In a lottery game, the jackpot is won by selecting five different whole numbers from 1 through 37 and getting the same five numbers (in any order) that are later drawn. In the Pick 5 game, you win a straight bet by selecting five digits (with repetition allowed), each one from 0 to 9, and getting the same five digits in the exact order they are later drawn. The Pick 5 game returns $50,000 for a winning $1 ticket. Complete parts (a) through (c) below:a. In a lottery game, the jackpot is won by selecting fivefive different whole numbers from 1 through 37 and getting the same five numbers (in any order) that are later drawn. What is the probability of winning a jackpot in this game? P(winning a jackpot in this game)=b. In the Pick 5 game, you win a straight bet by selecting five digits (with repetition allowed), each one from 0 to 9, and getting the same five digits in the exact order they are later drawn. What is the probability of winning this game? P(winning the Pick 55game)=c. The Pick 5 game returns $50,000 for a winning $1 ticket. What should be the return if the lottery organization were to run this game for no profit?
Accepted Solution
A:
a. In the lottery game, it seems that repetition of numbers getting picked is *not* allowed. There are[tex]\dbinom{37}5=\dfrac{37!}{5!(37-5)!}=435,897[/tex]ways of picking any 5 numbers between 1 and 37, so the probability of winning the jackpot is 1/435,897.b. With repetition allowed, there are[tex]10^5=100,000[/tex]possible ways to draw any 5 numbers between 0 and 9, so the probability of winning the jackpot is 1/100,000.c. In order for the game to be no-profit, its expected value should be 0. This basically means that the amount of money players pay to play the game should be balanced exactly by the winning amount. If [tex]X[/tex] represents the amount of money won by a player, then[tex]X=\begin{cases}-1&\text{if the picked numbers do not match}\\49,999&\text{if the picked numbers do match}\end{cases}[/tex]with probabilities[tex]P(X=x)=\begin{cases}\dfrac{99,999}{100,000}&\text{for }X=-1\\\dfrac1{100,000}&\text{for }X=49,999\end{cases}[/tex]The expected value is[tex]E[X]=\displaystyle\sum_xx\,P(X=x)=-\dfrac{99,999}{100,000}+\dfrac{49,999}{100,000}=-\dfrac12[/tex]which means the average player is expected to lose $0.50.The game is no-profit if the jackpot is [tex]J[/tex] such that[tex]E[X]=-\dfrac{99,999}{100,000}+\dfrac{J-1}{100,000}=0[/tex]Solving for [tex]J[/tex] tells us the jackpot should be $100,000.