SOLUTION: A raffle is being held at a benefit concert. The prizes are awarded as follows: 1 grand prize $8600.00 , 2 prizes of $660.00 , 4 prize of $56 and 10 prizes of $20 Suppose 15

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Question 1206326: A raffle is being held at a benefit concert. The prizes are awarded as follows: 1 grand prize $8600.00
, 2 prizes of $660.00 , 4 prize of $56 and 10 prizes of $20
Suppose 15000 raffle tickets are sold, if you buy one ticket for $2.00 then what is your expected value for this raffle?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52903)   (Show Source):
You can put this solution on YOUR website!
.
A raffle is being held at a benefit concert. The prizes are awarded as follows: 1 grand prize $8600.00,
2 prizes of $660.00, 4 prize of $56 and 10 prizes of $20.
Suppose 15000 raffle tickets are sold, if you buy one ticket for $2.00
then what is your expected value for this raffle?
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The probability to win 1 grand prize is ; the probability to win one of the two prizes of $660 is ; the probability to win one of the four prizes of $56 is ; the probability to win one of the ten prizes of $20 is . Therefore, the expected value of the raffle is E = - 2 = = -1.3104 (precise value !). (the typo is just fixed) It means that playing this game many times, the gamer will lose about $1.31 per game, in average.

Solved.

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Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

Answer: -1.31 dollars
It means you should expect to lose about $1.31 per ticket.

Explanation

Let's say you really wanted to win.
To fully guarantee a win, you'd have to buy all 15,000 tickets.
That would cost you 15000*2 = 30,000 dollars.
Buying all of the tickets would bring in 1*8600+2*660+4*56+10*20 = 10,344 dollars.

total cost = $30,000
winnings = $10,344

net = winnings - cost
net = $10,344 - $30,000
net = -19656 dollars
The negative amount means you'd lose money.

Divide that net amount over the total number of tickets.
-19656/15000 = -1.3104
You should expect to lose, on average, about $1.31 per ticket.

Tutor ikleyn has the right idea but made a slight typo.
The 600 should be 660.
Edit: The situation has been fixed.

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Another approach.
This is probably the more standard approach to expected value problems.

x = net winnings (after factoring in the cost of the ticket)
Example: x = 8600-2 = 8598 if you won the grand prize

x P(x) x*P(x)
8598 0.000067 0.576066
658 0.000133 0.087514
54 0.000267 0.014418
18 0.000667 0.012006
-2 0.998867 -1.997734
Sum -1.30773

Each decimal value is approximate to 6 decimal places.
1/15000 = 0.000067
2/15000 = 0.000133
4/15000 = 0.000267
10/15000 = 0.000667
14983/15000 = 0.998867

Spreadsheet software is strongly recommended. I used LibreOffice.
Add up the items in the x*P(x) column to get roughly -1.30773
There's some rounding error since the result should be exactly -1.3104

If we increase the rounding precision, say to 8 decimal places, then this is what the table would look like

x P(x) x*P(x)
8598 0.00006667 0.57322866
658 0.00013333 0.08773114
54 0.00026667 0.01440018
18 0.00066667 0.01200006
-2 0.99886667 -1.99773334
Sum -1.3103733

The result -1.3103733 is a bit more closer to -1.3104
Either way both results round to -1.31 dollars when rounding to the nearest cent.