Question

What is the expected value when a $1 lottery ticket is bought in which the purchaser wins exactly $10 million if the ticket contains the six winning numbers chosen from the set {1, 2, 3,…, 50} and the purchaser wins nothing otherwise?

Answer 1

We expect to lose $0.37 per lottery ticket

Explanation:

six winning numbers from = { 1, 2, 3, ....., 50}

So, the probability of winning:

$P(win) = \frac{ no of favorable outcomes}{no of possible outcomes}$

$P(win) = \frac{1}{^5^0C_6} \\\\P (win) = \frac{6! X (50 - 6)!}{50!} \\\\P(win) = \frac{6! X 44!}{50!} \\\\P(win) = \frac{1}{15,890,700}$

The probability of losing would be:

P(loss) = 1 - P(win)

$P(loss) = 1 - \frac{1}{15,890,700} \\\\P(loss) = \frac{15,890,699}{15,890,700}$

According to the question,

When we win, then we gain $10 million and lose the cost of the lottery ticket.

So,

$10,000,000 - 1 = $9,999,999

When we lose, then we lose the cost of the lottery ticket = $1

The expected value is the sum of the product of each possibility x with its probability P(x):

E(x) = ∑ xP(x)

$= 9,999,999 X \frac{1}{15,890,700} + ( -1 ) X \frac{15,890,699}{15,890,700} \\\\=- \frac{5,890,700}{15,890,700} \\\\= - \frac{58,907}{158,907} \\\\= - 0.37$

Thus, we expect to lose $0.37 per lottery ticket