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. It can also be used to analyze coincidences that happen in lottery drawings, such as repeated numbers appearing across different draws. ^[1]

In the following

• P is the number of balls in a pool of balls that the winning balls are drawn from, without replacement.
• W is the number of winning balls drawn from the pool.
• T is the number of balls listed on the lottery ticket. (It often equals W.)
• M is the number of balls that match on the lottery ticket and within the winning set.

Single pool of balls

Suppose there are P unique balls (such as P = 49) from which balls are to be drawn without replacement. Suppose a subset of W balls (such as W = 6) is drawn as the winning set. Suppose a subset of T balls (such as T = 6) is selected on a lottery ticket. Suppose M of the T balls from the lottery ticket are also among the W balls in the winning set. Out of the ${\displaystyle {P \choose W}}$ possible ways (see binomial coefficient) to draw the winning set, there are ${\displaystyle {T \choose M}}$ ways to have M of them come from the T on the lottery ticket and ${\displaystyle {P-T \choose W-M}}$ ways to have W − M of them come from the set of P − T not mentioned on the lottery ticket. That is, the probability of getting M matches is given by the following formula when there are P balls in the pool, each lottery ticket selects T balls, and W is the number of winning balls drawn for the lottery. $${\displaystyle \Pr[M\mid P,T,W]={\frac {{T \choose M}{P-T \choose W-M}}{P \choose W}}\,.}$$

Single pool examples

The chances of getting M matches when drawing W balls from a pool of P balls and lottery tickets with T balls each:

M matches	T=W=6 balls from P=49	T=W=5 balls from P=69
0	${\displaystyle {\frac {{6 \choose 0}{43 \choose 6}}{49 \choose 6}}={\frac {1\cdot 6{,}096{,}454}{13{,}983{,}816}}\approx 0.436}$	${\displaystyle {\frac {{5 \choose 0}{64 \choose 5}}{69 \choose 5}}={\frac {1\cdot 7{,}624{,}512}{11{,}238{,}513}}\approx 0.678}$
1	${\displaystyle {\frac {{6 \choose 1}{43 \choose 5}}{49 \choose 6}}={\frac {6\cdot 962{,}598}{13{,}983{,}816}}\approx 0.413}$	${\displaystyle {\frac {{5 \choose 1}{64 \choose 4}}{69 \choose 5}}={\frac {5\cdot 635{,}376}{11{,}238{,}513}}\approx 0.283}$
2	${\displaystyle {\frac {{6 \choose 2}{43 \choose 4}}{49 \choose 6}}={\frac {15\cdot 123{,}410}{13{,}983{,}816}}\approx 0.132}$	${\displaystyle {\frac {{5 \choose 2}{64 \choose 3}}{69 \choose 5}}={\frac {10\cdot 41{,}664}{11{,}238{,}513}}\approx 0.0371}$
3	${\displaystyle {\frac {{6 \choose 3}{43 \choose 3}}{49 \choose 6}}={\frac {20\cdot 12{,}341}{13{,}983{,}816}}\approx 0.0177}$	${\displaystyle {\frac {{5 \choose 3}{64 \choose 2}}{69 \choose 5}}={\frac {10\cdot 2{,}016}{11{,}238{,}513}}\approx 0.001\,79}$
4	${\displaystyle {\frac {{6 \choose 4}{43 \choose 2}}{49 \choose 6}}={\frac {15\cdot 903}{13{,}983{,}816}}\approx 0.000\,969}$	${\displaystyle {\frac {{5 \choose 4}{64 \choose 1}}{69 \choose 5}}={\frac {5\cdot 64}{11{,}238{,}513}}\approx 0.000\,0285}$
5	${\displaystyle {\frac {{6 \choose 5}{43 \choose 1}}{49 \choose 6}}={\frac {6\cdot 43}{13{,}983{,}816}}\approx 0.000\,0184}$	${\displaystyle {\frac {{5 \choose 5}{64 \choose 0}}{69 \choose 5}}={\frac {1\cdot 1}{11{,}238{,}513}}\approx 0.000\,000\,0890}$
6	${\displaystyle {\frac {{6 \choose 6}{43 \choose 0}}{49 \choose 6}}={\frac {1\cdot 1}{13{,}983{,}816}}\approx 0.000\,000\,0715}$

Power balls from a separate pool of balls

Some lotteries also have one or more power balls drawn from a separate pool of balls. For example, the first drawing may be for W₁ = 5 balls out of P₁ = 69 balls and then W₂ = 1 power ball may be drawn from P₂ = 26 balls. For both pools, the drawing is without replacement. Similarly a lottery ticket will indicate T₁ regular balls and T₂ power balls. Often the number of balls on the lottery ticket is the same as the winning set: T₁ = W₁ and T₂ = W₂. The probabilities of getting M₁ matches in the first drawing and M₂ matches in the power ball drawing is just the product of the individual probabilities: $${\displaystyle \Pr[M_{1},M_{2}\mid P_{1},T_{1},W_{1},P_{2},T_{2},W_{2}]={\frac {{T_{1} \choose M_{1}}{P_{1}-T_{1} \choose W_{1}-M_{1}}}{P_{1} \choose W_{1}}}{\frac {{T_{2} \choose M_{2}}{P_{2}-T_{2} \choose W_{2}-M_{2}}}{P_{2} \choose W_{2}}}\,,}$$ and similarly for three or more pools of balls.

For example, this formula could be used if one ball is drawn from a pool of balls numbered 0 to 9 and then a second and third ball are also drawn from their own pools of 10 balls: P₁ = P₂ = P₃ = 10, and T₁ = W₁ = T₂ = W₂ = T₃ = W₃ = 1. (Equivalently, all three balls could be drawn from the same pool, so long as each ball is returned to the pool before the next ball is drawn.) However, this case is more easily modeled as if a single ball, numbered from 0 (represented as "000") to 999 is drawn from a single pool of 1000 balls.

Separate pool examples

The chances of getting M₁ matches when drawing W₁ = 5 balls from a pool of P₁ = 69 balls, getting M₂ matches when drawing W₂ = 1 power balls from a separate pool of P₂ = 26 balls, and having T₁ = 5 regular balls and T₂ = 1 power balls selected on the lottery ticket:

M1 + M2 matches	T1=W1=5 balls from P1=69 and T2=W2=1 balls from P2=26
0 + 0	${\displaystyle {\frac {{5 \choose 0}{64 \choose 5}}{69 \choose 5}}{\frac {{1 \choose 0}{25 \choose 1}}{26 \choose 1}}={\frac {1\cdot 7{,}624{,}512\cdot 1\cdot 25}{11{,}238{,}513\cdot 26}}\approx 0.652}$
0 + 1	${\displaystyle {\frac {{5 \choose 0}{64 \choose 5}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {1\cdot 7{,}624{,}512\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.0261}$
1 + 0	${\displaystyle {\frac {{5 \choose 1}{64 \choose 4}}{69 \choose 5}}{\frac {{1 \choose 0}{25 \choose 1}}{26 \choose 1}}={\frac {5\cdot 635{,}376\cdot 1\cdot 25}{11{,}238{,}513\cdot 26}}\approx 0.272}$
1 + 1	${\displaystyle {\frac {{5 \choose 1}{64 \choose 4}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {5\cdot 635{,}376\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.001\,09}$
2 + 0	${\displaystyle {\frac {{5 \choose 2}{64 \choose 3}}{69 \choose 5}}{\frac {{1 \choose 0}{25 \choose 1}}{26 \choose 1}}={\frac {10\cdot 41{,}664\cdot 1\cdot 25}{11{,}238{,}513\cdot 26}}\approx 0.0356}$
2 + 1	${\displaystyle {\frac {{5 \choose 2}{64 \choose 3}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {10\cdot 41{,}664\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.001\,43}$
3 + 0	${\displaystyle {\frac {{5 \choose 3}{64 \choose 2}}{69 \choose 5}}{\frac {{1 \choose 0}{25 \choose 1}}{26 \choose 1}}={\frac {10\cdot 2{,}016\cdot 1\cdot 25}{11{,}238{,}513\cdot 26}}\approx 0.001\,72}$
3 + 1	${\displaystyle {\frac {{5 \choose 3}{64 \choose 2}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {10\cdot 2{,}016\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.000\,0690}$
4 + 0	${\displaystyle {\frac {{5 \choose 4}{64 \choose 1}}{69 \choose 5}}{\frac {{1 \choose 0}{25 \choose 1}}{26 \choose 1}}={\frac {5\cdot 64\cdot 1\cdot 25}{11{,}238{,}513\cdot 26}}\approx 0.000\,0274}$
4 + 1	${\displaystyle {\frac {{5 \choose 4}{64 \choose 1}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {5\cdot 64\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.000\,001\,10}$
5 + 0	${\displaystyle {\frac {{5 \choose 5}{64 \choose 0}}{69 \choose 5}}{\frac {{1 \choose 0}{25 \choose 1}}{26 \choose 1}}={\frac {1\cdot 1\cdot 1\cdot 25}{11{,}238{,}513\cdot 26}}\approx 0.000\,000\,0856}$
5 + 1	${\displaystyle {\frac {{5 \choose 5}{64 \choose 0}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {1\cdot 1\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.000\,000\,003\,42}$

Bonus balls from the same pool of balls

Some lotteries have one or more bonus balls drawn from the original pool of balls after the first round of balls is drawn. In this scenario each lottery ticket indicates T balls out of P possibilities, but the drawing is for W₁ balls plus W₂ bonus balls, without replacement. The probability that M₁ balls from the first drawing match the lottery ticket and M₂ balls from the bonus-ball drawing match the lottery ticket is given by $${\displaystyle \Pr[M_{1},M_{2}\mid P,T,W_{1},W_{2}]={\frac {{W_{1} \choose M_{1}}{W_{2} \choose M_{2}}{{P-W_{1}-W_{2}} \choose {T-M_{1}-M_{2}}}}{P \choose T}}\,.}$$

Example of balls and bonus balls from the same pool

When T = 6 of P = 49 numbers are on a lottery ticket but the winning set is W₁ = 6 numbers plus W₂ = 1 bonus ball then probabilities are as follows.

M1 + M2 matches	T=6 balls on ticket, W1=6 balls and W2=1 bonus ball drawn from P1=49 balls
0 + 0	${\displaystyle {\frac {{6 \choose 0}{1 \choose 0}{42 \choose 6}}{49 \choose 6}}={\frac {1\cdot 1\cdot 5{,}245{,}786}{13{,}983{,}816}}\approx 0.375}$
0 + 1	${\displaystyle {\frac {{6 \choose 0}{1 \choose 1}{42 \choose 5}}{49 \choose 6}}={\frac {1\cdot 1\cdot 850{,}668}{13{,}983{,}816}}\approx 0.0608}$
1 + 0	${\displaystyle {\frac {{6 \choose 1}{1 \choose 0}{42 \choose 5}}{49 \choose 6}}={\frac {6\cdot 1\cdot 850{,}668}{13{,}983{,}816}}\approx 0.365}$
1 + 1	${\displaystyle {\frac {{6 \choose 1}{1 \choose 1}{42 \choose 4}}{49 \choose 6}}={\frac {6\cdot 1\cdot 111{,}930}{13{,}983{,}816}}\approx 0.0480}$
2 + 0	${\displaystyle {\frac {{6 \choose 2}{1 \choose 0}{42 \choose 4}}{49 \choose 6}}={\frac {15\cdot 1\cdot 111{,}930}{13{,}983{,}816}}\approx 0.120}$
2 + 1	${\displaystyle {\frac {{6 \choose 2}{1 \choose 1}{42 \choose 3}}{49 \choose 6}}={\frac {15\cdot 1\cdot 11{,}480}{13{,}983{,}816}}\approx 0.0123}$
3 + 0	${\displaystyle {\frac {{6 \choose 3}{1 \choose 0}{42 \choose 3}}{49 \choose 6}}={\frac {20\cdot 1\cdot 11{,}480}{13{,}983{,}816}}\approx 0.0164}$
3 + 1	${\displaystyle {\frac {{6 \choose 3}{1 \choose 1}{42 \choose 2}}{49 \choose 6}}={\frac {20\cdot 1\cdot 861}{13{,}983{,}816}}\approx 0.00123}$
4 + 0	${\displaystyle {\frac {{6 \choose 4}{1 \choose 0}{42 \choose 2}}{49 \choose 6}}={\frac {15\cdot 1\cdot 861}{13{,}983{,}816}}\approx 0.000\,924}$
4 + 1	${\displaystyle {\frac {{6 \choose 4}{1 \choose 1}{42 \choose 1}}{49 \choose 6}}={\frac {15\cdot 1\cdot 42}{13{,}983{,}816}}\approx 0.000\,0451}$
5 + 0	${\displaystyle {\frac {{6 \choose 5}{1 \choose 0}{42 \choose 1}}{49 \choose 6}}={\frac {6\cdot 1\cdot 42}{13{,}983{,}816}}\approx 0.000\,0180}$
5 + 1	${\displaystyle {\frac {{6 \choose 5}{1 \choose 1}{42 \choose 0}}{49 \choose 6}}={\frac {6\cdot 1\cdot 1}{13{,}983{,}816}}\approx 0.000\,000\,429}$
6 + 0	${\displaystyle {\frac {{6 \choose 6}{1 \choose 0}{42 \choose 0}}{49 \choose 6}}={\frac {1\cdot 1\cdot 1}{13{,}983{,}816}}\approx 0.000\,000\,0715}$
6 + 1	Impossible

Ensuring to win the jackpot

There is only one way to ensure winning the jackpot; it 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 winning the jackpot in a 6/49 game. To be profitable, the cost of acquiring these tickets (including any overhead) must not exceed the total amount that those tickets will win, including jackpots and any smaller prizes. If it is likely that the jackpot (or any of the smaller prizes) will have to be split among several winners then only your likely share should be counted in the total amount that those tickets will win.

Minimum number of tickets for a match

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. ^[2]

Coincidences involving lottery numbers

Coincidences in lottery drawings often capture our imagination and can make news headlines as they seemingly highlight patterns in what should be entirely random outcomes. For example, repeated numbers appearing across different draws may appear on the surface to be too implausible to be by pure chance. For instance, on September 6, 2009, the six numbers 4, 15, 23, 24, 35, and 42 were drawn from 49 in the Bulgarian national 6/49 lottery, and in the very next drawing on September 10th, the same six numbers were drawn again. Lottery mathematics can be used to analyze these extraordinary events. ^[1]

References

1. M. Pollanen (2024) A Double Birthday Paradox in the Study of Coincidences, Mathematics 23(24), 3882. https://doi.org/10.3390/math12243882
2. Z. Füredi, G. J. Székely, and Z. Zubor (1996). "On the lottery problem". Journal of Combinatorial Designs. 4 (1): 5–10. doi:10.1002/(sici)1520-6610(1996)4:1<5::aid-jcd2>3.3.co;2-w.

External links

• Euler's Analysis of the Genoese Lottery – Convergence (August 2010), Mathematical Association of America
• Lottery Mathematics – INFAROM Publishing
• 13,983,816 and the Lottery – YouTube video with James Clewett, Numberphile, March 2012