In this post we will see how to calculate winning probabilities for the two most well-known games of the South African National Lottery: LOTTO and Powerball. We will also observe how changes in the rules of these games over time have had an impact on the probabilities. We will also learn why it does not matter which numbers you play, and yet in another sense it does.

LOTTO

The LOTTO game is the original South African lottery game and began when the South African National Lottery was introduced in 2000. The game rules and regulations can be read here. Briefly, there is a machine containing 52 individually numbered balls. (Prior to August 2017, the number of balls was 49, a point to which we will return.) Seven balls are successively drawn from the machine without replacement and their numbers noted. The first six numbers are called the ‘main numbers’ and the seventh is called the ‘bonus number.’ A player plays a game card by selecting six different numbers between 1 and 52. The idea is to correctly predict as many as possible of the numbers that will be drawn; the order does not matter except that the bonus number is treated separately. There are eight prize levels corresponding to the number of balls correctly predicted. A Level 1 (the ‘jackpot’) prize is won if the card selected all six main numbers correctly. A Level 2 prize is won if the player selected five main numbers and the bonus number. A Level 3 prize is won if the player selected five main numbers and not the bonus number. Prizes continue in this pattern (four main numbers with bonus number, four main numbers without bonus number, three main with bonus, three main without bonus), with the last prize (Level 8) being won by correctly picking two main numbers and the bonus number. Any other outcome results in no prize.

Let us refer to the numbers selected on the game card as ‘picked numbers.’ Which numbers are picked on the card has no effect on the winning probability, since all draw outcomes are equally likely. Thus, from a probability point of view, talk of ‘lucky numbers’ is much ado about nothing. (Your choice of numbers could, however, affect the prize amount in the unlikely event that you win the jackpot; we will see later why this is.)

Let \(N\) be the total number of balls in the game (in the case of LOTTO, \(N=52\)), and let \(n\) be the number of main balls drawn, which is also the number of balls picked on the game card (in LOTTO’s case, \(n=6\)). Let \(M\) be the number of drawn numbers that match picked numbers, and let \(B\) be the number of bonus numbers that match picked numbers. (Clearly, \(M\) can range between \(0\) and \(n\) and \(B\) can range between \(0\) and \(1\).) Achieving a certain prize level can be expressed as the event \(M=m \cap B=b\), where \(\cap\) denotes the intersection of two events; we can read it as ‘and’. Since \(M=m\) and \(B=b\) are dependent events (the probability that the seventh number drawn is a picked number changes depending on how many main picked numbers were drawn in the first six balls), we can use the multiplication rule for dependent events to write

\(\Pr(M=m \cap B=b)=\Pr(M=m) \Pr(B=b|M=m)\)

When \(n\) numbers between \(1\) and \(N\) are drawn without replacement, there are \({N \choose n}=\frac{N!}{n! (N-n)!}\) distinct possible draws (since order does not matter), all of which are equally likely. (We pronounce \({N \choose n}\) as ‘\(N\) choose \(n\)’, and it is sometimes written as \({}_N C_n\).) To find the probability that \(m\) of the \(n\) numbers on the game card are drawn, we simply need to count the number of draws that include \(m\) of the \(n\) picked numbers and divide this count by the total number of draws, \({N \choose n}\). A draw that includes \(m\) of the \(n\) picked numbers will also include \(n-m\) of the \(N-n\) numbers not picked, and so the number of ways the event \(M=m\) can happen is \({n \choose m}{N-n \choose n-m}\). Thus,

\(\Pr(M=m)=\displaystyle\frac{{n \choose m}{N-n \choose n-m}}{{N \choose n}}\) (This probability distribution has a name; it is called the hypergeometric distribution.)

All that remains is to calculate the conditional probability \(\Pr(B=b|M=m)\). The symbol \(|\) is read as ‘given that,’ so this is the probability of getting \(b\) bonus balls correct, assuming that we got \(m\) main numbers correct. \(B=1|M=m\) is the event that the bonus number is a picked number given that the main numbers included \(m\) picked numbers, while \(B=0|M=m\) is the event that the bonus number is not a picked number given that the main numbers included \(m\) picked numbers. Excluding the six main numbers already drawn, there are \(N-n\) possible outcomes for the bonus number. Of the six picked numbers, \(m\) cannot be the bonus number (because they were main numbers) and the other \(n-m\) could possibly be the bonus number. Hence, \(\Pr(B=1|M=m)=\frac{n-m}{N-n}\). By the complement rule, \(\Pr(B=0|M=m)=1-\Pr(B=1|M=m)=\frac{(N-n)-(n-m)}{N-n}=\frac{N-2n+m}{N-n}\). Thus, we can write

\(\Pr(B=b|M=m)=\left(\displaystyle\frac{n-m}{N-n}\right)^{b}\left(\displaystyle\frac{N-2n+m}{N-n}\right)^{1-b}\)

Hence, for the LOTTO Game in which \(N=52\) and \(n=6\),

\(\Pr(M=m \cap B=b)=\displaystyle\frac{{6 \choose m}{46 \choose 6-m}}{{52 \choose 6}}\left(\displaystyle\frac{6-m}{46}\right)^{b}\left(\displaystyle\frac{40+m}{46}\right)^{1-b}\)

To obtain the probability for any prize level from 1 to 8 we simply need to substitute the corresponding values of \(m\) and \(b\) into this formula. To obtain the probability of not winning any prize, we can either count up the remaining outcomes that do not win a prize, or we can take 1 minus the sum of probabilities of prize levels 1 to 8. The result is the following table.

Table 1: Winning Probabilities for LOTTO Game with 52 Balls
Prize Level Event Probability Prize Pool %
1 \(M = 6 \cap B = 0\) \(\frac{{6 \choose 6}{46 \choose 0}}{{52 \choose 6}}\left(\frac{46}{46}\right)=\text{1 in 20 358 520}\) 73%
2 \(M = 5 \cap B = 1\) \(\frac{{6 \choose 5}{46 \choose 1}}{{52 \choose 6}}\left(\frac{1}{46}\right)\approx \text{1 in 3 393 087}\) 2.3%
3 \(M = 5 \cap B = 0\) \(\frac{{6 \choose 5}{46 \choose 1}}{{52 \choose 6}}\left(\frac{45}{46}\right)\approx \text{1 in 75 402}\) 4%
4 \(M = 4 \cap B = 1\) \(\frac{{6 \choose 4}{46 \choose 2}}{{52 \choose 6}}\left(\frac{2}{46}\right)\approx \text{1 in 30 161}\) 5%
5 \(M = 4 \cap B = 0\) \(\frac{{6 \choose 4}{46 \choose 2}}{{52 \choose 6}}\left(\frac{44}{46}\right)\approx \text{1 in 1 371}\) 8.4%
6 \(M = 3 \cap B = 1\) \(\frac{{6 \choose 3}{46 \choose 3}}{{52 \choose 6}}\left(\frac{3}{46}\right)\approx \text{1 in 1 028}\) 7.3%
7 \(M = 3 \cap B = 0\) \(\frac{{6 \choose 3}{46 \choose 3}}{{52 \choose 6}}\left(\frac{43}{46}\right)\approx \text{1 in 72}\) R50 (fixed)
8 \(M = 2 \cap B = 1\) \(\frac{{6 \choose 2}{46 \choose 4}}{{52 \choose 6}}\left(\frac{4}{46}\right)\approx \text{1 in 96}\) R20 (fixed)
None \((M = 2 \cap B = 0)\cup(M = 1)\cup(M=0)\) \(\frac{{6 \choose 2}{46 \choose 4}}{{52 \choose 6}}\left(\frac{42}{46}\right)+\frac{{6 \choose 1}{46 \choose 5}}{{52 \choose 6}}+\frac{{6 \choose 0}{46 \choose 6}}{{52 \choose 6}}\approx 0.9739\) None


In the original LOTTO game from 2000 until July 2017, there were only 49 balls, and only seven prize levels (i.e., the event \(M=2 \cap B=1\) did not win a prize as it does now.) How did the changes that took effect from August 2017 change the probabilities? We need only substitute \(N=49\) into our probability formula instead of \(N=52\):

\(\Pr(M=m \cap B=b)=\displaystyle\frac{{6 \choose m}{43 \choose 6-m}}{{49 \choose 6}}\left(\displaystyle\frac{6-m}{43}\right)^{b}\left(\displaystyle\frac{37+m}{43}\right)^{1-b}\)

The resulting probabilities are displayed in the table below.

Table 2: Winning Probabilities for LOTTO Game with 49 Balls
Prize Level Event Probability
1 \(M = 6 \cap B = 0\) \(\frac{{6 \choose 6}{43 \choose 0}}{{49 \choose 6}}\left(\frac{43}{43}\right)=\text{1 in 13 983 816}\)
2 \(M = 5 \cap B = 1\) \(\frac{{6 \choose 5}{43 \choose 1}}{{49 \choose 6}}\left(\frac{1}{43}\right)\approx \text{1 in 2 330 636}\)
3 \(M = 5 \cap B = 0\) \(\frac{{6 \choose 5}{43 \choose 1}}{{49 \choose 6}}\left(\frac{45}{43}\right)\approx \text{1 in 55 491}\)
4 \(M = 4 \cap B = 1\) \(\frac{{6 \choose 4}{43 \choose 2}}{{49 \choose 6}}\left(\frac{2}{43}\right)\approx \text{1 in 22 197}\)
5 \(M = 4 \cap B = 0\) \(\frac{{6 \choose 4}{43 \choose 2}}{{49 \choose 6}}\left(\frac{44}{43}\right)\approx \text{1 in 1 083}\)
6 \(M = 3 \cap B = 1\) \(\frac{{6 \choose 3}{43 \choose 3}}{{49 \choose 6}}\left(\frac{3}{43}\right)\approx \text{1 in 812}\)
7 \(M = 3 \cap B = 0\) \(\frac{{6 \choose 3}{43 \choose 3}}{{49 \choose 6}}\left(\frac{43}{43}\right)\approx \text{1 in 61}\)
None \((M = 2)\cup(M = 1)\cup(M=0)\) \(\frac{{6 \choose 2}{43 \choose 4}}{{49 \choose 6}}+\frac{{6 \choose 1}{43 \choose 5}}{{49 \choose 6}}+\frac{{6 \choose 0}{43 \choose 6}}{{49 \choose 6}}\approx 0.9814\)


From the table, we can see that the addition of just three balls has decreased the probability of winning the jackpot from 1 in 13 983 816 to 1 in 20 358 520—a drop by over 45%. Meanwhile, despite the addition of a Level 8 prize, the probability of not winning any prize has increased slightly, from about 97.39% to about 98.14%. If there is a silver lining, it is that the jackpot’s share of the prize pool rolls over to the next draw if there is no jackpot winner, and with a lower jackpot win probability, the jackpot will tend to grow larger before someone wins. Perhaps the National Lottery’s operating company, iThuba, reasoned that bigger jackpots have greater marketing potential, even if jackpots happen less frequently.

Powerball

The Powerball game was introduced in 2009 and is very similar to the LOTTO game. The differences are as follows:

If we again let \(M\) be the number of drawn numbers (from the main draw) that match picked numbers, and let \(B\) be the number of bonus numbers (from the bonus draw) that match picked numbers, it follows that \(M\) can range between \(0\) and \(n=5\) and \(B\) can range between \(0\) and \(1\). A particular prize level corresponds to an event \(M=m \cap B=b\). In this case, however, the events \(M=m\) and \(B=b\) are independent, since they concern separate draws. Thus, \(\Pr(M=m \cap B=b)=\Pr(M=m)\Pr(B=b)\). To calculate \(\Pr(M=m)\) we use the same formula as before, changing \(N\) to \(50\) and \(n\) to \(5\). However, since the bonus ball draw is independent of the main draw, we have \(\Pr(B=1)=\frac{1}{20}\) \(\Pr(B=0)=\frac{19}{20}\). Thus, we can write

\(\Pr(M=m \cap B=b)=\displaystyle\frac{{5 \choose m}{45 \choose 5-m}}{{50 \choose 5}}\left(\displaystyle\frac{1}{20}\right)^{b} \left(\frac{19}{20}\right)^{1-b}\)

This results in the following table of probabilities for the different prize levels (again, substituting in the appropriate values of \(m\) and \(b\)):

Table 3: Winning Probabilities for Powerball Game with 50 Main Balls
Prize Level Event Probability Prize Pool %
1 \(M = 5 \cap B = 1\) \(\frac{{5 \choose 5}{45 \choose 0}}{{50 \choose 5}}\left(\frac{1}{20}\right)=\text{1 in 42 375 200}\) 70.73%
2 \(M = 5 \cap B = 0\) \(\frac{{5 \choose 5}{45 \choose 0}}{{50 \choose 5}}\left(\frac{19}{20}\right)\approx \text{1 in 2 230 274}\) 5.19%
3 \(M = 4 \cap B = 1\) \(\frac{{5 \choose 4}{45 \choose 1}}{{50 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 188 334}\) 3.25%
4 \(M = 4 \cap B = 0\) \(\frac{{5 \choose 4}{45 \choose 1}}{{50 \choose 5}}\left(\frac{19}{20}\right)\approx \text{1 in 9 912}\) 5.51%
5 \(M = 3 \cap B = 1\) \(\frac{{5 \choose 3}{45 \choose 2}}{{50 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 4 280}\) 6.23%
6 \(M = 3 \cap B = 0\) \(\frac{{5 \choose 3}{45 \choose 2}}{{50 \choose 5}}\left(\frac{19}{20}\right)\approx \text{1 in 225}\) 5.19%
7 \(M = 2 \cap B = 1\) \(\frac{{5 \choose 2}{45 \choose 3}}{{50 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 299}\) 3.90%
8 \(M = 1 \cap B = 1\) \(\frac{{5 \choose 1}{45 \choose 4}}{{50 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 57}\) R15 (fixed)
9 \(M = 0 \cap B = 1\) \(\frac{{5 \choose 0}{45 \choose 5}}{{50 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 35}\) R10 (fixed)
None \(M \le 2 \cap B = 0\) \(\left[\frac{{5 \choose 2}{45 \choose 3}}{{50 \choose 5}}+\frac{{5 \choose 1}{45 \choose 4}}{{50 \choose 5}}+\frac{{5 \choose 0}{45 \choose 5}}{{50 \choose 5}}\right]\left(\frac{19}{20}\right)\approx 0.9455\) None


We can see that one is less than half as likely to win the jackpot when playing Powerball as compared to LOTTO. However, the probability of winning no prize is also lower (0.9455 vs. 0.9739); one has roughly a 1 in 18 chance of winning a prize in Powerball compared with roughly a 1 in 38 chance of winning a prize in LOTTO.

Prior to June 2018 there were only 45 balls in Powerball rather than 50, but the prize categories were the same. Thus, replacing \(N=50\) with \(N=45\) in our probability formula, we arrive at the following table:

Table 4: Winning Probabilities for Powerball Game with 45 Main Balls
Prize Level Event Probability
1 \(M = 5 \cap B = 1\) \(\frac{{5 \choose 5}{40 \choose 0}}{{45 \choose 5}}\left(\frac{1}{20}\right)=\text{1 in 24 435 180}\)
2 \(M = 5 \cap B = 0\) \(\frac{{5 \choose 5}{40 \choose 0}}{{45 \choose 5}}\left(\frac{19}{20}\right)\approx \text{1 in 1 286 062}\)
3 \(M = 4 \cap B = 1\) \(\frac{{5 \choose 4}{40 \choose 1}}{{45 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 122 176}\)
4 \(M = 4 \cap B = 0\) \(\frac{{5 \choose 4}{40 \choose 1}}{{45 \choose 5}}\left(\frac{19}{20}\right)\approx \text{1 in 6 430}\)
5 \(M = 3 \cap B = 1\) \(\frac{{5 \choose 3}{40 \choose 2}}{{45 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 3 133}\)
6 \(M = 3 \cap B = 0\) \(\frac{{5 \choose 3}{40 \choose 2}}{{45 \choose 5}}\left(\frac{19}{20}\right)\approx \text{1 in 165}\)
7 \(M = 2 \cap B = 1\) \(\frac{{5 \choose 2}{40 \choose 3}}{{45 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 247}\)
8 \(M = 1 \cap B = 1\) \(\frac{{5 \choose 1}{40 \choose 4}}{{45 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 53}\)
9 \(M = 0 \cap B = 1\) \(\frac{{5 \choose 0}{40 \choose 5}}{{45 \choose 5}}\left(\frac{1}{20}\right)\approx \text{1 in 37}\)
None \(M \le 2 \cap B = 0\) \(\left[\frac{{5 \choose 2}{40 \choose 3}}{{45 \choose 5}}+\frac{{5 \choose 1}{40 \choose 4}}{{45 \choose 5}}+\frac{{5 \choose 0}{40 \choose 5}}{{45 \choose 5}}\right]\left(\frac{19}{20}\right)\approx 0.9438\)


We can see that increasing the number of balls by 5 resulted in the jackpot probability being nearly cut in half. Once again, this will tend to result in fewer but larger jackpots being won (because the jackpot prize money rolls over to the next draw if there is no winner.) The overall probability of winning a prize also decreased, but only marginally.

Conclusion

We have seen how to calculate winning probabilities for the LOTTO and Powerball games of the South African National Lottery using basic probability principles. We have also observed how changes in the rules of both games (particularly in the number of balls involved) during 2017-18 significantly reduced the probability of winning a jackpot. Of course, the probability was so tiny to begin with that the change is probably not noticeable to most players.

We mentioned in the introduction that, if you decide to play the lottery, it does not matter which numbers you choose, and yet it does. It is time to explain this paradoxical statement. We have already seen why it does not matter which numbers you choose: every combination of numbers has the same probability of being drawn. There are no ‘lucky numbers,’ and there is no way to improve one’s chances of winning per game card played.

However, while the numbers one picks has no effect on the probability of winning, it could affect the amount of prize money in the event that one does win. This is because, for the top few prize levels, the total prize money allocated to that level is shared equally among all winners at that level. For example, if the jackpot is R50 million and there is one winner, that person wins the whole R50 million. However, if there are two jackpot winners, each winner gets only R25 million. Thus, one should try to pick numbers that are less likely to be chosen by other players, in order to reduce the chances of having to split one’s winnings if one does hit the jackpot. How should one go about this? One foolproof technique is to use a pseudorandom number generator to generate one’s numbers, thus taking human psychology completely out of the equation. (In R software, one could generate a LOTTO play using the code sample(52, 6). ) Otherwise, at least avoid obvious patterns such as a sequence of consecutive numbers. Since it is well known that many people pick numbers based on the birthdays of loved ones, one may want to always pick at least one number greater than 31 (i.e., outside the range of possible birthday dates). This could be achieved in R using the following code:

picks <- -Inf
while (max(picks) <= 31) {
  picks <- sample(52, 6)
}
picks

It has been said that lotteries are a tax that is optional. Viewed that way, and in light of the very small probabilities of winning a large prize, the most sensible decision might be to avoid them. However, we should not forget that South African National Lottery revenues are used to fund many fantastic community projects around the country. The lottery is thus a good thing for the country. If you play, play responsibly, and maybe think of the R5.00 fee as a charitable donation with a twist.