# How To Calculate Winning Odds in California Lottery

Ever wonder how to calculate winning odds of lottery games? The winning odds of the top prize of Fantasy 5 in California Lottery are 1 in 575,757. The winnings odds of the top prize of SuperLOTTO plus are 1 in 41,416,353. The winnings odds of the top prize of Mega Millions are 1 in 175,711,534. In this post, we show how to calculate the odds for these games in the California Lottery. The calculation is an excellent combinatorial exercise as well as in calculating hypergeometric probability.

All figures and data are obtained from the California Lottery.

Update, April 27, 2017. The calculation in this post assumes certain background knowledge on combination and the multiplication principle (not explained here). For any reader who would like to further understand how lottery odds are calculated, see this blog post on Powerball. It is a self contained step by step explanation at the basic level on how to calculate winning odds in the Powerball game.

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Fantasy 5

The following figures show a playslip and a sample ticket for the game of Fantasy 5.

Figure 1

Figure 2

In the game of Fantasy 5, the player chooses 5 numbers from 1 to 39. If all 5 chosen numbers match the 5 winning numbers, the player wins the top prize which starts at $50,000 and can go up to$500,000 or more. The odds of winning the top prize are 1 in 575,757. There are lower tier prizes that are easier to win but with much lower winning amounts. The following figure shows the prize categories and the winning odds of Fantasy 5.

Figure 3

All 5 of 5
In matching the player’s chosen numbers with the winning numbers, the order of the numbers do not matter. Thus in the calculation of odds, we use combination rather than permutation. Thus we have:

$\displaystyle (1) \ \ \ \ \ \binom{39}{5}=\frac{39!}{5! \ (39-5)!}=575757$

Based on $(1)$, the odds of matching all 5 winning numbers is 1 in 575,757 (the odds of winning the top prize).

Any 4 of 5
To match 4 out of 5 winning numbers, 4 of the player’s chosen numbers are winning numbers and 1 of the player’s chosen numbers is from the non-winning numbers (34 of them). Thus the probability of matching 4 out of 5 winning numbers is:

$\displaystyle (2) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{34}{1}}{\displaystyle \binom{39}{5}}=\frac{5 \times 34}{575757}=\frac{1}{3386.8} \ \ \text{(1 out of 3,387)}$

Any 3 of 5
To find the odds for matching 3 out of 5 winning numbers, we need to find the probability that 3 of the player’s chosen numbers are from the 5 winning numbers and 2 of the selected numbers are from the 34 non-winning numbers. Thus we have:

$\displaystyle (3) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{34}{2}}{\displaystyle \binom{39}{5}}=\frac{10 \times 561}{575757}=\frac{1}{102.63} \ \ \text{(1 out of 103)}$

Any 2 of 5
Similarly, the following shows how to calculate the odds of matching 2 out of 5 winning numbers:

$\displaystyle (4) \ \ \ \ \ \frac{\displaystyle \binom{5}{2} \ \binom{34}{3}}{\displaystyle \binom{39}{5}}=\frac{10 \times 5984}{575757}=\frac{1}{9.6216} \ \ \text{(1 out of 10)}$

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SuperLOTTO Plus

Here are the pictures of a playslip and a sample ticket of the game of SuperLOTTO Plus.

Figure 4

Figure 5

Based on the playslip (Figure 4), the player chooses 5 numbers from 1 to 47. The player also chooses an additional number called a Mega number from 1 to 27. To win the top prize, there must be a match between the player’s 5 selections and the 5 winning numbers as well as a match between the player’s Mega number and the winning Mega number (All 5 of 5 and Mega in Figure 6 below).

Figure 6

All 5 of 5 and Mega
To find the odds of the match of “All 5 of 5 and Mega”, the total number of possibilities is obtained by choosing 5 numbers from 47 numbers and choose 1 number from 27 numbers. We have:

$\displaystyle (5) \ \ \ \ \ \binom{47}{5} \times \binom{27}{1}=41,416,353$

Thus the odds of matching “All 5 of 5 and Mega” are 1 in 41,416,353.

Any 5 of 5
To find the odds of matching “All 5 of 5” (i.e. the player’s 5 selections match the 5 winning numbers but no match with the Mega winning number), we need to choose 5 numbers from the 5 winning numbers, choose 0 numbers from the 42 non-winning numbers, choose 0 numbers from the 1 Mega winning number and choose 1 number from the 26 non-Mega winning numbers. This may seem overly precise, but will make it easier to the subsequent derivations. We have:

\displaystyle \begin{aligned}(6) \ \ \ \ \ \frac{\displaystyle \binom{5}{5} \ \binom{42}{0} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{1 \times 1 \times 1 \times 26}{41416353} \\&=\frac{1}{1592936.654} \\&\text{ } \\&=\text{1 out of 1,592,937} \end{aligned}

Any 4 of 5 and Mega
To calculate the odds for matching “any 4 of 5 and Mega”, we need to choose 4 out of 5 winning numbers, choose 1 out of 42 non-winning numbers, choose 1 out of 1 Mega winning number, and choose 0 out of 26 non-winning Mega numbers. We have:

\displaystyle \begin{aligned}(7) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{42}{1} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 42 \times 1 \times 1}{41416353} \\&=\frac{1}{197220.7286} \\&\text{ } \\&=\text{1 out of 197,221} \end{aligned}

Any 4 of 5
To calculate the odds for matching “any 4 of 5” (no match for Mega number), we need to choose 4 out of 5 winning numbers, choose 1 out of 42 non-winning numbers, choose 0 out of 1 Mega winning number, and choose 1 out of 26 non-winning Mega numbers. We have:

\displaystyle \begin{aligned}(8) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{42}{1} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 42 \times 1 \times 26}{41416353} \\&=\frac{1}{7585.412637} \\&\text{ } \\&=\text{1 out of 7,585} \end{aligned}

Any 3 of 5 and Mega
To calculate the odds for matching “any 3 of 5 and Mega”, we need to choose 3 out of 5 winning numbers, choose 2 out of 42 non-winning numbers, choose 1 out of 1 Mega winning number, and choose 0 out of 26 non-winning Mega numbers. We have:

\displaystyle \begin{aligned}(9) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{42}{2} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 861 \times 1 \times 1}{41416353} \\&=\frac{1}{4810.261672} \\&\text{ } \\&=\text{1 out of 4,810} \end{aligned}

The rest of the calculations for SuperLOTTO Plus should be routine. It is a matter to deciding how many to choose from the 5 winning numbers, how many to choose from the 42 non-winning numbers as well as how many to choose from the 1 winning Mega number and how many to choose from the 26 non-winning Mega numbers.

Any 3 of 5
\displaystyle \begin{aligned}(10) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{42}{2} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 861 \times 1 \times 26}{41416353} \\&=\frac{1}{185.0100643} \\&\text{ } \\&=\text{1 out of 185} \end{aligned}

Any 2 of 5 and Mega
\displaystyle \begin{aligned}(11) \ \ \ \ \ \frac{\displaystyle \binom{5}{2} \ \binom{42}{3} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 11480 \times 1 \times 1}{41416353} \\&=\frac{1}{360.7696254} \\&\text{ } \\&=\text{1 out of 361} \end{aligned}

Any 1 of 5 and Mega
\displaystyle \begin{aligned}(12) \ \ \ \ \ \frac{\displaystyle \binom{5}{1} \ \binom{42}{4} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 111930 \times 1 \times 1}{41416353} \\&=\frac{1}{74.00402573} \\&\text{ } \\&=\text{1 out of 74} \end{aligned}

None of 5 only Mega
\displaystyle \begin{aligned}(13) \ \ \ \ \ \frac{\displaystyle \binom{5}{0} \ \binom{42}{5} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{1 \times 850668 \times 1 \times 1}{41416353} \\&=\frac{1}{48.68685903} \\&\text{ } \\&=\text{1 out of 49} \end{aligned}

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Mega Millions

The following are a playslip, a sample ticket and the winning odds of the game of Mega Millions.

Figure 7

Figure 8

Figure 9

Based on the playslip (Figure 7), the player chooses 5 numbers from 1 to 56. The player also chooses an additional number called a Mega number from 1 to 46. To win the top prize, there must be a match between the player’s 5 selections and the 5 winning numbers as well as a match between the player’s Mega number and the winning Mega number. The calculation of the odds indicated in Figure 9 are left as exercises.

## 9 thoughts on “How To Calculate Winning Odds in California Lottery”

1. Excellent! For Fantasy 5, how would you calculate the odds of getting all 5 numbers if you purchased 100 ticket?

2. I reside in California and the lottery is about timing and guessing.I have played the daily 3,daily 4 and fantasy 5. I can say by timing the daily 3 and daily 4 and also by guessing I won 57 times in one year. Also I can say my 6 superlotto numbers came up but I did not play on that day. It’s only 39 million I lost out on. Where I reside in California the percent where I am in trying to win powerball.superlotto,mega and superlotto is 1% only,bay area is 4% and southern California is 95% that is the way it set up and it always stay that way. There is always a pattern if you look at the winning numbers.

3. Amazing..I will also try this method of calculating odds number.Thanks for sharing.

4. I don’t get the math. Sorry. Math isn’t my strong point. I want to play fantasy 5 because it is the easiest.

5. This doesn’t make sense. Using the SuperLoto as an example, it says the odds of drawing a MEGA number out of 27 numbers is 49 to 1. How can the odds of drawing one number out of a set of numbers be higher that the total numbers in the set? Simple math tells me it should be 27 to 1.

The number of permutations of a distinct objects n taken r at a time is n! / (n – r)!. Therefore, 27!/19! = 27.

Likewise, the same statistics formula shows the odds for matching 5 out of 47 without the MEGA number is over 184 MILLION to 1. Where did I go wrong?

• Hello John

Yes, in general picking one number of 27 numbers means that the odds of picking the winning number is 1 in 27. However, you have to realize that in playing SuperLottoPlus, you do not pick just one number. You pick 5 numbers (out of 47) for SuperLotto and pick 1 number (out of 27) for Plus (Mega). The outcome of “None of 5, Only Mega” means that you pick 5 losing numbers for the SuperLotto and the winning number for the Mega. With the 5 losing numbers and the one winning number for the Mega number, the odds are 1 in 49 (approximately). See calculation (13) for the details. The calculation done by California lottery also gives the odds of 1 in 49 (see Figure 6, which is a screen shot from their web site). So the odds of 1 in 49 is for winning the outcome of “None of 5 and Mega”. The odds of winning “Mega only” would be 1 in 27 but this outcome is not part of the game of SuperLottoPlus.

The whole purpose of this blog post is to demonstrate how to calculate the odds that are put out by the California lottery.

Dan Ma

6. What would be the formula to calculate the SuperLotto odds for not matching any of the 5 numbers or the mega number? Thanks!

• Hi, Arman

The odds for not matching the numbers and the mega number are pretty good, worked out to be roughly 1 in 1.3 (or 10 in 13) Here’s the calculation:

$\displaystyle \frac{\binom{42}{5} \times \binom{26}{1}}{\binom{47}{5} \times \binom{27}{1}}=\frac{31765734}{41416353}=\frac{1}{1.30380595}$

The numerator gives all the possible plays that completely miss the 5 winning numbers (from 1 to 47) and the Mega number. All such possibilities are from choosing 5 numbers from 42 numbers (1 to 47 minus the 5 winning numbers) and from choosing one number from 26 (27 choices for the Mega number minus 1). The denominator gives all the possible plays of SuperLotto Plus.

The notation $\displaystyle \binom{42}{5}$ is the number of ways to choose 5 numbers out of 42 numbers (also called binomial coefficient). You can compute this by using calculator or by the following formula.

$\displaystyle \binom{42}{5}=\frac{42!}{5! \ 37!}$

To do the above calculation, $n!$ is called n factorial and is defined to be n x (n-1) x (n-2) x … x 3 x 2 x 1. You can also use calculator.