The expected value of the game is simply the mean or average of the game
The expected value of a game is: [tex]-\frac {4}9x[/tex]
The given parameter is:
[tex]p = \frac 19[/tex] --- probability of winning
If the wager is x, then the profit is 4x
So, the expected value of the game is:
[tex]E(x) = \sum x \times P(x)[/tex]
The above formula is interpreted as:
Expected value = Probability of Winning x Profit + Probability of losing x Wager
Where
[tex]q= 1 - p[/tex] -- the probability of losing
Substitute [tex]p = \frac 19[/tex]
[tex]q= 1 - \frac 19[/tex]
[tex]q= \frac 89[/tex]
So, we have:
[tex]E(x) = \sum x \times P(x)[/tex]
[tex]E(x) = \frac 19 \times 4x + \frac 89 \times -x[/tex]
x is negative, because it represents loss
[tex]E(x) = \frac 49x - \frac 89x[/tex]
Take LCM
[tex]E(x) = \frac {4 - 8}9x[/tex]
[tex]E(x) = -\frac {4}9x[/tex]
Hence, the expected value of a game is: [tex]-\frac {4}9x[/tex]
Read more about expected values at:
https://brainly.com/question/13945225
