In the game of roulette, a player can place a $88 bet on the number 3232 and have a startfraction 1 over 38 endfraction 1 38 probability of winning. if the metal ball lands on 3232, the player gets to keep the $88 paid to play the game and the player is awarded an additional $280280. otherwise, the player is awarded nothing and the casino takes the player's $88. what is the expected value of the game to the player? if you played the game 1000 times, how much would you expect to lose?

The expected value of the game to the player is . A loss of $0.42105 is expected.

If you played the game 1000 times, the amount that you are expected to lose is

Step-by-step explanation:

From the information given we know:

There are 38 slots on a roulette wheel so the probability of losing is

We can find an expected value by multiplying each numerical outcome by the probability of that outcome and then summing those products together.

The expected value in dollars for this game is

A loss of $0.42105 is expected.

If you played the game 1000 times, the amount that you are expected to lose is

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!