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

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Jameson
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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:

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

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

What is the problem of the week #61 and when was it posted?

The problem of the week #61 was posted on May 27th, 2013. It is a mathematical problem that challenges individuals to find the missing number in a sequence of numbers.

How difficult is problem of the week #61?

The difficulty of the problem of the week #61 depends on an individual's mathematical abilities. Some may find it easy, while others may find it challenging. It is designed to be a fun and engaging problem for all levels of mathematical proficiency.

Can I get help or hints for solving problem of the week #61?

Yes, help and hints are often provided in the comments section of the problem of the week #61 post. You can also seek assistance from other mathematicians or use online resources to guide you through the problem.

Is there a time limit for solving problem of the week #61?

There is no official time limit for solving problem of the week #61. However, it is recommended to submit your answer within a week of the problem being posted to stay up-to-date with the challenge.

Are there any prizes for solving problem of the week #61?

At times, there may be prizes offered for solving problem of the week #61. However, the main goal of the problem is to exercise and improve your mathematical skills, rather than to win a prize.

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