The current Powerball jackpot is at roughly 675 million USD and the chances of winning with one random ticket is 1 in 292.2 million. Each ticket costs 2 USD.
From a general perspective, it appears that the Powerball lottery tickets have a positive expected value but we didn't include all the other factors yet.
If we decide to take the entire winnings at once instead of getting paid over 30 years, we should be expecting to receive 428 million before taxes. Then we have to include both state and (25%) federal tax which shaves off at least another 100 million for the United States.
Then we would also have to include the possible of the Powerball having multiple winners. We could estimate how many people will purchase tickets for the next drawing based on how many purchased for the last one.
Lastly, we would also have to factor in the cost of maintenance for this large amount of money.
How can we go about calculating our expected value when purchasing a ticket in hopes of hitting the jackpot? Do these tickets truly give us a positive return at its current price?
Best Answer
This is a tricky problem. The lottery people would love for you to think of the problem simplistically so you arrive at the wrong answer. However, a careful analysis shows why the lottery people really know why they will make tons of money from even a huge payout.
Let $p$ be the probability of winning, $C$ be the cost of a ticket, and $V$ be the value of the winnings. Then the expected value $E$ of a ticket, assuming one winner, would be approximately
$$E = p V - (1-p) C$$
However, this is really not correct. In reality, there will be more than one winner. Or none. Who knows? But when there are more than one winner, the value of the the winnings to each person are reduced, as the winnings are split evenly among the winners.
It may be assumed that the number of winners follows a binomial distribution. Assume a population of $N$ possible tickets. The probability of $k$ tickets, including yours, being winners is
$$\binom{N-1}{k-1} p^k (1-p)^{N-k} $$
where $k =1$ corresponds to the ideal case in which you alone are the winner, and k may vary between $1$ and $N-1$. so that the actual expected value of your winning ticket is equal to
$$E = \sum_{k=1}^{N} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} \frac{V}{k} - (1-p) C $$
which may be simplified to
$$E = \frac{1-(1-p)^N}{N} V - (1-p) C $$
Note this takes into account the number of tickets purchased and will reduce the expected value of the ticket from the simpler assumption.
Given the numbers: $p$ being $1$ in $282.2$ million, $V=\$700$ million, and $C=\$2$, this distinction is crucial. The expected value for the simple case $N=1$ is positive (about $ \$0.396 $); people who understand the expected value at this level may be induced to buy a ticket, thinking that each ticket has positive value. However, one may show that, when there are more than about $108.8$ million tickets sold, the expected value goes negative. My guess is that the number of tickets sold will certainly exceed this number and that the lottery people will make a profit.