﻿ 赵森烽-克勤概率的赌本分配研究与期望值定理
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 智能系统学报  2017, Vol. 12 Issue (5): 608-615  DOI: 10.11992/tis.201604020 0

### 引用本文

ZHAO Keqin, ZHAO Senfeng. Distribution of gambling capital and expectation value theorem for Zhao Senfeng-Keqin probability[J]. CAAI Transactions on Intelligent Systems, 2017, 12(5): 608-615. DOI: 10.11992/tis.201604020.

### 文章历史

1. 诸暨市联系数学研究所, 浙江 诸暨 311811;
2. 浙江大学 非传统安全与和平发展研究中心, 浙江 杭州 310058

Distribution of gambling capital and expectation value theorem for Zhao Senfeng-Keqin probability
ZHAO Keqin1,2, ZHAO Senfeng1
1. Zhuji Institute of Connection Mathematics, Zhuji 311811, China;
2. Center for Non-traditional Security and Peaceful Development Studies, Zhejiang University, Hangzhou 310058, China
Abstract: With respect to the reasonable distribution of gambling capital in the developmental history of probability theory, Zhao Senfeng-Keqin probability has been used to investigate the minimum number of gambling times necessary for the rational allocation of the minimum amount of gambling capital. Results have shown that the mathematical expectation for this problem, based on classical probability, failed to occur in practice. What appeared instead are two extreme values of "mathematical expectation" based on the Zhao Senfeng-Keqin probability, which can objectively reflect the gambling results within the smallest and largest number of gambling times for a given rule. In addition, it describes both the classic expectation value and the actual value, thereby providing a basis for formulating or amending specific gambling tactics and the reasonable allocation of gambling capital. The result is an uncertainty theorem for the expectation value. In this paper, we illustrate the practical significance of this theorem by giving an example of service charging on a robot.
Key words: distribution of gambling capital    mathematical expectation    Zhao Senfeng-Keqin probability (contact probability)    uncertainty    expectation value theorem

1 如何合理分配赌本 1.1 问题描述

1.2 帕斯卡解法

1.3 惠更斯解法

 $\frac{{ua + vb}}{{u + v}} = ap + b\left( {1 - p} \right)$ (1)

 ${\rm{E}}\left( X \right) = ap + b\left( {1 - p} \right) = \frac{{ua + vb}}{{u + v}}$ (2)

2 赵森烽-克勤概率(联系概率)在分赌本中的应用 2.1 赵森烽-克勤概率

 $P\left( {A,\bar A} \right) = P\left( A \right) + P\left( {\bar A} \right)i$ (3)

 ${P_c}\left( A \right) = P\left( A \right) + P\left( {\bar A} \right)i$ (4)

 $P\left( A \right) + P\left( {\bar A} \right) = 1$ (5)

 $P\left( {\bar A} \right) = 1 - P\left( A \right)$ (6)

2.2 基于赵森烽-克勤概率的分赌本分析

 ${\rm{E}}\left( X \right) = 100 \times \left( {\frac{3}{4} + \frac{1}{4}i} \right) + 0 \times \left( {\frac{1}{4} + \frac{3}{4}i} \right) = 75 + 25i$

i=1时，甲赢得赌本100法郎；

i=-1时，甲赢得赌本50法郎；

i=0时，甲赢得赌本75法朗，这是第1节中帕斯卡法的结果。

W1出现时，甲共赢了3次，这时甲得100法郎；当W2出现时，甲、乙各赢2次，这时如果终止赌博，甲只能得50法郎，也就是不存在甲得75法郎这种情况(表 5)。

3 基于赵森烽-克勤概率的数学期望 3.1 数学期望

1) 算术平均

 $\bar X = \frac{1}{n}\sum\limits_{k = 1}^n {xk,k = 1,2, \cdots ,n}$ (7)

2) 加权平均

 $\bar X = \frac{1}{n}\sum\limits_{t = 1}^k {{n_t}{X_t}} = \sum\limits_{t = 1}^k {\frac{{{n_t}}}{n}{X_t}} ,\;\;\;\;t = 1,2, \cdots ,k$ (8)

 $p\left( {{X_t}} \right) = P\left( {X = {X_t}} \right),t = 1,2, \cdots ,n, \cdots$

 ${\rm{E}}\left( X \right) = \sum\limits_{t = 1}^\infty {{X_t}P\left( {{X_t}} \right)}$ (9)

 $\int_{ - \infty }^\infty {\left| x \right|p\left( x \right){\rm{d}}x < \infty }$ (10)

 ${\rm{E}}\left( X \right) = \int_{ - \infty }^\infty {xp\left( x \right){\rm{d}}x}$ (11)

3.2 基于赵森烽-克勤概率的数学期望

 $p\left( {{X_t},{{\bar X}_t}} \right) = P\left( {X = {X_t},{{\bar X}_t}} \right),t = 1,2, \cdots ,n, \cdots$

 $\sum\limits_{t = 1}^\infty {\left| {{X_t}} \right|P\left( {{X_t}} \right) < \infty }$
 $\sum\limits_{t = 1}^\infty {\left| {{{\bar X}_t}} \right|P\left( {{X_t}} \right) < \infty }$

 ${\rm{E}}\left( {X,\bar X} \right) = \sum\limits_{t = 1}^\infty {\left( {{X_t},{{\bar X}_t}} \right)P\left( {{X_t},{{\bar X}_t}} \right)}$ (12)

 ${\rm{E}}\left( {X,\bar X} \right) = \sum\limits_{t = 1}^\infty {\left( {{X_t},{{\bar X}_t}} \right)\left( {P\left( {{X_t}} \right) + P\left( {{{\bar X}_t}} \right)i} \right)}$ (13)

 ${{\rm{E}}_{\rm{c}}}\left( X \right) = \sum\limits_{t = 1}^\infty {{{\left( {{X_t}} \right)}_{\rm{c}}}\left[ {{P_{\rm{c}}}\left( {{X_t}} \right)} \right]}$ (14)

 ${\rm{E}}\left( {X,\bar X} \right) = {\rm{E}}\left( X \right) + {\rm{E}}\left( {\bar X} \right)i$ (15)

 ${{\rm{E}}_{\rm{c}}}\left( X \right) = {\rm{E}}\left( X \right) + {\rm{E}}\left( {\bar X} \right)i$ (16)

3.3 基于赵森烽-克勤概率的数学期望性质

 ${\rm{E}}\left( C \right) = C$ (17)

 ${\rm{E}}\left( {kX} \right) = k{\rm{E}}\left( X \right)$ (18)

 ${\rm{E}}\left[ {{g_1}\left( x \right) \pm {g_2}\left( x \right)} \right] = {\rm{E}}\left[ {{g_1}\left( x \right)} \right] \pm {\rm{E}}\left[ {{g_2}\left( x \right)} \right]$ (19)

 ${\rm{E}}\left( {c,d} \right) = c + di$ (20)

 $\begin{array}{*{20}{c}} {{\rm{E}}\left( {X,\bar X} \right) = {\rm{E}}\left( X \right) + {\rm{E}}\left( {\bar X} \right)i}\\ {{\rm{E}}\left( {c,d} \right) = {\rm{E}}\left( c \right) + {\rm{E}}\left( d \right)i + c + di} \end{array}$ (21)

 ${{\rm{E}}_c}\left( X \right) = c + di$ (22)

4 数学期望值定理

 ${\rm{E}}\left( {X,\bar X} \right) = {\rm{E}}\left( X \right) + {\rm{E}}\left( {\bar X} \right)i$ (23)

 ${{\rm{E}}_{\rm{c}}}\left( X \right) = {\rm{E}}\left( X \right) + {\rm{E}}\left( {\bar X} \right)i$ (24)

 ${{\rm{E}}_{\rm{c}}}\left( {\bar X} \right) = {\rm{E}}\left( {\bar X} \right) + {\rm{E}}\left( X \right)i$ (25)

5 应用

 ${\rm{E}}\left( X \right) = ap + b\left( {1 - p} \right) = \frac{{ua + vb}}{{u + v}}$ (26)

 ${\rm{E}}\left( {\bar X} \right) = bp + a\left( {1 - p} \right) = \frac{{ub + va}}{{u + v}}$ (27)

 ${\rm{E}}\left( {X,\bar X} \right) = \left[ {ap + b\left( {1 - p} \right)} \right] + \left[ {bp + a\left( {1 - p} \right)} \right]i$ (28)

 ${{\rm{E}}_{\rm{c}}}\left( X \right) = \left[ {ap + b\left( {1 - p} \right)} \right] + \left[ {bp + a\left( {1 - p} \right)} \right]i$ (29)

 ${\rm{E}}\left( {X,\bar X} \right) = 75 + 25i$ (30)
 ${{\rm{E}}_{\rm{c}}}\left( X \right) = 75 + 25i$ (31)

i=－1时，得

 $当\;i = - 1\;时,得\;{{\rm{E}}_{\rm{c}}}\left( X \right) = 75 + 25i = 50$ (32)

i=1时，得

 $当\;i = 1\;时,得\;{{\rm{E}}_{\rm{c}}}\left( X \right) = 75 + 25i = 100$ (33)

 ${{\rm{E}}_{\rm{c}}}\left( X \right) = 75 + 25i = 75$ (34)

 $\begin{array}{*{20}{c}} {{\rm{E}}\left( X \right) = 3000 \times \left( {0.6 + 0.4i} \right) + 5000 \times }\\ {\left( {0.4 + 0.6i} \right) = 3800 + 4200i} \end{array}$ (35)

 $\left\{ \begin{array}{l} x + y = 30\\ \frac{{5000}}{{30}}x + \frac{{3000}}{{30}}y = 4800 \end{array} \right.$ (36)

6 结束语

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