temp2 - 幻想境界「醉生夢死」 - 兔 @ ウィキ - atwiki（アットウィキ）

幻想境界「醉生夢死」 - 兔 @ ウィキ

幻想境界「醉生夢死」 - 兔 @ ウィキ

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temp2

$\verb|Assumptions: |$

$P \equiv \verb|chance of winning|$

$\verb|Every win gives you b*A golds, while every lose gives you nothing.|$
$\verb|where A is the investment, and b is an arbitrary constant.|$

$\verb|Analysis of Snake's Gambling method :|$

$\verb|1st round(A invested): we have a P chance of getting bA. |$
$\verb|If not, we invest c*A in the next round, where c is an arbitrary constant.|$

$\verb|Following this procedure, by wining at n-th round we have:|$

$\verb|Total investment = | \sum^{n-1}_{k=0} A*c^{k} = \frac{A(c{^n}-1)}{c-1} \verb|(assuming c>1)|$

$\verb|Total income = | bc^{n-1}A$

$\verb|Worth it? Total income - Total investment = | \frac{bc^{n-1}A(c-1)-A(c^n-1)}{c-1}$

$= \frac{A}{c-1}(c^n(b-1)-bc^{n-1}+1)$

$\verb|Chance of getting this result is | P(1-P)^{n-1}$

$\verb|Expectation value | \equiv <X> \equiv \sum_{i} X_iP(X_i)$

$= \sum^{\infty}_{k=0} P(1-P)^{k}\frac{A}{c-1}(c^k(b-1)-bc^{k-1}+1)$

$= \frac{PA}{c-1}((b-1)\sum^{\infty}_{k=0}c^k(1-P)^k - \frac{b}{c}\sum^{\infty}_{k=0}c^k(1-P)^k + \sum^{\infty}_{k=0}(1-P)^k)$

$\verb|(assuming | \sum^{\infty}_{k=0}c^k(1-P)^k \verb| exists)|$

$= \frac{PA}{c-1}(\frac{(b-1)(c-1)-1}{c}\frac{1}{1-c(1-P)}+\frac{1}{1-(1-P)})$

$= \frac{PA}{c-1}(\frac{(b-1)(c-1)-1}{c(1-c(1-P))} + \frac{1}{P})$

$= \frac{PA}{c-1}\frac{ P[(b-1)(c-1)-1]-c[1-c(1-P)] }{Pc(1-c(1-P))}$

$= \frac{A}{c(c-1)(1-c+cP)}(P(bc-b-c)-c(1-c+cP))$

$= \frac{A}{c(c-1)(1-c+cP)}(Pbc-Pb-Pc-c+c^2-Pc^2)$

$\verb|Reference: expectation value = | PbA - A = A(Pb-1)$

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最終更新：2012年01月27日 18:50