# Winning Powerball Tickets in Arizona and Missouri: How to Calculate the (Slim) Odds

Do you have a Powerball Ticket? Image by Upupa4me

The winning Powerball tickets were sold in Arizona and Missouri – if you didn’t win, you may be kicking yourself for wasting money on a ticket, but the odds of winning are always better if you buy in.

What are those odds, anyway, and how do we calculate them?

## Calculating the Powerball Odds

It’s easy to learn the poor odds of becoming incredibly wealthy by winning the Powerball lottery, but just how do they calculate the odds of winning the ultimate get-rich-quick scheme?

The current rules for a Powerball lotto’s grand prize win are fairly simple. The winning ticket must predict the values of five white balls chosen without replacement from a group of 59, and also predict the value of one red ball chosen from a group of 35.

The official Powerball site gives the probability of winning the \$2 play as “1 in 175,223,510” to match all 5 white plus 1 red ball.

They don’t show how they calculate the number of possible draw combinations:

• There are ( (59*58*57*56*55)*35 )/ (5!) = ( (59*58*57*56*55)*35 )/ (5*4*3*2*1)  combinations.
• According to my computer’s calculator, this equation equals 175,223,510.

The probability of winning is therefore one out of all the possible combinations, or 1/175,223,510.

What does this calculation of combinations mean?

## The Formula for the Probability of a Lucky Draw in a Lottery

The general  formula for the number of combinations when drawing ‘m‘ correct items from a set of ‘n‘, without replacement and regardless of the order, is:

C = n*(n – 1)*(n – 2)*…(n – m – 1)/(m!), where…

• ‘C’ is the number of combinations;
• ‘n’ is the number of items in the set;
• ‘m’ is the number of items to be selected.

So we have ‘n’ choices for the first item, but only (n-1) choices for the second selection. Why? We’ve already taken away one item.

When we take away the m-th item, there are only (m – 1) items remaining for the selection. Thus the second factor is (m-1). This continues until all  ‘m’ factors have been multiplied together.

Then we divide by “m!”, or “m factorial”, which was discussed in “Introducing the Factorial: the Exclamation Mark of Math“.

The  five white balls in Powerball count down from 59 through 55 and the “5!” divisor in the main calculation. The number 35 accounts for the solitary red ball.

Click to Read Page Two: Simple Example of Lottery Probability

Pages: 1 2

• H.M.Martens

oddly enough the odds do not stay fixed if the amount of tickets is not fixed!

• Andres

The authar has a typo in calc – ” There are ( (59*58*57*56*55)*39 )/ (5!) = ( (59*58*57*56*55)*35 )/ (5*4*3*2*1)
In the first calc the 39 should be a 35, as seen on the right hand side

• Jon

Good mathematic demo of probability! Our college students should read this!

• J.S.

One thing that always goes unnoticed by most news outlets about this jackpot is that the advertised amount is the estimated “annuitized value to a single winner”. (Annuitized meaning paid out over the next 30 years.) Since there were 2 perfect tickets sold, the prize would be split between the two tickets. Even if there was only 1 perfect ticket, if the winners decided to take their share of the prize in a lump sum, that value would be about only 60% of the advertised jackpot, before taxes.

• robert

should m! be 6

• bliddel

The odds of selecting the winning number are easy to calculate. The value of an entry in a raffle are easy to caluclate: Odds of winning, times the value of the prize awarded. The value of a ticket in a paramutual lottery are much harder to calculate. Even when one makes educated assumptions about the estimated payout and the number of entries people will purchase, the complexity of the problem exceeds the reach of most actuarial students, and I know several Fellows in the Society of Actuaries who don’t want to tackle this question either.

• http://decodedscience.com Mark B.

This analysis is fine as far as it goes, but it is incomplete. Since a player must buy a ticket with after-tax cash, the calculation should be based on after-tax cash payout. Given that Pwerball sets the cash payout at half of the prize, and assuming 43% taxes (federal, state, local), the prize would need to be \$1,168,156,733 (\$350,447,020 times 2 divided by .6) for this to be a rational gamble.

• Decoded Science

Editor Note: Thanks, Andres – Typo fixed.

• http://decodedscience.com Mark B.

Correction: \$1,229,638,667 ((\$350,447,020 times 2 divided by ..57)

• Mike DeHaan

Thanks for the kinds remarks and especially for all the corrections. I’d thought I’d fixed the “35 not 39″, which started with Turpin; did Powerball change the number of red balls at some point? Nonetheless, that’s my faulty midnight proof-reading to blame.
Many noticed that I ignored [a] taxes and [b] “annuitized value to a single winner”, when calculating an economic break-even point. Again it’s “mea culpa” for allowing my Canadian complacency about full & immediate, tax-free payout from other lotteries such as Lotto Max or the 6/49 lottery to blind me to American realties.
Again, thanks much, and I hope to improve.

• brett

Great explanation. Shockingly, those precise odds are actually printed on the back of the ticket, too!
The lottery used to be called the “stupidity tax” until the Power/MegaBall era, where the pot is rolled over. In that case, you can have a non-negative expected value ticket provided the pot exceeds the denominator, i.e., \$175m. However, i think it’s still hard to determine the true odds in an event like this week because you don’t know how many tickets are outstanding.

• Walter McLaughlin

Very interesting article, Mike!