How to calculate probability of lottery numbers

How to calculate probability of lottery numbers DEFAULT

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.


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 notpicked, 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 thatthe main numbers included \(m\)picked numbers, while \(B=0|M=m\)is the event that the bonus number is nota 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


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.

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.

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.


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

  • The total number of balls is \(N=50\) (note: prior to June 2018 it was \(N=45\))
  • The bonus ball comes from a completely separate draw consisting of \(20\) balls
  • A player picks \(n=5\) numbers between \(1\) and \(N\) for the main draw and separately picks one number between \(1\) and \(20\) for the bonus draw
  • There are nine prize levels, as follows: level 1 (the jackpot) consists of matching \(5+1\) (five main numbers and one bonus number), level 2 consists of matching \(5+0\), then \(4+1\), \(4+0\), \(3+1\), \(3+0\), \(2+1\), \(1+1\), and \(0+1\).
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 \(b\)):

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:

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.


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 . ) 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:

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.


How to Calculate Lottery Probability

About the Author

Dez has been a mathematician since grade school and has a master's degree in Applied Mathematics.

As a mathematician, I have never purchased a lottery ticket. I find the odds depressing and have never had luck in winning anything from these kinds of games.

This hub is all about calculating lottery probability or odds. In order to make it more relevant to me, I decided to base it on the Grandlotto 6/55, the lottery game with the biggest prize money here in the Philippines. There will be two different cases discussed in the hub: the probability of winning the game with all six numbers matching, and the probability of having n numbers matching.

Rules of the Lottery Game

It is always important to find out the rules of any game before participating in it. For the Grandlotto 6/55, in order to win the jackpot prize, you have to match six numbers from a pool of 55 numbers ranging from 1-55. The initial payout is a minimum of P20 (or around $0.47). It is also possible to win some money if you are able to match three, four, or five numbers of the winning combination. Note that the order of the winning combination here does not matter.

Here is a table for the prizes you can obtain:

No. of Matching Nos.Prize Money (in Php)Prize Money (in $)


minimum of 30 million











Some Probability Concepts

Before we start with the calculations, I would like to talk about Permutations and Combinations. This is one of the basic concepts you learn in Probability Theory. The main difference being that permutations consider order to be important, while in combinations, order isn't important.

In a lottery ticket, permutation should be used if the numbers in your ticket have to match the order of the draw for the winning string of numbers. In the Grandlotto 6/55, order is not important because so long as you have the winning set of numbers, you can win the prize.

The next formulas only apply for numbers without repetition. This means that if the number x is drawn, it cannot be drawn again. If the number drawn from the set is returned before the next draw, then that has repetition.

, where n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.

Note that based on the formulas given, C(n,k) is always less than or equal to P(n,k). You will see later on why it is important to make this distinction for calculating lottery odds or probability.

How to Calculate Lottery Probability for 6 Matching Numbers

So now that we know the basic concepts of permutations and combinations, let us go back to the example of Grandlotto 6/55. For the game, n = 55, the total number of possible choices. k = 6, the number of choices we can make. Because order is not important, we will use the formula for combination:

These are the odds or the total number of possible combinations for any 6-digit number to win the game. To find the probability, just divide 1 by the number above, and you will get: 0.0000000344 or 0.00000344%. See what I mean by depressing odds?

So what if we're talking about a different lottery game where order does matter. We will now use the permutation formula to get the following:

Compare these two results and you will see that the odds for getting the winning combination where order matters has 3 additional zero's! It's going from about 28 million:1 odds to 20 billion:1 odds! The probability of winning for this case is 1 divided by the odds which equals to 0.0000000000479 or 0.00000000479%.

As you can see, because the permutation is always greater than or equal to the combination, the probability of winning a game where order matters is always less than or equal to the probability of winning a game where order does not matter. Because the risk is greater for games where order is required, this implies that the reward must also be higher.

How to Calculate Lottery Probability with Less Than 6 Matching Numbers

Because you can also win prizes if you have less than 6 matching numbers, this section will show you how to calculate the probability if there are x matches to the winning set of numbers.

First, we need to find the number of way to choose x winning numbers from the set and multiply it by the number of ways to choose the losing numbers for the remaining 6-x numbers. Consider the number of ways to choose x winning numbers. Because there are only 6 possible winning numbers, in essence, we are only choosing x from a pool of 6. And so, because order does not matter, we get C(6, x).

Next, we consider the number of ways to choose the remaining 6-x balls from the pool of losing numbers. Because 6 are winning numbers, we have 55 - 6 = 49 balls to choose the losing numbers from. So, the number of possibilities for choosing a losing ball can be obtained from C(49, 6 - x). Again, order does not matter here.

So, in order to calculate the probability of winning with x matching numbers out of a possible 6, we need to divide the outcome from the previous two paragraphs by the total number of possibilities to win with all 6 matching numbers. We get:

If we write this in a more general form, we get:

, where n = total number of balls in the set, k = total number of balls in the winning combination for the jackpot prize, and x = total numbers of balls matching the winning set of numbers.

If we use this formula to calculate the probability (and the odds) of winning the Grandlotto 6/55 with only x matching numbers, we get the following:

x matchesCalculationProbabilityOdds (1/Probability)


C(6,0) * C(49,6)/C(55,6)




C(6,1) * C(49,5)/C(55,6)




C(6,2) * C(49,4)/C(55,6)




C(6,3) * C(49,3)/C(55,6)




C(6,4) * C(49,2)/C(55,6)




C(6,5) * C(49,1)/C(55,6)




C(6,6) * C(49,0)/C(55,6)



How to Choose the Winning Numbers in Lottery

As you can see from the math in this hub, the probability of winning the lottery is the same for any 6-number combination available in the Grandlotto 6/55 game. This is also applicable for other lottery games out there.

As I was researching for this hub, I came across links that said never choose numbers that are sequential, like from 1-6 or some such nonsense. There is no such secret to winning the lottery! Each number is as equally likely to come up in the draw as the next number.

If you are willing to face the very little probability of winning the lottery, I say go choose any number you want. You can base it on your birthdays, special days, anniversaries, lucky numbers, etc. Just remember that with great risk comes great reward!


Brandon on February 13, 2020:

In a game of lotto, balls are numbered 1 through to 44. They are placed in a barrel and six balls are drawn without replacement. The balls are of the same size and are equally likely to be drawn. The first five balls drawn out are numbered 34, 2, 15, 29 and 42. what is the probability that the next ball draw out will be number 26?

Rocco on July 16, 2019:

I want to determine some odds. There are 4 different variables. Column one has a 1 in 9 chance of getting what you want, column 2 has a 1 in 8 chance of getting what you want, column 3 has a 1 in 5 chance of getting what you want, and column 4 has a 1 in 5 chance of getting what you want. How do I determine the odds of getting what I want from each column in one random roll/event?

iegsaan on April 19, 2019:

i just feel this person is trying to make lottery predictions to difficult because she hasn't figured out all lotteries strategies across the globe. any game of chance whether it be lotto , dice, cards its all a con. if you go into a casino and you good at memorizing cards they kick your ass out saying you cheating but bottom line any betting house crook the games. lotteries are the worse. think of it like this the lottery is a business and to make a profit it needs to eat your money. if they played fairly then we would have a lot more winners and lottery boards would go bankrupt . that is why they need time before the draw to run through all the tickets and choose the numbers with the least amount to payout. so predicting numbers is easy try to crook a crook much more difficult. funny how government allow this kind of white collar crimes but end up sending other people to jail for the exact same things.

Zack on January 29, 2019:

I am looking at a lottery game here in Brazil that says something abouts the odds that sounds "odd" to me. Pick 50 numbers out of 100. They say the odds of getting all 20 drawn numbers correct is the same odds of getting zero numbers correct (1:11,372,635). This doesn't sound possible. What do you think?

Andy on January 17, 2019:

In-fact even in close proximity they did well enough.


Andy on January 17, 2019:

if any of you had used a combi like this you would of done ok

01-06-12-13-54-55 it's all comes don't to brilliance!!!

Andy Martin on January 17, 2019:

Play the lottery hotpicks because the odds are 800/1 but only use 3 number sequences. The possibilities should prove an overall win return.

dean martin on January 17, 2019:

The lottery itself is designed from patterns then created into numbers also restricting and removing numbers also... reducing your odds even further.

So mathematics can't be applied but in someway can help with sequences or possibilities. It is not even a 50/50 odds simply because we can't use a 50.50 odds system if numbers are removed.

paul smith on January 17, 2019:

You can't use mathematics to win the lottery simple because the lottery removes numbers. The most simplest way to calculate your odds are 1/59 then 1/58 1/57 1/56 1/55 1/54

Gabriel on January 08, 2019:

According to quantum theory and the theory of everything, we are actually able to calculate the exact lottery winning numbers of a particular day with using the theory of time and space, if we know the exact day, time and location that the numbers are withdraw from the machine. nanosecond is 10 -9 s; one picosecond is 10 -12 s. So this means if our time are exactly picosecond accurate with the lottery machine we can predict the numbers. Which would be near impossible but not impossible.

Brigid on January 06, 2019:

@ Patel Ali how can permutations help with a pick 4 draw ?? Have you tried it and has it worked for you?

Patel ali on December 23, 2018:

4 digits number quite easy to predict for draw..

Example... 9269 =2057(permutation)

Are u agree with me?

roby on October 21, 2018:

what is the formula of choosing 2 numbers where these 2 number will appear in the 6 draw numbers of a lottery

Ralph on October 07, 2018:

No, probability will not be able to predict any future draw, patterns should be searched in sequences, try Tesla...

anna on August 24, 2018:

The Powerball lottery is decided every Wednesday and Saturday night by drawing five white balls out of a drum with 69 balls and one red ball out of a drum with 26 red balls. The Powerball jackpot is won by matching all five white balls in any order and the red Powerball. Each ticket costs $2.

a. [2 pts] How many possible different Powerball tickets can be purchased?

b. [1 pts] How many possible different winning Powerball tickets are there?

Sean Castleton on August 06, 2018:

Unfortanetly, the Law of Probability simply goes out the window if a lottery system is rigged, which many of them are.

Nardo on March 26, 2018:

Probability is one part of the whole thing which is true by itself. However there are also many factors(known and many unknown) which produces lottery results that happen to produce patterns, which may help in striking like few matching numbers but still ultra difficult in winning 6 matching numbers. It's still a combination of luck, for now i conclude it's combination of probability+luck (L.O.A or whatever u wanna call it)...cheers n good luck hehe

dezalyx (author) from Philippines on January 15, 2018:

Hi, Subs.

From the definition of classical probability, every statistical outcome will contain elements that are equally likely to happen. This means that if you roll a 6-sided dice, it is equally probable that the outcome will be 1, 2, 3, 4, 5, or 6. Or in the case of flipping a coin, the probability of heads will be equal to the probability of tails. No magic coins, no loaded dice, all equally probable to happen.

When calculating the lottery probability in this article, this assumption is already used. This is usually the case when calculating probabilities theoretically. Having no affiliation to companies who hold the lottery, it would be difficult for me to assume that certain balls are more likely to surface because they are loaded.

In my opinion, historical data does not have any bearing on lottery outcomes. Statistically speaking, working on other probability approaches will not work because of this. Just because the number N appears frequently in lottery drawings, does not mean that the ball is more likely to surface than the number N+1 in the next drawing.

Subs on January 14, 2018:

Can the classical probability theory be used here? And please give an example

Subs on January 14, 2018:

What other probability approaches can be used?

dezalyx (author) from Philippines on September 01, 2017:

Hi, Odds. Thanks for commenting.

I did not say choosing one's birthday would contribute to one's luck. I just said you can choose any number you want if you're willing to face the fact that there's only a very small chance that you can win the jackpot prize. Heck, you could choose 1, 2, 3, 4, 5, 6 if you wanted to. The basis of this article is that the probability of each ball getting drawn is the same as any other ball in the lottery.

While I kept getting comments on how likely certain numbers will get picked over other numbers, mathematically speaking, there is no such thing as having weighted balls for lottery drawings.

While you might not have studied mathematics, you certainly were on point when you said that you should never try your luck in gambling. That's actually a mathematician's saying. Never gamble, because the house always wins.

ODDS on August 30, 2017:

I didn't study mathematics, but you were wrong when you mentioned that the numbers of the player's birthday would contribute to his luck, rather it would add up to the improbability, because certainly the player would not live enough to wait his birthday's numbers to come up!

And yeah you certainly can top up your chances even if mathematically every number may pop up randomly.

If i were to buy a ticket, i would consider not selecting numbers that are altogether evens/odds, divisible by 5 altogether etc. If the order does not matter, you will get more chances!

But you know what, never try your luck in gambling, it's made to steal your money and you end up depressed.

dezalyx (author) from Philippines on August 13, 2017:

Hi, jazer. I've been receiving a ton of comments lately asking for what the winning number will be for the next lottery draw, but the article is just about the probability of winning the lottery. I don't know about the specifics of the composition of lotto balls, since it probably differs for every company that makes them, but looking for trends when there is no factual basis will just result in you losing money. And given the odds of winning the lottery, you would be better off investing that money in something else with a higher chance of getting a profit. While I agree with the idea that maybe the composition of the balls would get it drawn more than the others, you can never bank on that, since it could be changed without you ever knowing about it (repainting the balls numbers, changing the balls entirely, having different sets of balls for different drawings, etc.).

jazer on August 13, 2017:

Probability is probability. We cannot predict the future. But for the purpose of conversation, some people used they called a "trend". A history of past result. From trend you may "guess" which number can be with higher probability. Is this true in nature? Actually, this is only my opinion. My reasoning with the trend is somewhat related to the composition of the lotto balls e.g. weight, etc, that's why some numbers were drawn more that the others. What is your opinion on this? Thanks.

Kgomotso Comfort on June 19, 2017:

I just want know how can I calculate the probabilities of 6/49

dezalyx (author) from Philippines on May 08, 2017:

Edgard, it seems to me that your question is still about the gambler's fallacy. Just because a certain number comes out approximately 6 times out of x number of drawings will not change the probability of the number being drawn again in the future.

Imagine changing the game into a simple heads or tails coin flip. The probability of getting either outcome is 1/2. Just because you get 6 heads in a row does not mean the next result would be a tail. Even if you get 100 heads in a row, the probability of getting either outcome is still going to be 1/2.

There won't be any probability equations to predict what you are looking for. If there is, the people who have discovered such equations would already be taking advantage of the lottery system.

The only thing you can do with the study is plot historical graphs on how many times a certain number has come out previously and marvel at how some numbers seem to appear more than others.

Edgard on May 05, 2017:

Dear dezalix,

We really appreciate your time and answers

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How to Calculate Lotto Odds

Image titled Calculate Lotto Odds Step 1

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How to Calculate the Odds of Winning Lotto - Step by Step Instructions - Tutorial - Probability

Lottery Calculator

The lottery formula

What kind of formula does our calculator use? It finds the lottery odds according to the following equation:


  • — Number of balls in the pool;
  • — Number of balls to be drawn;
  • — Expected number of matches;
  • — Number of combinations (in how many ways you can choose out of numbers?); and
  • and — Number of combinations to choose out of and out of respectively.

For example, in a typical Lotto game, where you choose 6 numbers out of 59, the chance of winning the jackpot (matching all six numbers) is 1 in 45,057,474. That means that statistically, you would have to play over 45 million times to win just once. Chances for matching five numbers are bigger, but also not too optimistic — 1 to 141,690. On the other hand, the odds for matching three numbers are 1 to 96 — a bit over 1%. It means that matching three numbers is fairly common. Too bad it doesn't really pay off…

Bonus balls

Some lotteries include a bonus ball that is drawn after the initial balls. Our calculator allows you to take into consideration two types of bonus balls:

  1. Bonus balls taken from the remaining pool. In this case, an additional ball is chosen from the pool to give players who hit all but one number a chance for a better prize. For example, in a game where you pick 6 numbers of 59, a seventh ball is randomly selected. Let's assume that you matched 5 out of 6 numbers. Our calculator tells you what are the odds of the sixth number you chose being on the bonus ball.
    • Number of matches is set by default to m-1. That means that if you initially wanted to match 6 balls, the lottery calculator will find the odds of matching 5 balls and a bonus ball.
  2. Bonus balls taken from a bonus pool. In this case, bonus balls are randomly picked from a separate pool. That means that one number can come up twice. Our lottery odds calculator finds what are the chances of you getting m matches and picking a bonus ball number. For example, if you want to hit the jackpot (6 matches out of 6 numbers), this calculator will find the odds of matching six numbers and a bonus ball.
    • Balls to be drawn — Number of balls drawn from the bonus pool.
    • Matches with bonus pool — How many numbers from your lottery ticket should come up in the bonus pool?
    • Balls in the bonus pool — Total number of balls in the bonus pool. It can, but does not have to, be equal to the number of balls in the initial pool.

Calculate of numbers lottery how to probability

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Probability using combinatorics

Video transcript

How to Calculate the Odds of Winning Lotto - Step by Step Instructions - Tutorial - Probability

Lottery mathematics

Lottery mathematics is used to calculate probabilities of winning or losing a lottery game. It is based primarily on combinatorics, particularly the twelvefold way and combinations without replacement.

Probabilities of winning or losing a lottery game

Choosing 6 from 49[edit]

In a typical 6/49 game, each player chooses six distinct numbers from a range of 1-49. If the six numbers on a ticket match the numbers drawn by the lottery, the ticket holder is a jackpot winner—regardless of the order of the numbers. The probability of this happening is 1 in 13,983,816.

The chance of winning can be demonstrated as follows: The first number drawn has a 1 in 49 chance of matching. When the draw comes to the second number, there are now only 48 balls left in the bag, because the balls are drawn without replacement. So there is now a 1 in 48 chance of predicting this number.

Thus for each of the 49 ways of choosing the first number there are 48 different ways of choosing the second. This means that the probability of correctly predicting 2 numbers drawn from 49 in the correct order is calculated as 1 in 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but of course we could have arrived at this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49, again in the correct order, is 1 in 49 × 48 × 47. This continues until the sixth number has been drawn, giving the final calculation, 49 × 48 × 47 × 46 × 45 × 44, which can also be written as {49! \over (49-6)!} or 49 factorial divided by 43 factorial. This works out to 10,068,347,520, which is much bigger than the ~14 million stated above.

However; the order of the 6 numbers is not significant. That is, if a ticket has the numbers 1, 2, 3, 4, 5, and 6, it wins as long as all the numbers 1 through 6 are drawn, no matter what order they come out in. Accordingly, given any set of 6 numbers, there are 6 × 5 × 4 × 3 × 2 × 1 = 6! or 720 orders in which they could be drawn. Dividing 10,068,347,520 by 720 gives 13,983,816, also written as {\displaystyle {49! \over 6!*(49-6)!}}, or more generally as

{n \choose k}={n! \over k!(n-k)!}, where n is the number of alternatives and k is the number of choices. Further information is available at binomial coefficient and multinomial coefficient.

This function is called the combination function. For the rest of this article, we will use the notation {n \choose k}. "Combination" means the group of numbers selected, irrespective of the order in which they are drawn.

An alternative method of calculating the odds is to note that the probability of the first ball corresponding to one of the six chosen is 6/49; the probability of the second ball corresponding to one of the remaining five chosen is 5/48; and so on. This yields a final formula of

{n \choose k}={49 \choose 6}={49 \over 6}*{48 \over 5}*{47 \over 4}*{46 \over 3}*{45 \over 2}*{44 \over 1}

The range of possible combinations for a given lottery can be referred to as the "number space". "Coverage" is the percentage of a lottery's number space that is in play for a given drawing.

Odds of getting other possibilities in choosing 6 from 49[edit]

One must divide the number of combinations producing the given result by the total number of possible combinations (for example, {49 \choose 6}=13,983,816 ). The numerator equates to the number of ways to select the winning numbers multiplied by the number of ways to select the losing numbers.

For a score of n (for example, if 3 choices match three of the 6 balls drawn, then n = 3), {6 \choose n} describes the odds of selecting n winning numbers from the 6 winning numbers. This means that there are 6 - n losing numbers, which are chosen from the 43 losing numbers in {43 \choose 6-n} ways. The total number of combinations giving that result is, as stated above, the first number multiplied by the second. The expression is therefore {{6 \choose n}{43 \choose 6-n} \over {49 \choose 6}}.

This can be written in a general form for all lotteries as:

{\displaystyle {K \choose B}{N-K \choose K-B} \over {N \choose K}}

where N is the number of balls in lottery, K is the number of balls in a single ticket, and B is the number of matching balls for a winning ticket.

The generalisation of this formula is called the hypergeometric distribution.

This gives the following results:

Score Calculation Exact Probability Approximate Decimal Probability Approximate 1/Probability
0 {{6 \choose 0}{43 \choose 6} \over {49 \choose 6}}435,461/998,844 0.436 2.2938
1 {{6 \choose 1}{43 \choose 5} \over {49 \choose 6}}68,757/166,474 0.413 2.4212
2 {{6 \choose 2}{43 \choose 4} \over {49 \choose 6}}44,075/332,948 0.132 7.5541
3 {{6 \choose 3}{43 \choose 3} \over {49 \choose 6}}8,815/499,422 0.0177 56.66
4 {{6 \choose 4}{43 \choose 2} \over {49 \choose 6}}645/665,896 0.000969 1,032.4
5 {{6 \choose 5}{43 \choose 1} \over {49 \choose 6}}43/2,330,636 0.0000184 54,200.8
6 {{6 \choose 6}{43 \choose 0} \over {49 \choose 6}}1/13,983,816 0.0000000715 13,983,816

When a bonus number is included, the adjusted odds are:[1]

Score Calculation Exact Probability Approximate Decimal Probability Approximate 1/Probability
5, bonus not won {\displaystyle {6 \choose 5}{(43-1) \choose 1} \over {49 \choose 6}}0.0000180208 55,491.33
5, bonus won {\displaystyle {6 \choose 5}{(43-1) \choose (1-1)} \over {49 \choose 6}}0.0000004291 2,330,636

Ensuring to win the jackpot[edit]

There is only one known way to ensure winning the jackpot. That is to buy at least one lottery ticket for every possible number combination. For example, one has to buy 13,983,816 different tickets to ensure to win the jackpot in a 6/49 game.

Lottery organizations have laws, rules and safeguards in place to prevent gamblers from executing such an operation. Further, just winning the jackpot by buying every possible combination does not guarantee to break even or make a profit.

If p is the probability to win; c_{t}the cost of a ticket; c_{l} the cost for obtaining a ticket (e.g. including the logistics); c_{f} one time costs for the operation (such as setting up and conducting the operation); then the jackpot m_{j} should contain at least

{\displaystyle m_{j}\geq c_{f}+{\frac {c_{t}+c_{l}}{p}}}

to have a chance to at least break even.

The above theoretical "chance to break-even" point is slightly offset by the sum {\displaystyle \sum _{i}{}m_{i}} of the minor wins also included in all the lottery tickets:

{\displaystyle m_{j}\geq c_{f}+{\frac {c_{t}+c_{l}}{p}}-\sum _{i}{}m_{i}}

Still, even if the above relation is satisfied, it does not guarantee to break even. The payout depends on the number of winning tickets for all the prizes n_{x}, resulting in the relation

{\displaystyle {\frac {m_{j}}{n_{j}}}\geq c_{f}+{\frac {c_{t}+c_{l}}{p}}-\sum _{i}{}{\frac {m_{i}}{n_{i}}}}

In probably the only known successful operations[2] the threshold to execute an operation was set at three times the cost of the tickets alone for unknown reasons

{\displaystyle m_{j}\geq 3\times {\frac {c_{t}}{p}}}


{\displaystyle {\frac {n_{j}p}{c_{t}}}\left(c_{f}+{\frac {c_{t}+c_{l}}{p}}-\sum _{i}{}{\frac {m_{i}}{n_{i}}}\right)\ll 3}

This does, however, not eliminate all risks to make no profit. The success of the operations still depended on a bit of luck. In addition, in one operation the logistics failed and not all combinations could be obtained. This added the risk of not even winning the jackpot at all.

Powerballs and bonus balls[edit]

Further information: Twelvefold way

Many lotteries have a Powerball (or "bonus ball"). If the powerball is drawn from a pool of numbers different from the main lottery, the odds are multiplied by the number of powerballs. For example, in the 6 from 49 lottery, given 10 powerball numbers, then the odds of getting a score of 3 and the powerball would be 1 in 56.66 × 10, or 566.6 (the probability would be divided by 10, to give an exact value of {\textstyle {\frac {8815}{4994220}}}). Another example of such a game is Mega Millions, albeit with different jackpot odds.

Where more than 1 powerball is drawn from a separate pool of balls to the main lottery (for example, in the EuroMillions game), the odds of the different possible powerball matching scores are calculated using the method shown in the "other scores" section above (in other words, the powerballs are like a mini-lottery in their own right), and then multiplied by the odds of achieving the required main-lottery score.

If the powerball is drawn from the same pool of numbers as the main lottery, then, for a given target score, the number of winning combinations includes the powerball. For games based on the Canadian lottery (such as the lottery of the United Kingdom), after the 6 main balls are drawn, an extra ball is drawn from the same pool of balls, and this becomes the powerball (or "bonus ball"). An extra prize is given for matching 5 balls and the bonus ball. As described in the "other scores" section above, the number of ways one can obtain a score of 5 from a single ticket is {\textstyle {6 \choose 5}{43 \choose 1}=258}. Since the number of remaining balls is 43, and the ticket has 1 unmatched number remaining, 1/43 of these 258 combinations will match the next ball drawn (the powerball), leaving 258/43 = 6 ways of achieving it. Therefore, the odds of getting a score of 5 and the powerball are {\textstyle {6 \over {49 \choose 6}}={1 \over 2,330,636}}.

Of the 258 combinations that match 5 of the main 6 balls, in 42/43 of them the remaining number will not match the powerball, giving odds of {\textstyle {{258\cdot {\frac {42}{43}}} \over {49 \choose 6}}={\frac {3}{166,474}}\approx 1.802\times 10^{-5}} for obtaining a score of 5 without matching the powerball.

Using the same principle, the odds of getting a score of 2 and the powerball are {\textstyle {6 \choose 2}{43 \choose 4}=1,\!851,\!150} for the score of 2 multiplied by the probability of one of the remaining four numbers matching the bonus ball, which is 4/43. Since {\textstyle 1,851,150\cdot {\frac {4}{43}}=172,\!200}, the probability of obtaining the score of 2 and the bonus ball is {\textstyle {\frac {172,200}{49 \choose 6}}={\frac {1025}{83237}}=1.231\%}, approximate decimal odds of 1 in 81.2.

The general formula for B matching balls in a N choose K lottery with one bonus ball from the N pool of balls is:

{\displaystyle {\frac {{\frac {K-B}{N-K}}{K \choose B}{N-K \choose K-B}}{N \choose K}}}

The general formula for B matching balls in a N choose K lottery with zero bonus ball from the N pool of balls is:

{\displaystyle {N-K-K+B \over N-K}{K \choose B}{N-K \choose K-B} \over {N \choose K}}

The general formula for B matching balls in a N choose K lottery with one bonus ball from a separate pool of P balls is:

{\displaystyle {1 \over P}{K \choose B}{N-K \choose K-B} \over {N \choose K}}

The general formula for B matching balls in a N choose K lottery with no bonus ball from a separate pool of P balls is:

{\displaystyle {P-1 \over P}{K \choose B}{N-K \choose K-B} \over {N \choose K}}

Minimum number of tickets for a match[edit]

It is a hard (and often open) problem to calculate the minimum number of tickets one needs to purchase to guarantee that at least one of these tickets matches at least 2 numbers. In the 5-from-90 lotto, the minimum number of tickets that can guarantee a ticket with at least 2 matches is 100.[3]

Information theoretic results[edit]

Further information: Quantities of information

As a discreteprobability space, the probability of any particular lottery outcome is atomic, meaning it is greater than zero. Therefore, the probability of any event is the sum of probabilities of the outcomes of the event. This makes it easy to calculate quantities of interest from information theory. For example, the information content of any event is easy to calculate, by the formula

{\displaystyle \operatorname {I} (E):=-\log {\left[\Pr {\left(E\right)}\right]}=-\log {\left(P\right)}.}

In particular, the information content of outcomex of discrete random variable X is

{\displaystyle \operatorname {I} _{X}(x):=-\log {\left[p_{X}{\left(x\right)}\right]}=\log {\left({\frac {1}{p_{X}{\left(x\right)}}}\right)}.}

For example, winning in the example § Choosing 6 from 49 above is a Bernoulli-distributed random variable  X with a 1/13,983,816 chance of winning ("success") We write {\textstyle X\sim \mathrm {Bernoulli} \!\left(p\right)=\mathrm {B} \!\left(1,p\right)} with {\textstyle p={\tfrac {1}{13,983,816}}} and {\textstyle q={\tfrac {13,983,815}{13,983,816}}}. The information content of winning is

{\displaystyle \operatorname {I} _{X}({\text{win}})=-\log _{2}{p_{X}{({\text{win}})}}=-\log _{2}\!{\tfrac {1}{13,983,816}}\approx 23.73725}
shannonsor bitsof information. (See units of informationfor further explanation of terminology.) The information content of losing is
{\displaystyle {\begin{aligned}\operatorname {I} _{X}({\text{lose}})&=-\log _{2}{p_{X}{({\text{lose}})}}=-\log _{2}\!{\tfrac {13,983,815}{13,983,816}}\\&\approx 1.0317\times 10^{-7}{\text{ shannons}}.\end{aligned}}}

The information entropy of a lottery probability distribution is also easy to calculate as the expected value of the information content.

{\displaystyle {\begin{alignedat}{2}\mathrm {H} (X)&=\sum _{x}{-p_{X}{\left(x\right)}\log {p_{X}{\left(x\right)}}}\ &=\sum _{x}{p_{X}{\left(x\right)}\operatorname {I} _{X}(x)}\\&{\overset {\underset {\mathrm {def} }{}}{=}}\ \mathbb {E} {\left[\operatorname {I} _{X}(x)\right]}\end{alignedat}}}

Oftentimes the random variable of interest in the lottery is a Bernoulli trial. In this case, the Bernoulli entropy function may be used. Using  X representing winning the 6-of-49 lottery, the Shannon entropy of 6-of-49 above is

{\displaystyle {\begin{aligned}\mathrm {H} (X)&=-p\log(p)-q\log(q)=-{\tfrac {1}{13,983,816}}\log \!{\tfrac {1}{13,983,816}}-{\tfrac {13,983,815}{13,983,816}}\log \!{\tfrac {13,983,815}{13,983,816}}\\&\approx 1.80065\times 10^{-6}{\text{ shannons.}}\end{aligned}}}


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