次数的期望值。
例如， $E_{A} (0.25, 0.25, 0.5) \approx 0.585786$ ， $E_{A} (0.47, 0.48, 0.001) \approx 377.471736$ ，均保留小数点后 $6$ 位小数。

记 $H (n) = \sum_{k = 3}^{n} E_{A} (\frac{1}{\sqrt{k + 3}}, \frac{1}{\sqrt{k + 3}} + \frac{1}{k^{2}}, \frac{1}{k^{3}})$
例如， $H (3) \approx 6.8345$ ，保留小数点后 $4$ 位小数。

求 $H (50)$ 并保留小数点后 $4$ 位小数。