Lottery mathematics

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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]

Suppose there are N unique balls (such as N = 49) to be drawn from. Suppose a subset of K balls (such as K = 6) is drawn as the winning set. The value of K also is the number of balls selected on a lottery ticket. Suppose B of the K balls from the lottery ticket are also in the winning set. Out of the ${\displaystyle {N \choose K}}$ possible ways (see binomial coefficient) to draw the winning set, there will ${\displaystyle {K \choose B}}$ ways to have B of them come from the K on the lottery ticket and ${\displaystyle {N-K \choose K-B}}$ ways to have K − B of them come from the set of N − K not mentioned on the lottery ticket. That is, when choosing K balls from a pool of N balls, the probability that there will be B matches between the lottery ticket and the winning set is given by $${\displaystyle \Pr[B\mid N,K]={\frac {{K \choose B}{N-K \choose K-B}}{N \choose K}}\,.}$$

The chances of getting B matches when drawing K balls from a pool of N balls.

B matches	K=6 balls from N=49	K=5 balls from N=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	${\displaystyle {\frac {{6 \choose 6}{43 \choose 0}}{49 \choose 6}}={\frac {1\cdot 1}{13{,}983{,}816}}\approx 0.000\,000\,0715}$

Some lotteries also have one or more power balls drawn from a separate pool of balls. For example the first drawing may be for K₁ = 5 balls out of N₁ = 69 balls and then K₂ = 1 power ball may be drawn from N₂ = 26 balls. In this case, the probabilities of getting B₁ matches in the first drawing and B₂ matches in the power ball drawing is just the product of the individual probabilities: $${\displaystyle \Pr[B_{1},B_{2}\mid N_{1},K_{1},N_{2},K_{2}]={\frac {{K_{1} \choose B_{1}}{N_{1}-K_{1} \choose K_{1}-B_{1}}}{N_{1} \choose K_{1}}}{\frac {{K_{2} \choose B_{2}}{N_{2}-K_{2} \choose K_{2}-B_{2}}}{N_{2} \choose K_{2}}}\,,}$$ and similarly for three or more pools of balls.

The chances of getting B₁ matches when drawing K₁ = 5 balls from a pool of N₁ = 69 balls and getting B₂ matches when drawing K₂ = 1 power balls from a separate pool of N₂ = 26 balls:

B1 + B2 matches	K1=5 balls from N1=69 and K2=1 balls from N2=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	${\displaystyle {\frac {{5 \choose 5}{64 \choose 0}}{69 \choose 5}}{\frac {{1 \choose 1}{25 \choose 0}}{26 \choose 1}}={\frac {1\cdot 1\cdot 1\cdot 1}{11{,}238{,}513\cdot 26}}\approx 0.000\,000\,003\,42}$

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 L balls out of N possibilities, but the drawing is for K₁ balls plus K₂ bonus balls. The probability that B₁ balls from the first drawing match the lottery ticket and B₂ balls from the bonus-ball drawing match the lottery ticket is given by $${\displaystyle \Pr[B_{1},B_{2}\mid N,L,K_{1},K_{2}]={\frac {{K_{1} \choose B_{1}}{K_{2} \choose B_{2}}{{N-K_{1}-K_{2}} \choose {L-B_{1}-B_{2}}}}{N \choose L}}\,.}$$

When L = 6 of N = 49 numbers are on a lottery ticket but the winning set is K₁ = 6 numbers plus K₂ = 1 bonus number then we have 49!/6!/1!/42!.=combin(49,6)*combin(49-6,1)=601304088 different possible drawing results.

Score	Calculation	Exact Probability	Approximate Decimal Probability	Approximate 1/Probability
5 + 0	${\displaystyle {6 \choose 5}{1 \choose 0}{42 \choose 1} \over {49 \choose 6}}$	252/13983816	0.0000180208	55,491.33
5 + 1	${\displaystyle {6 \choose 5}{1 \choose 1}{42 \choose 0} \over {49 \choose 6}}$	6/13983816	0.0000004291	2,330,636

You would expect to score 3 of 6 or better once in around 36.19 drawings. Notice that It takes a 3 if 6 wheel of 163 combinations to be sure of at least one 3/6 score.

1/p changes when several distinct combinations are played together. It mostly is about winning something, not just the jackpot.

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 that one will break even or make a profit.

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

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

I.e.

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

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.

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), (i.e., 1 in 566.6), with the 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 ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with one bonus ball from the ${\displaystyle N}$ 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 ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with zero bonus ball from the ${\displaystyle N}$ 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 ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with one bonus ball from a separate pool of ${\displaystyle P}$ balls is:

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

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

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

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]

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]

As a discrete probability 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 outcome ${\displaystyle x}$ of discrete random variable ${\displaystyle X}$ 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 ${\displaystyle X}$ 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

$${\displaystyle \operatorname {I} _{X}({\text{win}})=-\log _{2}{p_{X}{({\text{win}})}}=-\log _{2}\!{\tfrac {1}{13,983,816}}\approx 23.73725}$$shannons or bits of information. (See units of information for 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 ${\displaystyle X}$ representing winning the 6-of-49 lottery, the Shannon entropy of 6-of-49 above is

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

• 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