I have a problem where I'm supposed to determine if a lottery system is profitable. I solved the problem and found it to be profitable, but I am not 100% sure about all of my calculations. Below is the rules of the game, and my calculations.
There is a total of $30$ numbers to draw from and $6$ will be drawn (order does not matter). There's also $3$ colored balls and all three will be drawn (order does matter). Each ticket sold is unique, and every ticket is sold. Tickets cost $\$1$. The jackpot does not play a role in the payout cost. Assume the jackpot is $\$0$.
- Any ticket with $4$ matching numbers gets $\$10$.
- Any ticket with $4$ matching numbers and the matching color sequence gets $\$50$.
- Any ticket with $5$ matching numbers gets $\$1,000$.
- Any ticket with $5$ matching numbers and the matching color sequence gets $\$5,000$.
- Any ticket with $6$ matching numbers gets $\$100,000$.
- Any ticket with $6$ matching numbers and the matching color sequence gets the jackpot.
First step was to find the total possible tickets sold. The problem states that every ticket is sold, and every ticket is unique, so:
Combinations of numbers is $\left( {\begin{array}{*{20}c} 30 \\ 6 \\ \end{array}} \right)$ which is $593775$.
For color balls, order matters, so it's $3!$ which is 6. So, the total possible number of tickets is $3!\left( {\begin{array}{*{20}c} 30 \\ 6 \\ \end{array}} \right)$ which is $3562650$ meaning the lottery will bring in a total of \$3,562,650.00.
Now determine the money that must be paid out. Since every ticket is sold, every possible winning ticket must be added to the total payout. These are the calculations I'm not entirely sure about.
4 number winners:
From Wikipedia, the formula to find the particular number of winning tickets with 4 correct numbers is: $$\left( {\begin{array}{*{20}c} 6 \\ 4 \\ \end{array}} \right)\left( {\begin{array}{*{20}c} 24 \\ 2 \\ \end{array}} \right)$$
The result is 4140, however 6 of these also have the correct color combination, so the total of \$10 winning tickets is 4134, costing \$41,340. The remaining 6 tickets with correct color combinations will cost \$300.
5 number winners:
The formula for this case would be: $$\left( {\begin{array}{*{20}c} 6 \\ 5 \\ \end{array}} \right)\left( {\begin{array}{*{20}c} 24 \\ 1 \\ \end{array}} \right)$$
The result is 144, but again 6 of these have a correct color combination so the total number of \$1,000 tickets is 138 costing \$138,000 and the reaming 6 tickets will cost \$30,000.
Finally, due to the nature of combinations there can only be one ticket with all 6 numbers, however plugging into the formula before also results in 1. The payout for this ticket is \$100,000. This is the part that's confusing me though and makes me think I did the other parts wrong. If every ticket is sold and every ticket is unique, doesn't that mean there must be a jackpot winning ticket? The math shows that there can only be one ticket with all 6 numbers, but that doesn't guarantee that it will have the correct color combination. So, someone can win but not win the jackpot, but the question originally states that every ticket is sold...Are there multiple tickets with correct numbers?
Anyways, you get the total payout by adding the above payouts which comes to \$309,640 which is far less than the total income of \$3,562,650 so the lottery is profitable.
If anyone spots an error I'd appreciate it as I am not 100% sure I did this correctly. There might also be a better way to calculate it.
Your calculation of $4140$ ways to choose exactly four correct numbers is fine, but you will then sell $6 \cdot 4140$ tickets with four correct numbers. $4140$ will have the correct order of balls and $5 \cdot 4140=20700$ will not. You have a similar problem with five and six correct numbers.