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(52776)   (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?
~~~~~~~~~~~~~~~~~~~~~~ 

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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If you want to see other similar and different problems on mathematical expectations
playing games, for entertainment or for learning, look into these lessons
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    - Math expectation of winning in lottery problems
    - Math expectation of winning in games with rolling pair of dice
in this site.




Answer by math_tutor2020(3816)   (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
xP(x)x*P(x)
85980.0000670.576066
6580.0001330.087514
540.0002670.014418
180.0006670.012006
-20.998867-1.997734
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
xP(x)x*P(x)
85980.000066670.57322866
6580.000133330.08773114
540.000266670.01440018
180.000666670.01200006
-20.99886667-1.99773334
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.
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