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# Calculate the Expected Value?

Calculate the expected value for playing mega millions with $1000 and winning a prize. Each ticket costing $1

PRIZE LEVEL- CHANCES OF WINNING PER- GAMEPRIZE

First- 1 in: 175,711,536- $88,000,000

Second- 1 in: 3,904,701- $250,000*

Third- 1 in: 689,065- $10,000*

Fourth- 1 in: 15,313- $150*

Fifth- 1 in: 13,781- $150*

Sixth - 1 in: 844- $10

Seventh- 1 in: 306- $7

Eighth- 1 in: 141- $3

Ninth- 1 in: 75- $2

### 2 Answers

- No BodyLv 410 years agoFavorite Answer
First find the expected value of one ticket, then multiply that value by 1000. Note that we should expect a negative value because the people/entities who fund these games like money:

EV = (1/175711536)(88000000) + (1/3904701)(250000) + (1/689065)(10000) + (1/15313)(150) + (1/13781)(150) + (1/844)(10) + (1/306)(7) + (1/141)(3) + (1/75)(2) + N

N is messy, so I'm leaving it separate. Its value is:

(1 - 1/75 - 1/141 - 1/306 - 1/844 ... etc)(-1)

So N will be slightly greater than -1, while all of the terms before N should sum to a positive number whose magnitude is slightly less than N's. Add everything together, then multiply the result by 1000 to get the expected value.

Note: The expected value is actually lower in reality since for larger winnings you have to pay tax, but we're ignoring that for this problem.

- ?Lv 45 years ago
expected value is the average possible value of x or the mean of a probablity distribution. if the probablity of x1 is p1 and x2 is p2 etc the formula is x1*p1+x2*p2+...xn*pn (the 1s, 2s and ns should be a subscript). to find the varience, find the expected value then use the formula (x1-expected value)^2*p1+(x2-expected value)^2*p2+...(xn-expected value)^2*pn the standard deviation is the square root of the varience. i hope the way i typed the formulas isnt too confusing.