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[组合] Random Walk: Gambler's Ruin - Infinite Fortune

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hbghlyj posted 2023-6-3 00:58 |Read mode
Last edited by hbghlyj 2023-6-6 08:22Random Walk: Gambler's Ruin - Infinite Fortune
Given that one starts at $i$ dollars, how does one find the probability of getting infinitely rich/getting ruined (the complement)?

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original poster hbghlyj posted 2023-6-3 01:01
我想,1减去(破产的概率)不等于(无限富有的概率)
因为很可能既不会破产也不会变得无限富有,只是在小范围内徘徊

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original poster hbghlyj posted 2023-6-3 01:06
A8LectureNotes_MT22_24Sep2022.pdf page 51
$p>q$的情况有可能无限富有
Screenshot 2023-06-02 at 18-05-55 A8LectureNotes_MT22_24Sep2022.pdf.png

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tommywong posted 2023-6-3 06:54
佢應該係$\displaystyle P_i(x)=\sum_{n=0}^\infty \binom{2n+i}{n}x^n$嘅
寫漏嘢啫
现充已死,エロ当立。
维基用户页:https://zh.wikipedia.org/wiki/User:Tttfffkkk
Notable algebra methods:https://artofproblemsolving.com/community/c728438
《方幂和及其推广和式》 数学学习与研究2016.

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original poster hbghlyj posted 2023-6-6 15:11
tommywong 发表于 2023-6-2 23:54
佢應該係$\displaystyle P_i(x)=\sum_{n=0}^\infty \binom{2n+i}{n}x^n$嘅
寫漏嘢啫
已修改。修改的同时把这个帖子顶了起来,而且4小时前,有人评论原回答是错的,有人发了一个新回答

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