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No links please A gambler rolls two unbiased dice and stands to loss 2 if he fails to throw a six, to win 4 if he throws one six and to win 10 if he throws two sixes. Is the game fair?

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{A} \mathrm{game} \mathrm{is} \mathrm{considered} \mathrm{fair} \mathrm{if} \mathrm{the} \mathrm{expected} \mathrm{change} \mathrm{of} \mathrm{wealth} \mathrm{on} \mathrm{each} \mathrm{play} \mathrm{is} \mathrm{zero}. \\ \mathrm{Here} \\ \mathrm{Probabilty} \mathrm{of} \mathrm{one} \mathrm{a} \mathrm{six}{=}^2{\mathrm{C}}_1\times \frac{1}{6}\times \frac{5}{6}=\frac{10}{36} \\ \mathrm{Probablity} \mathrm{of} \mathrm{getting} \mathrm{two} \mathrm{sixes}=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36} \\ \mathrm{Probability} \mathrm{of} \mathrm{getting} \mathrm{no} \mathrm{six}=\frac{5}{6}\times \frac{5}{6}=\frac{25}{36} \\ \mathrm{Expected} \mathrm{value}=\sum _{}^{}\left(\mathrm{Probability} \times \mathrm{Money}\right) \\ =10\times \frac{1}{36}+4\times \frac{10}{36}-2\times \frac{25}{36} \\ =\frac{10+40-50}{36} \\ =\frac{50-50}{36} \\ =0 \\ \mathrm{Hence} \mathrm{game} \mathrm{is} \mathrm{fair}. \end{gathered}$$

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