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Powerball Lottery: Odds of Winning a Prize

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Date: 09/02/2001 at 16:16:57
From: Joshua A. Cippel
Subject: Odds of the Powerball Lottery

Dr. Math,

Hello. In the last few weeks, with all of the attention given to it
in the news media, I have become interested in the Powerball lottery.
More specifically, the odds of winning. This is not out of any desire
to win money, but out of simple mathematical curiosity.

As you may know, the object of the game is to choose 5 numbers from 1
to 49, and then the "Powerball" number from 1 to 42. After a number is
drawn to become one of the 5 from 1 to 49, it cannot be drawn again.
However, any number from 1 to 42 can be drawn as the "Powerball." Of
course, the person who picks all 6 numbers correctly wins the jackpot,
but one can also win a prize by picking some of the correct numbers.
What has me baffled is the statistics of winning. Quoted from
www.powerball.com, the odds are:

Match 5+ Powerball = 1:80,089,128
Match 5            = 1:1,953,393
Match 4+ Powerball = 1:364,042
Match 4            = 1:8,879
Match 3+ Powerball = 1:8,466
Match 3            = 1:207
Match 2+ Powerball = 1:605
Match 1+ Powerball = 1:118
Match Powerball    = 1:74

OVERALL ODDS OF WINNING A PRIZE: 1:35

Trying all of the statistics that I know to manipulate these numbers,
I simply cannot determine how these odds are established. I would be
very interested if you could help with this puzzle.

Thank you.
```

```
Date: 09/02/2001 at 21:33:39
From: Doctor Paul
Subject: Re: Odds of the Powerball Lottery

Let's start with the odds of winning the big prize. To do this, you
have to pick all five numbers correctly and you also have to pick the
powerball number correctly.

The answer depends on whether or not order is important and whether or
not the numbers are "replaced" after they are drawn. Clearly, we are
working "without replacement" here since a number cannot be drawn
twice. That is, once a number has been drawn, it is not placed back
into the pool of eligible numbers.

In lottery games, the order in which the numbers are chosen is
unimportant.  All that matters is that you have the same numbers on
your card as were drawn by the lottery officials.

Since order doesn't matter, we're going to use a combination.  We
would use a permutation if order mattered.

In general, the number of ways to choose r objects from n when order
is not important and we are drawing without replacement is given by:

C(n,r) = n!/(r! * (n-r)!)

The probability of winning the big jackpot is given by the total
number of ways to win divided by the possible number of outcomes. The
number of possible outcomes (the denominator) is computed as follows:

There are C(49,5) = 49!/(5! * (49-5)!) = 1,906,884 ways to pick your
five numbers. And there are C(42,1) = 42 ways to pick the powerball
(you verify the computation using the formula above). Thus there are
1,906,884 * 42 = 80,089,128 total number of ways that the drawing can
occur.

To figure out how many ways you can win (the answer is obviously one,
but we go through the formalities because we'll need them to do the
next part), think of the balls as being partitioned as follows:

Among the 49 numbers, 5 are "winners" and 44 are "losers." Similarly,
among the 42 powerball numbers, 1 is a "winner" and 41 are "losers."

If you want to pick all five numbers correctly and pick the powerball
correctly, then what you want to do is pick 5 out of the 5 winners and
0 out of the 44 losers on the non-powerball side, and you want to pick
1 out of 1 winners and 0 out of 41 losers on the powerball side.

This makes the numerator:

C(5,5) * C(44,0) * C(1,1) * C(41,0) = 1 * 1 * 1 * 1 = 1

Hence the probablity is 1/80,089,128

Now let's look at the odds of the "Match 5." Here you have to get all
five numbers right on the non-powerball side but the number on the
powerball is not important. Using the tools we developed above, you
need to pick 5 of the 5 winners on the non-powerball side and 0 of the
44 losers. But this time, we pick 0 of the 1 winners on the powerball
side and 1 of the 41 losers.

Hence the numerator is given by:

C(5,5) * C(44,0) * C(1,0) * C(41,1) = 1 * 1 * 1 * 41 = 41

The denominator remains the same for all of these problems. The number
of possible powerball combinations never changes.

Thus the probability of winning a match 5 is: 41/80,089,128, which is
pretty close to 1/1,953,393.

For the "Match 4 + Powerball" the numerator would be:

C(5,4) * C(44,1) * C(1,1) * C(41,0) = 5 * 44 * 1 * 1 = 220

Thus the odds here are 220/80,089,128, which is pretty close to
1/364,042.

You should be able to establish the rest of the formulas pretty easily
just by following the pattern that has been established. If you have

The overall odds of winning (1:35) can be verified by adding up all of
the individual probabilities of winning:

1/74 + 1/118 + 1/605 + ... + 1/80,089,128 = .028706, which is very
close to 1/35.

I hope this has helped to take some of the mystery out of the odds.
Again - write back if you need further clarification. If you aren't
familiar with combinations and permutations in probability, please
consult a Probability/Statistics textbook or the Dr. Math FAQ:

http://mathforum.org/dr.math/faq/faq.comb.perm.html

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/21/2001 at 14:03:32
From: G Saenger
Subject: Question on your powerball odds calcs

I read your analysis on the calculation of the Powerball odds, but I
have a real problem with the odds of 1:74 that are calculated for
selecting the correct powerball number only. I realize you have to
take into account selecting 5 incorrect numbers on the non-powerball
side, and include that calculation, but it still defies logic:

No matter what 5 numbers you select on the non-powerball side (out of
the field of 49 numbers), you still select one number on the
powerball side, which has a field of 42 numbers. You are therefore
selecting 1 number out of a field of 42, and by definition, your odds
should be 1:42 of selecting the correct powerball number.  It defies
logic to think you would have worse odds (1:74) than 1:42 in
selecting the correct powerball number.

Thanks!
```

```
Date: 11/21/2001 at 14:40:10
From: Doctor Paul
Subject: Re: Question on your powerball odds calcs

Perhaps it would help to recall what is meant by the probability of an
event. The probability that an event occurs is given by:

the total number of ways the event can occur
--------------------------------------------
the total number of possible outcomes

In the case above (picking only the powerball number correctly), you
need to count the number of ways to draw only the powerball. This
means you pick the powerball number correctly but it also means that
you don't pick any of the 5 numbers from the non-powerball side
correctly.

As mentioned above, it is helpful to think of the numbers as being
partitioned in the following way:

Among the 49 non powerball numbers, there are 5 "winners" and 44
"losers."

Similarly, among the 42 powerball numbers, there are 1 "winner" and
41 "losers."

In order to draw only the powerball, you have to pick:

0 of the 5 winners on the non-powerball side
5 of the 44 losers on the non-powerball side
1 of the 1 winners on the powerball side
0 of the 41 losers on the powerball side

In your argument above, you seem to be making the assumption that you
can pick any 5 numbers from the 49 available on the non-powerball side
as long as you pick the powerball number correctly. This is in fact
not the case. What if one of the five you chose on the powerball side
was a winner? Then you wouldn't be in the scenario of picking only the
powerball anymore. You would be in the "Match 1 + Powerball" scenario.

Using the method explained in the link above, the number of ways to
pick:

0 of the 5 winners on the non powerball side
5 of the 44 losers on the non powerball side
1 of the 1 winners on the powerball side
0 of the 41 losers on the powerball side

is:

C(5,0) * C(44,5) * C(1,1) * C(41,0) = 1 * 1086008 * 1 * 1 = 1086008

So the odds of picking only the powerball correctly are:

1086008 / 80089128 = .013559992812

which is very close to

1 / 74 = .013513513514

Now - if we wanted to know the odds of getting the powerball correct
(and we didn't care about what happened on the other side) then we
would just lump all 49 non-powerball numbers together and pick any
five of them. This would give:

C(49,5) * C(1,1) * C(41,0) = 1906884 * 1 * 1 = 1906884

ways of picking our numbers in such a way that guarantees that we get
the powerball number correct.

The probability of this happening is:

1906884 / 80089128 = .02380952381

and this is exactly 1/42.

So the odds of picking the powerball number correctly are in fact 1/42
(as you mentioned should intuitavely be the case). But you have to
realize that the odds of picking only the powerball correctly are in
fact lower because in picking only the powerball correctly, you must
guarantee that you don't pick any of the non-powerball numbers
correctly.

Finally, notice that 1/42 is the sum of the probabilities given above
that are associated with winning the powerball. Here is the chart from
above:

Match 5+ Powerball = 1:80,089,128
Match 5            = 1:1,953,393
Match 4+ Powerball = 1:364,042
Match 4            = 1:8,879
Match 3+ Powerball = 1:8,466
Match 3            = 1:207
Match 2+ Powerball = 1:605
Match 1+ Powerball = 1:118
Match Powerball    = 1:74

If you sum all of the ones that mean correctly picking the powerball,

1/74 + 1/118 + 1/605 + 1/8466 + 1/364042 + 1/80089128 = .023761861306

and 1/42 = .02380952381

These numbers are not exactly the same because the probabilities
listed in the chart above are rounded off.

For example - notice that when we computed the "Match Powerball"
probability above, we got a number that was close to 1/74 but it
wasn't exactly 1/74.

some more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations
High School Probability

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