1. ## Expected Value

Confused in this section.. help appreciated..

If anyone can help with this... is there not a calculator function that will help??

Two examples:

In a lottery 3,000,000 tickets are sold at a dollar a piece. Out of these 12,006 are drawn at random without replacement and awarded prizes of 12,000 $25, 4$10,000 prize, 1 $50,000 prize and 1$200,0000 prize. What is the expected value of 1 per week.

Next one,

In a state lottery a three-digit integer is selected randomly. If a player bets $1, on a number, the payoff is$500 minus $1 paid for ticket. Let X equal the payoff of -1 - 499 and find E(X) Just a heads up I may need help on Bernoulli next. 2. Originally Posted by skyslimit Confused in this section.. help appreciated.. If anyone can help with this... is there not a calculator function that will help?? Two examples: In a lottery 3,000,000 tickets are sold at a dollar a piece. Out of these 12,006 are drawn at random without replacement and awarded prizes of 12,000$25, 4 $10,000 prize, 1$50,000 prize and 1 $200,0000 prize. What is the expected value of 1 per week. Hi there You should consider each prize and the probability of recieving thast prize. For$25: probabilty = 12,000/3,000,000

For $10,000: probabilty = 4/3,000,000 For$50,000: probabilty = 1/3,000,000

For $200,000: probabilty = 1/3,000,000 and finally if you don't win the prize is$-1 so

For \$-1: probabilty = (3,000,000-12,006)/3,000,000 = 2,987,994/3,000,000

Now multiply each return by its respective probabilty and sum the products together. This is the expected value.

$E(X) = 25\times\frac{12000}{3000000}+$

$10000\times\frac{4}{3000000}+$

$50000\times\frac{1}{3000000}$

$-1\times\frac{2987994}{3000000}$

$=-0.80$

On the second, I understand I must simply multiply the odds of of winning and losing (+- one dollar) by the winnings and loss. However, I can't seem to remember the odds of a random 3 digit integer... could it be simply would. Obvously 3! is too small... would it be, however, 27 * 27 * 27??

4. Originally Posted by cjmac87
So $E(X) = (500) \frac{1}{900} + (-1) \frac{899}{900} = -0.44$
So $E(X) = (500) \frac{1}{999} + (-1) \frac{899}{999} = -0.40$