An archive of Mark's Fall 2017 Intro Stat course.

# Expected value #2

ren98christy

While working on the study guide for tomorrow’s quiz I came across question #2, which says:

Suppose I can buy a lottery ticket for $1 to get 1 chance in 10,000 to win$10,000 and 99 chances in 10,000 to win $100. What is the expected value of that ticket? I was a bit confused as how to go about finding the expected value of the ticket. Could anyone help lead me in the right direction?? mark Considering the fact that we pay$1 for the ticket, there are three possible net outcomes:

• We net $9999 with probability 1/10000 . • We net$99 with probability 99/10000
• We net -\$1 with probability
1-\frac{1}{10000}-\frac{99}{10000} = \frac{99}{100}.

Thus, the expected value is

\frac{9999}{10000}+\frac{99\times99}{10000}-\frac{99}{100} = \frac{99}{100} = 0.99.
ren98christy

I’m confused because the second equation ( 9999/10000+ 9999/10000-99/100) equals 1.0098 and not 0.99.

mark

@ren98christy
It’s not

\frac{9999}{10000}+\frac{9999}{10000}-\frac{99}{100},

it’s
\frac{9999}{10000}+\frac{99\times99}{10000}-\frac{99}{100}.

I added the explicit multiplication symbol; it was just a space before.

asiarenee5

1- (1/10,000 * 10,000 + 99/10,000 * 100)= 0.99