Question

In a dice game if you roll a 2, 4, or 6 you get the value of the die. If you roll as 1, 3, or 5 you lose $5. What is the expected value of the game?

Answer 1

multiply the amount you win times the probability you win it and add it up

Answer 2

the probability you get any of those outcome is \(\frac{1}{6}\) so compute via
\[2\times \frac{1}{6}+4\times \frac{1}{6}+6\times \frac{1}{6}-5\times \frac{1}{6}-5\times \frac{1}{6}-5\times \frac{1}{6}\]
\[=\frac{2+4+6-5-5-5}{6}\]

Answer 3

i get
\[\frac{12-15}{6}=\frac{-3}{6}=-\frac{1}{2}\] meaning on average you lose half a dollar every time you play

Answer 4

if {2,4,6} you win you can expect to win $4 as 4 is the average of {2,4,6}

if{1,3,5} you loose you loose $5

since there are as many terms in each case each is al likely,

so you can expect to win as much as you loose,
the average of {4,-5}=(4-5)/2 =-1/2