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Lottery Probability for Dummies

Updated on March 14, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

Learn lottery math for with this easy guide for dummies!

Do you consider yourself a math dummy but still want to understand how lottery odds work? With a little patience and a willingness to learn, even dummies can become more aware of what lottery probabilities mean and even how to calculate them. Rather than present a list of combinatorial and probability formulas with realistic but complicated examples, this article will explain how the formulas are derived using extremely simple examples even a dummy can understand. From there, you can figure out how to calculate probabilities and odds for any lotto game.


Counting the Total Number of Outcomes (Possible Ticket Combinations) and the Jackpot Probability

The cornerstone of calculating lottery probabilities is learning how to count all the possible lottery ticket number combinations. Combinatorial formulas provide shortcuts since enumerating all the ticket combinations would be an onerous task for a lottery such as Powerball, which has 175,223,510 different combinations.

However, we will use a very simple lottery with a small number of ticket combinations as an example where enumerating all the possible combinations is not such a hard task. Let's call this simple lotto game Pick 3 Out of 8. In this imaginary lottery, three distinct numbers are selected from the set {1, 2, 3, 4, 5, 6, 7, 8}. If you match all three numbers, you win the jackpot. The order of the three numbers does not matter. To figure out the probability of winning the jackpot, we first need to figure out how many different three-number selections there are. Enumerating them all, we find there are 56.

 
 
 
 
 
 
 
123
136
157
237
258
356
458
124
137
158
238
267
357
467
125
138
167
245
268
358
468
126
145
168
246
278
367
478
127
146
178
247
345
368
567
128
147
234
248
346
378
568
134
148
235
256
347
456
578
135
156
236
257
348
457
678

Since only one out of the 56 possible ticket combinations can be the winner, the probability of winning the jackpot is 1/56 ≈ 0.017857, or about 1.78%.

How do we go from counting out all the possible ways to pick three numbers out of a set of eight to using a shortcut formula?

If we consider that there are 8 possibilities for the first number drawn, then 7 possibilities for the second number drawn, and then 6 possibilities for the third number drawn, we should get a total of 8 x 7 x 6 = 336 different lottery tickets. But what is wrong with this calculation is that we have counted the numbered of ordered sets of three numbers. In the lottery, the order doesn't matter. For any three elements A, B, and C there are 6 different ways of ordering them:

ABC - ACB - BAC - BCA - CAB - CBA

There are 6 different orderings because there are three possible choices for the first number, two for the second, and one for the last, and 3 x 2 x 1 = 6. This is called a permutation. Therefore, to get the correct count we have to divide 336 by 6. This gives us the correct value of 56 which is demonstrated in the table above.

The expression that directly counts the number of tickets is 8! / (5! x 3!), where ! is the factorial function. Working this out gives you

8! / (5! x 3!) =
[8 x 7 x 6 x 5 x 4 x 3 x 2 x 1] / [5 x 4 x 3 x 2 x 1 x 3 x 2 x 1]
= [8 x 7 x 6] / [3 x 2 x 1]
= 336 / 6
= 56

For any lottery in which you must pick K numbers from 1 to N, the total number of different ticket combinations is

N! / [(N-K)! x K!]

And the probability of winning the jackpot in such a lottery is the reciprocal of this number

[(N-K)! x K!] / N!

In this example, we applied the formula with N = 8 and K = 3.


How to Calculate Odds When the Lottery Has an Extra Ball

Many lotteries are structured such that you pick K distinct numbers from 1 to N, and then an extra number from between 1 and P. For example, Powerball and MegaMillions are structured this way. In that case, the total number of ticket combinations is

P x N! / [(N-K)! x K!]

In other words, you simply multiply the standard formula by P. As an example, let's imagine another lottery called Super Cash Money in which you have to pick 3 distinct numbers from 1 to 8, and then a fourth number from between 1 and 13. We already know there are 56 ways to choose 3 out of 8, and for each of these 56 lottery ticket combinations there are 13 ways to choose the fourth extra number. This means there are 13 x 56 = 728 different lottery tickets in Super Cash Money.

The probability of winning the jackpot in this game is the reciprocal of 728, which is 1/728 ≈ 0.0013736, or about 0.137%. The formula for the probability is

[(N-K)! x K!] / [N! x P]

In this example, we used the formula with the values N = 8, K = 3, and P = 13.


How to Calculate the Chance of Winning a Smaller Prize for a Partial Match

Let's use our example of the Super Cash Money lottery again, where the goal is to match three distinct numbers selected from the set {1, 2, 3, 4, 5, 6, 7, 8} and one number from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Suppose that this lottery offers two lower prize tiers for two different partial matches:

  • 2nd Prize Tier: Win $200 if you match 3 out of 8 but not the extra (fourth) number
  • 3rd Prize Tier: Win $100 if you match 2 out of 8 and the extra (fourth) number

To figure the odds of winning one of these prizes let's pretend that the lottery drawing has already occurred and the winning ticket was {1, 3, 8} & 11.

First we enumerate the number of tickets that are eligible to win the 2nd tier prize. This means all the tickets that have the {1, 3, 8} part correct but the bonus number wrong. There are only 12 such tickets, and they are

 
 
 
 
{1, 3, 8} & 1
{1, 3, 8} & 4
{1, 3, 8} & 7
{1, 3, 8} & 10
{1, 3, 8} & 2
{1, 3, 8} & 5
{1, 3, 8} & 8
{1, 3, 8} & 12
{1, 3, 8} & 3
{1, 3, 8} & 6
{1, 3, 8} & 9
{1, 3, 8} & 13

One way of deducing that there are 12 without writing them all out is to realize that there is only one way to get the {1, 3, 8} part correct, but 12 ways to get the other number wrong, since out of 13, only one can be right.

Since there are 728 different possible tickets in total and only 12 that can win the 2nd prize, the probability of winning the second prize is 12/728 ≈ 0.016484, or about 1.65%.

Now let's enumerate all the tickets that can win the 3rd tier prize. This is a little more difficult because we need to find all the tickets that have exactly two out of the set {1, 3, 8} correct with the other number wrong, while the bonus number 11 remains unchanged. There are 15 tickets than can win this prize level and they are

 
 
 
{1, 2, 3} & 11
{1, 2, 8} & 11
{2, 3, 8} & 11
{1, 3, 4} & 11
{1, 4, 8} & 11
{3, 4, 8} & 11
{1, 3, 5} & 11
{1, 5, 8} & 11
{3, 5, 8} & 11
{1, 3, 6} & 11
{1, 6, 8} & 11
{3, 6, 8} & 11
{1, 3, 7} & 11
{1, 7, 8} & 11
{3, 7, 8} & 11

How can we deduce that there are 15 without writing them all out? Out of the set {1, 3, 8} there are three ways to remove one of the correct numbers -- 1, 3, or 8 -- and five ways to replace it with awrong number -- 2, 4, 5, 6, or 7. There is only one way to pick the bonus number correctly. Since 3 x 5 x 1 = 15, you get 15 possibilities for winning the 3rd lottery prize.

The probability of winning the 3rd prize is 15/728 ≈ 0.020604, or about 2.06%.

No matter what the winning ticket happens to be, this method of enumeration by example will always work. For a lottery with lots of combinations, it will take more time to write them all out, but luckily there are shortcut formulas for these calculations as well.

What are the formulas for computing the 2nd and 3rd tier prize probabilities for an arbitrary lottery where players choose K distinct numbers from 1 to N, and an extra number between 1 and P?

The number of ways to win the second tier prize is P - 1. The probability of winning the second tier prize is P - 1 divided by the total number of different ticket combinations. This works out to

2nd Tier Prize Probability =
[(P-1) x (N-K)! x K!] / [P x N!]

The number of ways to win the third tier prize is K x (N-K) = NK - K^2. This is derived from the fact that there are K ways to choose one of the numbers wrong, and N-K ways to replace it with a non-winning number. The probability of winning the 3rd tier prize is K x (N-K) divided by the total number of tickets, which works out to

3rd Tier Prize Probability =
[K x (N-K) x K! x (N-K)!] / [P x N!]

You can check if you substitute N = 8, K = 3 and P = 13 you get the same answers as with enumeration.


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