MHB Problem of the week #61 - May 27th, 2013

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The Powerball lottery requires players to match 6 numbers drawn from two ranges: 1-59 for the first five numbers and 1-35 for the final number. The odds of winning the grand prize are extremely low, and a ticket costs $2. For playing to be a rational investment, the grand prize must exceed the expected return, which is calculated based on the total number of tickets sold. Additionally, if 10 million tickets are sold, the probability of being the sole winner is significantly low, further emphasizing the game's risk. Overall, the discussion highlights the mathematical challenges and considerations involved in playing the lottery.
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The Powerball lottery game pays a large cash prize to anyone who correctly matches all 6 winning numbers that are drawn at random each week. For the first 5 spaces you can choose numbers from 1-59 (no repeats) and then there is 1 space where you can choose a number from 1-35. The order of the 6 numbers you choose this way doesn't change anything. 123456 is the same ticket as 213456.

1) What are the chances of winning the grand prize by correctly guessing all 6 numbers?
2) Assuming that each ticket costs $2, how big must the grand prize be in order for the ticket to not be an investment with an expected negative return? In other words, when does it become logical to play the lottery (assuming just 1 grand prize winner)?

Bonus: If it is known that 10,000,000 lottery tickets will be purchased in addition to your ticket, what is the probability that you are the only winner of the grand prize and get to keep all the money for yourself? :)
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone

Partial solution:

1) mathworker

Solution (from MarkFL):
1.) In order to win, one must correctly guess the random number (event A) and correctly guess the 5 numbers which are not repeated (event B). Let X denote the event of winning. Since both event A and event B must occur for event X to take place, we may write:

$P(X)=P(A)\cdot P(B)$

For event A, there is one favorable outcome and 35 total outcomes, hence:

$\displaystyle P(A)=\frac{1}{35}$

For event B, there is one favorable outcome, and the total number of outcomes is the number of ways to choose 5 from 59, hence:

$\displaystyle P(B)=\frac{1}{{59 \choose 5}}$

Thus, we may conclude:

$\displaystyle P(X)=\frac{1}{35{59 \choose 5}}=\frac{1}{175223510}$

2.) Let J be the jackpot amount in dollars. In order for the expected profit to be non-negative, we require:

$\displaystyle -2\cdot\frac{175223509}{175223510}+J\frac{1}{175223510}\ge0$

$\displaystyle J\ge350447018$

Thus, we find it becomes logical to play the lottery (assuming just 1 grand prize winner) when the jackpot amount is at least \$350,447,018.

Bonus:

In order for the other 10,000,000 tickets to be different from yours (event Y), we observe each of the other tickets can contain any combination except for the one you have chosen, hence:

$\displaystyle P(Y)=\left(\frac{175223509}{175223510} \right)^{10^7}\approx0.9445279805064648$ [/I]

Thus, the probability that you win AND the other 10,000,000 tickets are different from yours is:

$\displaystyle P(X)\cdot P(Y)=\frac{1}{175223510}\cdot\left(\frac{175223509}{175223510} \right)^{10^7}\approx5.390418103749119\,\times\,10^{-9}$
 
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