Throughout these calculations, you can presume that the first five numbers are selected independently from the Mega Ball number. However, the first five numbers are selected without replacement so computing probabilities with those does not allow for independence. This part is hypergeometric with the $n_1 = 5$ numbers you selected being the "desired" numbers and the Lottery Commission picking a subset of size r = 5 from the 75 possible numbers. So, your likelihood of matching all five would be

\begin{equation*} \frac{\binom{5}{5} \cdot \binom{70}{0}}{\binom{75}{5}} = \frac{1}{17259390}. \end{equation*}

Multiplying this by the 1 chance in 15 that you also match the Mega Ball gives

\begin{equation*} P(\text{Match 5 plus Mega Ball}) = \frac{1}{17259390} \cdot \frac{1}{15} = \frac{1}{258,890,850}. \end{equation*}

To match only 5 means you also MUST miss the Mega Ball which has probability 14/15 to give

\begin{equation*} \frac{1}{17259390} \cdot \frac{14}{15} = \frac{1}{17259390 \cdot \frac{15}{14}} \approx \frac{1}{18492204}. \end{equation*}

Continue in this manner to determine the other odds.

For the expected earnings, first determine a value function corresponding to each outcome and apply the discrete expected value process. This gives

\begin{align*} & \$32600000 \cdot \frac{1}{258,890,850} + \$1000000 \cdot \frac{1}{18,492,204} \\ & + \$5000 \cdot \frac{1}{739,688} + \$500 \cdot \frac{1}{52,835}\\ & + \$50 \cdot \frac{1}{10,720} + \$5 \cdot \frac{1}{766} \\ & + \$5 \cdot \frac{1}{473} + \$2 \cdot \frac{1}{56} + \$1 \cdot \frac{1}{21}\\ & \approx \$0.3013. \end{align*}

So, the expected payout is approximately 30 cents. Subtracting the cost of playing ($1) indicates that the average winnings per play of the Louisiana Lottery would be -70 cents. So, you would be better off to take, say, 50 cents and just give it to the local school system every time you consider playing this game rather than actually playing.

To determine the Jackpot A needed to make this a fair game means to solve the equation

\begin{align*} & A \cdot \frac{1}{258,890,850} + \$1000000 \cdot \frac{1}{18,492,204} \\ & + \$5000 \cdot \frac{1}{739,688} + \$500 \cdot \frac{1}{52,835}\\ & + \$50 \cdot \frac{1}{10,720} + \$5 \cdot \frac{1}{766} \\ & + \$5 \cdot \frac{1}{473} + \$2 \cdot \frac{1}{56} + \$1 \cdot \frac{1}{21}\\ & = 1 \end{align*}

for A.

Finally, to deal with the multiplier, note that all but the Jackpot payouts would be increased by the multiplier m where $m \in \{1,2,3,4,5\}\text{.}$ For the cost of an extra $1 (total cost of $2 per bet) the expected payout increases as the multiplier increases but each of these decreases likelihood of winning that payout by a factor of 1/5. In general, let x = 1, 2, ..., 9 indicate the various winning options in order listed above, f(x) the corresponding probabilities listed for each option, and u(x) the listed payouts. Then the expected payout is given by

\begin{equation*} \$32600000 \cdot \frac{1}{258,890,850} + \sum_{m=1}^5 \sum_{x=2}^9 m \cdot u(x) f(x)/5 \end{equation*}

\begin{align*} \$32600000 \cdot \frac{1}{258,890,850} & + \sum_{m=1}^5 \frac{m}{5} \sum_{x=2}^9 u(x) f(x) \\ & = \frac{\$32600000}{258890850} + \sum_{m=1}^5 \frac{m}{5} 0.17539\\ & = 0.12592 + 3 \cdot 0.17539\\ & = 0.65209 \end{align*}

Therefore, the expect value of spending another dollar to get the multiplier effect is about -$1.35. Since this is slightly less than doubling the expected loss of 70 cents for playing without the multiplier with $1, it make more sense to bet $2 once rather than betting $1 twice. Or, you can send the extra nickel to this author of this text and call it quits.