Expected value help

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susumandrai

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The 20 tickets in a raffle with one cash prize cost $2.00 each.
(a) What should be the prize if the raffle is fair?
(b) What should each ticket cost if the prize is $80.00 and the expected gain to the organiser is $30.00?

Answers:
(a) For a fair game E(X) = 0 ; let prize be p
(p - 2) x (1/20) + (-2 x 19/20) = 0
40 = p
winning prize should be $40.

(b) - I dont know where to start :(
 
Where does the money go to? $30 goes to the organiser and $80 goes to the prize winner. If the raffle is fair, how much shout should the of the 20 raffle tickets cost?
 
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Jomo said:
Where does the money go to? $30 goes to the organiser and $80 goes to the prize winner. If the raffle is fair, how much shout the each of the 20 raffle tickets cost?
Click to expand...
Is it okay then to say that 80 + 30 =$110
$110 / 20 tickets = $5.50 :)

How about if I were to do this problem with a probability distribution table?
How would the values of the random variable look like, what what would their probabilities be?

Many thanks :)
 
Yes, $5.50/ticket to break even

How much do the 19 losers lose? What is the probability that you lose? How much do you get if you win? What is the probability that you win. Just use the usual formula.
 
Jomo said:
Yes, $5.50/ticket to break even

How much do the 19 losers lose? What is the probability that you lose? How much do you get if you win? What is the probability that you win. Just use the usual formula.
Click to expand...
19 losers lose -> 19 times price per ticket -> 19p ; and prob of losing = 19/20
If you win you get 80 - price of ticket -> 80 - p ; and prob of winning = 1/20

Does this seem correct?
Cheers
 
You did not get $5.50/ticket. In fact you get got any answer for the price of the ticket. Continue on and see if you can get $5.50/ticket.
 
Jomo said:
You did not get $5.50/ticket. In fact you get got any answer for the price of the ticket. Continue on and see if you can get $5.50/ticket.
Click to expand...
Thanks a lot, i think I got it :)

Please see here:

Since the prize is $80.00 and the expected gain to the organiser is $30.00, the ticket price has to compensate for $30 and $80 both.
Let the price of a ticket be 'p'.
Let the random variable X represent the gains for a player. Then this random variable will take on two values:
#1 -> (80 + 30) - p -> 110 - p ; with probability 1/20 [gain = winnings take away purchase price]
#2 -> -p ; with probability 19/20 [the player loses the money he bought the ticket for]

(110 - p) x 1/20 + (-p x 19/20) = 0 [for the game to be fair]
(110 - p) / 20 - 19p / 20 = 0
110 - p - 19p = 0
110 - 20p = 0
110 = 20p
110/20 = p
5.5 = p

Price of ticket has to be $5.50 for the game to be fair in which case the expected gain to the organiser is $30.

Hopefully my reasoning and commentary is correct.

Many thanks :)
 
I see 1 flaw in your work. You claim 110 - p ; with probability 1/20 [gain = winnings take away purchase price] . So if you win, you win 110-$5.50? Not true.

Please try again. Every line you write must be true!
 
How about now:

[gain = winnings take away purchase price]

Total gains = gain for the player + gain for the organiser

Since gain for the organiser is negative gain for winner:
gain for the player + gain for the organiser = 80 – p
gain for the player – 30 = 80 – p
gain for the player = 80 + 30 – p

(80 + 30) - p -> 110 - p ; with probability 1/20

what do you think :unsure:
 
Jomo said:
Continue working it out.

You do seem to have many definitions for gain?
Click to expand...
I think I got it now :)
Please see revised version here
[gain = winnings (for there to be a winner, $80 has to go to winner and $30 to organiser) take away purchase price]
#1 -> (80 + 30) - p -> 110 - p ; with probability 1/20

What about now :unsure:
 
Lets start here. Let X be the random variable which states how much you can win (lose).
Let p = amount you pay for a ticket.
There are two values for X. X = 80-p or X=-p.
So what is the expected value for X. Just use the expected value formula, set it equal to 0 and solve for p.
 
Jomo said:
Lets start here. Let X be the random variable which states how much you can win (lose).
Let p = amount you pay for a ticket.
There are two values for X. X = 80-p or X=-p.
So what is the expected value for X. Just use the expected value formula, set it equal to 0 and solve for p.
Click to expand...

(80 - p) × 1/20 + -p × 19/20 = 0
(80 - p)/20 - 19p/20 = 0
80 - p = 19p
80 = 20p
4 = p

But this will give me only $4.
How do I factor in $30 within the probability distribution.
Thanks :)
 
susumandrai said:
(80 - p) × 1/20 + -p × 19/20 = 0
(80 - p)/20 - 19p/20 = 0
80 - p = 19p
80 = 20p
4 = p

But this will give me only $4.
How do I factor in $30 within the probability distribution.
Thanks :)
Click to expand...
That is a good point! I need to think about this one! I can tell you that the missing$30/20 gives you the missing $1.50
 
Let p + $1.50 = cost of a ticket.
x1 = $80 - (p + $1.50)
x2 = -(p+1.50).

Now try it!
 
Jomo said:
Let p + $1.50 = cost of a ticket.
x1 = $80 - (p + $1.50)
x2 = -(p+1.50).

Now try it!
Click to expand...

$80 - (p + $1.50) × 1/20 + -(p+1.50) × 19/20 = 0
80 - p -1.5 × 1/20 + (-p - 1.5) × 19/20 = 0
78.5 - p + 19(-p - 1.5) = 0
78.5 - p - 19p - 28.5 = 0
50 = 20p
2.5 = p

cost of ticket = 2.5 + 1.5 = $4 :(
 
susumandrai said:
??
Click to expand...
Keep it simple. Start with the basic equation:
Gain = Tickets - Prize

Apply Expected Operator:
E[Gain] = E[Tickets] - E[Prize]
80 = 20p - 30
p = 110/20 = 5.50

Same idea for (a).
E[Gain] = E[Tickets] - E[Prize]
0 = 2(20) - E[Prize]
E[Prize] = 40
 
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