Question

A raffle has a grand prize of a Caribbean cruise valued at $10,500 with a second prize of a weekend in San Diego valued at $1,300. If each ticket costs $2 and 9600 tickets are sold, what are the expected winnings for a ticket buyer? Express to at least three decimal place accuracy in dollar form (as opposed to cents).

Answer 1

This question is confusing. First, "Express to at least three decimal place accuracy in dollar form (as opposed to cents)" does not make sense. If you go to 3 decimal places not only are you expressing in dollars AND cents, but factions of cents as well. Also, I'm not sure what is meant by "expected winnings." This would ultimately depend on how many tickets a buyer bought. We could assume 1 per person, but it does not say that, I wish I had a better answer, but I don't understand the question. Let me try to get a few more minds on this.

Answer 2

It's how my teacher phrased it which has me confused. Expected winnings is for statistics or probability I don't know if that would help and I think it is assumed that each person only buys 1 ticket.

Answer 3

Well at least that part makes more sense now. Still don't know about dollars expressed to the 3rd decimal place though.

Answer 4

yes, the phrasing is horrible

Answer 5

Yeah, I know he's really horrible at phrasing which makes everything more confusing

Answer 6

Here I think I found something that you will find helpful. http://www.statisticshowto.com/articles/how-to-figure-out-an-expected-value-in-statistics/

Answer 7

this is my idea:

considering that there is no third prize, therefor, we can claim that from the given, all ticket buyers will win consolation prices, except for the winner of first price and second price. .

but, do you think the promoter will to distribute all the income to the ticket buyers as winnings?

Answer 8

@jagatuba I used that! and got -0.770 out of it but when I entered it it claimed to be wrong.

Answer 9

We cannot claim that because it is not explicitly stated in the problem that there are any other prizes other than the two listed.

Answer 10

Hmm. I stink at statistics.lol

Answer 11

@jagatuba ,then how would you explain this," what are the expected winnings for a ticket buyer?"

Answer 12

I guess it's about probability

Answer 13

@jerwyn gayo it's statistics so it's all about probability. You can win and gain x or you could lose and lose y.

Answer 14

Okay let me think about this. (please don't make me pull out my stats book. please don't make me pull out my stats book.)

Answer 15

therefor, 9600(2) = x : where 2 is the price for every ticket and 9600 is total sold ticket

9600-2 (the two winners for 1 and 2 prize) = y

the expected winning for each ticket buyer is x/y=?


lol, my stat sucks,. correct me if i'm wrong. .

Answer 16

so you would just subtract 2 from 9600?

Answer 17

Hmm. I get -0.770 too. there must be something we are missing.

Answer 18

10500*(1/9600)+1300*(1/9600)+(-2*(9598/9600))=-0.770

Answer 19

If I do it another way with another x value equalling 0-1 with the probability outcome being 9598/9600 it'll change to -0.76 but that was wrong too

Answer 20

But according to the definition of expected value, this should be correct.

Answer 21

Exactly which is why I'm confused. Let me try putting it in with the negative value

Answer 22

Yes it should be negative

Answer 23

What if we times the -.770 by the number of ticket holders?

Answer 24

no that wouldn't be right. -7392? No I don't think so.

Answer 25

I just had a message pop up and it says Message: Your answer isn't a number (it looks like a list of numbers)

Answer 26

Hiding away lol

Answer 27

lol

Answer 28

if it looks like "a list of numbers" it could be for each value instead of the total value?

Answer 29

If we take it one prize at a time we get an expected gain of
$1.093 for the first prize
AND
$0.135 for the second
FOR
$1.228, but I know this is not how you figure it.

Answer 30

It's almost quitting time. When I get home I'm going to crack open the statistics book (BLEH) and take another crack at it.

Answer 31

I have to go also, thank you for your help though!

Answer 32

Well hopefully I or someone much greater than I will have an answer to this question by tomorrow.

Answer 33

Hopefully.

Answer 34

Okay. Consulted my statistics text book and it explained how to figure this a little bit different than that website.

Take your prize amounts, add them then subtract the number of tickets sold minus the number of prizes times the cost per ticket and divide the whole thing by the total number of ticket sold, which gives you the equation:
\[\frac{10500+1300-(9598\times2)}{9600}=-0.770416667 \]OR
-$0.770, which is still the same answer that we have been coming up with.

So, the best I can figure is that either the answer key is wrong or it is looking for a specific format to be entered. Perhaps this has to do with the confusing "Express to at least three decimal place accuracy in dollar form (as opposed to cents)" part.

Suggestion: Try entering every conceivable form of the answer and see if any work:
-0.770
-0.77
-$0.770
-$0.77
$-0.770
$-0.77
If those don't work you can try all of those without the leading zero (-.770, etc.), but I really don't think those will work. Other than that, the next best thing to do is ask your instructor for help. There has to be something we are missing.

Answer 35

Hi! I went to a tutor and we worked it out and it's -0.771 not 0 in case you guys were wondering!

Answer 36

That's weird. Why round to -0.771 because to the 4th digit it is -0.7704?