P(甲贏) = \(\displaystyle\frac{1}{4} + \sum_{n=1}^\infty\left(\frac{1}{4}\left(C^n_n\left(\frac{1}{4}\right)^n+C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)+\frac{1}{2}\left(C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)\right)\cdot\left(C^n_n\left(\frac{1}{4}\right)^n+C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)^2\)
P(乙贏) = \(\displaystyle\left(\frac{1}{4}+\frac{1}{2}\right)\cdot\frac{1}{4} + \sum_{n=1}^\infty\left(\frac{1}{4}\left(C^n_n\left(\frac{1}{4}\right)^n+C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)+\frac{1}{2}\left(C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)\right)\cdot\left(C^n_n\left(\frac{1}{4}\right)^n+C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)\cdot\left(C^{n+1}_{n+1}\left(\frac{1}{4}\right)^{n+1}+C^{n+1}_1\left(\frac{1}{4}\right)^{n}\left(\frac{1}{2}\right)\right)\)
P(丙贏) = \(\displaystyle\left(\frac{1}{4}+\frac{1}{2}\right)^2\cdot\frac{1}{4} + \sum_{n=1}^\infty\left(\frac{1}{4}\left(C^n_n\left(\frac{1}{4}\right)^n+C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)+\frac{1}{2}\left(C^n_1\left(\frac{1}{4}\right)^{n-1}\left(\frac{1}{2}\right)\right)\right)\cdot\left(C^{n+1}_{n+1}\left(\frac{1}{4}\right)^{n+1}+C^{n+1}_1\left(\frac{1}{4}\right)^{n}\left(\frac{1}{2}\right)\right)^2\)
以上這一長串並非不能算,乘開之後雖然有點眼花撩亂,但還是可以算的~
(乘開後是"等比級數"或"等差×等比"或是"等差^2 ×等比"或是"等差^3 ×等比"形式的雜級數~都是可以算的~)
眼睛很花~就讓我稍微偷懶一下,有請 wolfram alpha 幫我計算一下
P(甲贏) = \(\displaystyle\frac{301552}{583443}\),請點
http://goo.gl/cBQ65
P(乙贏) = \(\displaystyle\frac{173444}{583443}\),請點
http://goo.gl/mBxFs
P(丙贏) = \(\displaystyle\frac{36149}{194481}\),請點
http://goo.gl/obil0
而且無聊還可以驗證一下~~~ \(\displaystyle\frac{301552}{583443}+\frac{173444}{583443}+\frac{36149}{194481}=1\)
然後我說明一下~
P(甲贏) = 要嘛甲在第一局就贏,或是甲在經過 n 局之後,繼續丟~以兩正面獲勝~或以一正一反面獲勝~
如果甲最後是以兩正面獲勝,則甲擲的前 n 局有可能全部都是兩反~或是~恰一次的一正一反&n-1次兩反,
如果甲最後是以一正一反獲勝,則甲擲的前 n 局恰一次的一正一反,
當然不管甲是以何種方式獲勝,乙、丙兩人擲的前 n 局必須要「全部都是兩反~或是~恰一次的一正一反&n-1次兩反」。
乙贏~或丙贏~請同理類推。:)