An unfair coin lands on heads with probability $\frac35$, on tails with probability $\frac15$, and on its edge with probability $\frac15$. I

Hi Mellie.

expected value,  E(x) = 0.6*4 + 0.2*-1 + 0.2*-10 

= 2.18

≈ $2

Probabilty carries no certainty. Knowing my luck, if I was given 10 throws, I'd come out $100 poorer!

$${\mathtt{0.6}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{10}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\mathtt{0.2}}$$

 Expected win = $0.20