Question Number 174611 by MathsFan last updated on 05/Aug/22
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{expected}\:\mathrm{payback}\:\mathrm{for}\:\mathrm{a}\: \\ $$$$\mathrm{game}\:\mathrm{in}\:\mathrm{which}\:\mathrm{you}\:\mathrm{bet}\:\$\mathrm{8}.\mathrm{00}\:\mathrm{on}\:\mathrm{any} \\ $$$$\mathrm{number}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{99}\:\mathrm{and}\:\mathrm{if}\:\mathrm{your} \\ $$$$\mathrm{number}\:\mathrm{comes}\:\mathrm{up},\:\mathrm{you}\:\mathrm{win}\:\$\mathrm{2},\mathrm{000}.\mathrm{00} \\ $$
Commented by shikaridwan last updated on 06/Aug/22
![Probability of getting x (x∈[0,99]) =(1/(99)) not getting =1−(1/(99))=((98)/(99)) Expected payback =$(((2000)/(99))+((98)/(99)).8) =$28.12](Q174638.png)
$${Probability}\:{of}\:{getting}\:{x}\:\left({x}\in\left[\mathrm{0},\mathrm{99}\right]\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{99}} \\ $$$${not}\:{getting}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{99}}=\frac{\mathrm{98}}{\mathrm{99}} \\ $$$${Expected}\:{payback} \\ $$$$=\$\left(\frac{\mathrm{2000}}{\mathrm{99}}+\frac{\mathrm{98}}{\mathrm{99}}.\mathrm{8}\right)\:\:=\$\mathrm{28}.\mathrm{12} \\ $$
Commented by MathsFan last updated on 09/Aug/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$