Practice Exam - Lottery question (expected winnings)

justjuice

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Hi everyone! First post here :) I'm going through a practice exam without solutions at the moment, and came across a statistics question I couldn't quite figure out:

3,000 numbers are drawn from a pool of 200,000 for a lottery - and each of the 3000 is drawn without replacing any, and recorded on a piece of paper. One ticket is then drawn from the pool of 200,000 after the 3,000 are returned, and if this 1 matches any of the previously drawn 3000, the prize is awarded.If no prize is won (the one drawn is not any of the initial 3000), and another is drawn. This is repeated until there is a winner.

The question has four parts, and I can't get the last part: What is the expected winning of this jackpot lottery? I believe is related to E(X) (expected values) or Expected Value of Binomial distribution. If someone could point me in the right direction, that would be greatly appreciated.

There are three sub-questions before it, which may provide some context:
What is the probability than a prize is won for any draw? 3,000/200,000

Jackpot is initially $15,000, and goes up by 15,000 each time there fails to be a winner. What is the chance that the prize will exceed winnings of $375,000?
There must be no winner 25 times - and thus, there will be 26 draws. (197/200)^25 * (3/200)

Suppose X is a random variable signifying the number of draws until there is a winner. What is P (X = k)? What is the total prize money won in this instance? P(X=k) = (197/200)^ (k-1) * (3/200)

Thanks in advance!
 
Last edited:
Hi everyone! First post here :) I'm going through a practice exam without solutions at the moment, and came across a statistics question I couldn't quite figure out:

3,000 numbers are drawn from a pool of 200,000 for a lottery - and each of the 3000 is drawn without replacing any, and recorded on a piece of paper. One ticket is then drawn from the pool of 200,000 after the 3,000 are returned, and if this 1 matches any of the previously drawn 3000, the prize is awarded.If no prize is won (the one drawn is not any of the initial 3000), and another is drawn. This is repeated until there is a winner.

The question has four parts, and I can't get the last part: What is the expected winning of this jackpot lottery? I believe is related to E(X) (expected values) or Expected Value of Binomial distribution. If someone could point me in the right direction, that would be greatly appreciated.

There are three sub-questions before it, which may provide some context:
What is the probability than a prize is won for any draw? 3,000/200,000

Jackpot is initially $15,000, and goes up by 15,000 each time there fails to be a winner. What is the chance that the prize will exceed winnings of $375,000?
There must be no winner 25 times - and thus, there will be 26 draws. (197/200)^25 * (3/200)

Suppose X is a random variable signifying the number of draws until there is a winner. What is P (X = k)? What is the total prize money won in this instance? P(X=k) = (197/200)^ (k-1) * (3/200)

Thanks in advance!
1) you have made the assumption that 3000 << 200000, so that probability p=3/200 is constant. That approximation allows you to use the binomial distribution - if you never compute N any larger than 26, that MAY be ok...

2) the word "exceeds" means "at least 26", so all you need is for the first 25 to be false. Evaluate (197/200)^25.

3) Here might be a place to show off using factorials to simplify the product of all probabilities up to k.
P(k) = P(not1) * P(not2) *. . . * P(not(k-1)) * P(k)
......= (197000/200000) * (197000/1999999) * . . . * (197000/(200000-k)) * 3000/(200001-k)
......= . . .
oops - I used P(k) to mean two different things! All probabilities on the right side of the equation are conditional, with the assumption that all preceding conditions are met. The P(k) on the left is unconditional.

PrizeMoney(k) = $15,000 * k

4) E[PrizeMoney] = $15,000 * Sum(k * P(k))
I don't think you can use the binomial distribution because each term in the sum is from a different number of trials, k.
The APPROXIMATION will be a geometric series, with the first term being 3/200 and each successive term being multiplied by (197/200). You can find the formula for the sum of k terms of such a series.
The more precise version of that procedure uses a varying multiplier. Could be calculated numerically in a spreadsheet. Keep a running sum for E[k], till the terms become insignificant.
 
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