Finding the minimal price of a lottery ticket

Question

In (simplified) lottergy game a number between 0 - 99 is drawn uniformly at random. If this number has exactly one digit as the corresponding digit on a lottery ticket then the organizers give 10 dollars to the person with the ticket. If the drawn number is the exact match, then the reward is 100 dollars. What minimal price the organizers should set per ticket?

Answer 1

OK, so I'm going to answer the question using several different options.

First off, note that there are ten $0$'s from $0-99$, and twenty of every other digit. This means that the maximum amount of money given away will occur when the number drawn is greater than $9$, does not contain a zero, and when the two digits are not the same.

For this case, there are $36$ numbers that have a matching digit.

From here, there are a couple options, for demonstration I'm going to use $98$ as the number drawn. The matches in this case are $$\left \{8,9,18,19,28,29,38,39,48,49,58,59,68,69,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99 \right \}$$

Option 1

Option 1 is that if someone holds a ticket with double number, such as $99$ or $88$, they win $\$20$ for having matched $8$ or $9$ twice. This would mean that someone with ticket $88$, $89$, or $98$ wins $\$20$. In this case, the amount of money given away ($B$) is $$B = (3*\$20)+(1*\$100)+(32*\$10) = \$480$$ So the amount they would have to set a ticket at to break even is $$\frac{\$480}{100} = \$4.80$$

Anything over $\$4.80$ would net a profit.

Option 2

In option 2, a person with $88$ or $99$ wins $\$10$ for matching only one of the digits of $98$. However, a person would receive $\$20$ for a ticket of $89$ since they matched both $9$ and $8$ once. In this case, the total given away is: $$B = (1*\$20)+(1*\$100)+(34*\$10) = \$460$$

So the breakeven ticket price in this case is $$\frac{\$460}{100} = \$4.60$$

Option 3

In the final option, a person receives $\$10$ for having a ticket with $88$, $89$, or $99$.

The amount given away is $$B = (1*\$100)+(35*\$10) = \$450$$

So the breakeven ticket price for option 3 is $$\frac{\$450}{100} = \$4.50$$