Lottery mathematics

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This article discusses the way to calculate various probabilities in a lottery game in which one selects 6 numbers from 49, and hopes that as many of those 6 as possible match the 6 that are randomly selected from the same pool of 49 numbers in the "draw".

1 Calculation explained in choosing 6 from 49
2 Odds of getting other possibilities in choosing 6 from 49
3 Powerballs And Bonus Balls
4 See also
5 External links

[edit] Calculation explained in choosing 6 from 49

In a typical 6/49 lotto, 6 numbers are drawn from a range of 49 and if the 6 numbers on a ticket match the numbers drawn, the ticket holder is a jackpot winner—this is true no matter the order in which the numbers appear. The odds of this happening are 1 in 14 million (13,983,816 to be exact).

The relatively small chance of winning can be demonstrated as follows:

Starting with a bag of 49 differently-numbered lottery balls, there is clearly a 1 in 49 chance of predicting the number of the 1st ball out of the bag. Accordingly, there are 49 different ways of choosing that first number. When the draw comes to the second number, there are now only 48 balls left in the bag (because the balls already drawn are not returned to the bag), so there is now a 1 in 48 chance of predicting this number.

Thus, each of the 49 ways of choosing the first number has 48 different ways of choosing the second. This means that the odds of correctly predicting 2 numbers drawn from 49 is calculated as 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but of course someone picking numbers would have gotten to this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49 is calculated as 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 ${49!\over (49-6)!}$ . This works out to a very large number, 10,068,347,520, which is however much bigger than the 14 million stated above.

The last step is to understand that 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 ways they could be drawn. Dividing 10,068,347,520 by 720 gives 13,983,816, also written as 49! / (6! × (49 - 6)!), or more generally as

${n\choose k}={n!\over k!(n-k)!}$ .

In most popular spreadsheets, this function, called the combination function is denoted COMBIN(n, k). For example, COMBIN(49, 6) (the calculation shown above), would return 13,983,816. For the rest of this article, we will use the notation ${n\choose k}$

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

One must calculate the total number of lottery combinations (for example, ${49\choose 6} = 13,983,816$ , as explained in the section above), and divide it by the number of those combinations which give the desired result. This equates to the number of ways one can select the winning numbers multiplied by the number of ways one can select the losing numbers.

For a score of n (for example, if 3 of your numbers match the 6 balls drawn, then n = 3), there are ${6\choose n}$ ways of selecting the n winning numbers from the 6 drawn balls. For losing numbers, there are ${43\choose 6-n}$ ways to select them from the 43 losing lottery numbers. The total number of combinations giving that result is, as stated above, the first number multiplied by the second. The expression is therefore ${49\choose 6}\over {6\choose n}{43\choose 6-n}$ .

This gives the following results:

Score	Calculation	Exact Probability	Approximate Decimal Probability
0	${49\choose 6}\over {6\choose 0}{43\choose 6}$	435,461/998,844	0.436
1	${49\choose 6}\over {6\choose 1}{43\choose 5}$	68,757/166,474	0.413
2	${49\choose 6}\over {6\choose 2}{43\choose 4}$	44,075/332,948	0.132
3	${49\choose 6}\over {6\choose 3}{43\choose 3}$	8,815/499,422	0.0177
4	${49\choose 6}\over {6\choose 4}{43\choose 2}$	645/665,896	0.000969
5	${49\choose 6}\over {6\choose 5}{43\choose 1}$	43/2,330,636	0.0000184
6	${49\choose 6}\over {6\choose 6}{43\choose 0}$	1/13,983,816	0.0000000716

[edit] Powerballs And Bonus Balls

Many lotteries have a "powerball" (or "bonus ball"). If the powerball is drawn from a pool of numbers different from the main lottery, then simply multiply the odds by the number of powerballs. For example, in the 6 from 49 lottery, if there were 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, of course, be divided by 10, to give an exact value of 8815/4994220).

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 should be calculated using the method shown in the "other scores" section above (in other words, treat the powerballs 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, one must calculate the number of winning combinations, including the powerball. For games based on the Canadian lottery (such as Lotto, the United Kingdom's lottery), 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"), and there is an extra prize 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 ${6\choose 5}{43\choose 1}$ or 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). So, there are 258/43 = 6 ways of achieving it. Therefore, the odds of getting a score of 5 and the powerball are ${49\choose 6}\over {6}$ = 1 in 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 ${49\choose 6}\over {258 \cdot {{42}\over {43}}}$ = 166,474/3 (approximately 55,491.33) for obtaining a score of 5 without matching the powerball.

Using the same principle, to calculate the odds of getting a score of 2 and the powerball, calculate the number of ways to get a score of 2 as ${6\choose 2}{43\choose 4}$ = 1,851,150 then multiply this by the probability of one of the remaining four numbers matching the bonus ball, which is 4/43. SInce 1,851,150 × (4/43) = 172,200, the probability of obtaining the score of 2 and the bonus ball is ${172,200}\over {49\choose 6}$ = 1025/83237. This gives approximate decimal odds of 81.2.