# 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.

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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 27 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.