Expected Net Winnings from 1,000 Raffle Tickets

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The discussion revolves around calculating the expected net winnings from purchasing one raffle ticket in a scenario where 1,000 tickets are sold for $100 each, with a prize of $75,000. The expected value calculation indicates that if a ticket is purchased, the average loss is $24.90, considering the probabilities of winning and losing. However, the net winnings for the actual winner would be $74,900 after deducting the ticket cost. Clarification on the wording of the question highlights the distinction between expected values for all participants versus the winner's net winnings. Ultimately, the expected net winnings for the winner, based on the calculations, is determined to be -$25 when factoring in the probabilities.
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Homework Statement


1,000 raffle tickets are being sold for $100 each. The winning prize is worth $75,000. Find the expected net winnign for the winner if he/she buys one ticket.


Homework Equations





The Attempt at a Solution



I'm not exactly sure what the "expected" net winnings mean. I've been searching on the internet, but came up empty.
Thanks for any help you can give me on this.
 
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So I'm thinking this is the equation:
(-$100 * 999/1000) + ($75,000 * 1/1000)

That makes the answer: -24.9

I'm not sure if this is correct.
 
This gives that a player on average will lose $24.90 on a $100 dollar bet.

Your questions asks for the expected net winnings for the winner so this is not the same thing. I was assuming this question was really about expected values and as you had had difficulty in finding anything about this I gave you the link.

Actually as worded the winner wins $75,000 on a $100 bet so his/her net winnings are $74,900.

Suggest you look again at the wording of the question wherever it came from.
 
What? The problem says expected net winnings. If he wins, he wins 75000 so his net winnings are $75000- 100= $74900. If he loses, his net "winnings" are -$100. The probability he will win is 1/1000 (one ticket out of 1000) and the probability he will lose is 999/1000. His expected net winnings are $74900(1/1000)- $100(999/1000)= (74900- 99900)/1000= -25000/1000= -$25.
 
HallsofIvy said:
What? The problem says expected net winnings.

The problem as stated by OP is the expected net winnings for the winner .

I was querying whether this was what was actually asked.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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