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?

Source for Powerball statistics

  • $\begingroup$ Federal tax on that prize would be more like either 39.6% or 43.1% depending on whether or not the Net Investment Income surcharge applies to lottery winnings. It is definitely not 25%. $\endgroup$ Commented Jan 8, 2016 at 16:21
  • $\begingroup$ In Georgia, they tax/deduct directly on payout: 6% state and 25% federal when you claim your ticket. Then, you are hit with another 15% federal when you do your taxes next spring. So basically 46% total state and federal when all is said and done. $\endgroup$
    – nanonerd
    Commented Apr 17, 2018 at 19:33
  • $\begingroup$ what is the "cost of maintenance for this large amount of money"? $\endgroup$
    – nanonerd
    Commented Apr 17, 2018 at 19:34

1 Answer 1


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.

  • $\begingroup$ Interesting. In 2012, a 550 million jackpot had sold 130,000+ tickets. Assuming there is only 1 winner, the range of what N could realistically be is negligible. I switched my V to be $525 million to account for federal tax and the expected value becomes negative 20 cents :( $\endgroup$
    – krikara
    Commented Jan 8, 2016 at 12:37
  • $\begingroup$ The point is that you cannot assume that there will be only one winner. History affirms this - there have been no single winners in the past several mega-jackpots. So you have to take this into account, and the result is a dependence on the number of tickets sold. And as you would expect, the expected value decreases markedly with the number of tickets sold. Your data point of 2012 of only $130,000$ tickets sold is very curious - the lottery would lose a lot of money unless the prob of winning were decreased to the point of ludicrousness. $\endgroup$
    – Ron Gordon
    Commented Jan 8, 2016 at 12:41
  • $\begingroup$ Yes many lotteries have multiple winners. I just found it sad that even with 1 winner, the expected value was still negative. Ahh yes, I misread the Google caption. In the waning hours of the past lottery, 130,000 tickets were being sold by the minute. $\endgroup$
    – krikara
    Commented Jan 8, 2016 at 12:54
  • 1
    $\begingroup$ (+1) Although you say "Assume a population of $N$ possible tickets", you seem rather to treat $N$ as the number of tickets sold (each acting like a Bernoulli trial). The number of possible (distinct) tickets, on the other hand, is of course a constant -- viz., ${69\choose{5}}26 = 292,201,338$ for the current PowerBall game. (Note your typo in "$282.2$ million".) More than $10^9$ dollars was reported in ticket sales, so presumably $N$ would be somewhat less than half of that number (due to the "PowerPlay" option of paying $3$ (instead of $2$) dollars per ticket, for enhanced winnings). $\endgroup$
    – r.e.s.
    Commented Jan 10, 2016 at 18:25
  • 1
    $\begingroup$ One issue with this analysis is that it assumes the EV for the ticket is only for the main jackpot. The chances of winning the smaller prizes are not insignificant and these are never shared. So the calculation must be done on all the other prizes too and summed in order to get the true EV. $\endgroup$
    – davmp
    Commented Oct 22, 2018 at 19:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .