Lotteries have long captivated the imagination of hopeful participants, offering the possibility of instant wealth. However, for a select few, the experience becomes a perplexing journey through a winless lottery streak. In this work, we delve into the mathematics underlying such streaks, exploring the concepts of probability, randomness, and the elusive nature of winning. Specifically, we give an upper bound estimate of the streaks and apply it to two real-world examples.

If p is the probability of winning the jackpot, then $q=1-p$ is the probability of not winning. Let X be the number of draws when a winning ticket is found. We denote the number of tickets sold at the k^th draw by $n_k$.

Theorem 1 On average, the number of draws when the winning ticket is found is

$E\left[X\right]=1+q^{n_1}+q^{n_1+n_2}+q^{n_1+n_2+n_3}+\cdots \mathrm{.}$

Proof The probability mass function of X is

$X=(\begin{array}{ll}\mathrm{1,}\hfill & \text{with}\text{probability}1-q^{n_1}\hfill \\ \mathrm{2,}\hfill & \text{with}\text{probability}{q}^{n_1}\left(1-q^{n_2}\right)\hfill \\ \mathrm{3,}\hfill & \text{with}\text{probability}{q}^{n_1}{q}^{n_2}\left(1-q^{n_3}\right)\hfill \\ \hfill & ⋮\hfill \\ k\mathrm{,}\hfill & \text{with}\text{probability}{q}^{n_1}{q}^{n_2}\cdots q^{n_{k-1}}\left(1-q^{n_k}\right)\hfill \\ \hfill & ⋮\hfill \end{array}$

This is a generalized geometric distribution and the expectation of X is given by [1]

$E\left[X\right]=1-q^{n_1}+2q^{n_1}\left(1-q^{n_2}\right)+3q^{n_1}{q}^{n_2}\left(1-q^{n_3}\right)+\cdots \mathrm{.}$

We rewrite this sum and add each row in Table 1 and observe that the sum in each row is a telescopic series.

Therefore,

$E\left[X\right]=1+q^{n_1}+q^{n_1+n_2}+q^{n_1+n_2+n_3}+\cdots$ $■$

If there is no winning ticket that matches all the lottery numbers in the current draw, the lottery authority raises the jackpot amount. As a result, more tickets will be sold in the next draw. It is reasonable to assume the number of tickets sold is increasing: $n_1<n_2<\cdots$.

Corollary 2 $E\left[X\right]$ is at most $\frac{1}{1-q^{n_1}}$.

Proof

$\begin{array}{c}E\left[X\right]=1+q^{n_1}+q^{n_1+n_2}+q^{n_1+n_2+n_3}+\cdots \\ \le 1+q^{n_1}+q^{2n_1}+q^{3n_1}+\cdots \\ =\frac{1}{1-q^{n_1}}\end{array}$ $■$

Example The longest streak of non-winning draws in Mega Millions® [2] was 36 consecutive drawings that started on September 18, 2020 and ended on January 22, 2021. During this period, no ticket matched all the numbers drawn to win the jackpot. The number of the tickets sold on September 18, 2020 was 8,517,586. For each jackpot, between seven and nine million Mega Millions® tickets are sold initially.

The probability of winning the jackpot in Mega Millions® is $p=\frac{1}{\left(\begin{array}{c}70\\ 5\end{array}\right)\times 25}$ and the probability of not winning the jackpot is $q=1-p=0.99999999669504$.

From Table 2 [3] , the number of tickets sold is steadily increasing since the first draw: $n_1<n_2<\cdots$. By Corollary 2,

Continued

$\begin{array}{c}E\left[X\right]\le (\begin{array}{ll}\frac{1}{1-{0.99999999669504}^{7000000}}\mathrm{,}\hfill & \text{if}{n}_1=7000000\hfill \\ \frac{1}{1-{0.99999999669504}^{9000000}}\mathrm{,}\hfill & \text{if}{n}_1=9000000\hfill \end{array}\\ =(\begin{array}{ll}\mathrm{43.7,}\hfill & \text{if}{n}_1=7000000\hfill \\ \mathrm{34.1,}\hfill & \text{if}{n}_1=9000000\hfill \end{array}\end{array}$

Example According to Powerball [4] , when the jackpot is low, initially, around six to ten million tickets are sold. The longest streak of non-winning draws in Powerball was 41 consecutive drawings that started on August 6, 2022 and ended on November 7, 2022. The probability of winning the jackpot in

Powerball is $p=\frac{1}{\left(\begin{array}{c}69\\ 5\end{array}\right)\cdot 26}$ and the probability of not winning the jackpot is $q=1-p=0.9999999965777$. By Corollary 2,

$\begin{array}{c}E\left[X\right]\le (\begin{array}{ll}\frac{1}{1-{0.9999999965777}^{6000000}},\hfill & \text{if}{n}_1=6000000\hfill \\ \frac{1}{1-{0.9999999965777}^{10000000}},\hfill & \text{if}{n}_1=10000000\hfill \end{array}\\ =(\begin{array}{ll}49.2,\hfill & \text{if}{n}_1=6000000\hfill \\ 29.7,\hfill & \text{if}{n}_1=10000000\hfill \end{array}\end{array}$

Corollary 3 If $\left\{n_k\right\}$ is an geometric sequence $n_k={\alpha }^{k-1}{n}_1$, for $k\ge 1$ and $\alpha >1$, then $E\left[X\right]=q^{-\frac{1}{\alpha -1}{n}_1}f\left(q^{\frac{n_1}{\alpha -1}},\alpha \right)$, where $f\left(x\mathrm{,}y\right)=x+x^y+x^{y^2}+x^{y^3}+\cdots$.

Proof

$\begin{array}{c}E\left[X\right]=1+q^{n_1}+q^{n_1+n_2}+q^{n_1+n_2+n_3}+\cdots \\ =1+q^{n_1}+q^{\left(1+\alpha \right)n_1}+q^{\left(1+\alpha +{\alpha }^2\right)n_1}+\cdots \\ =1+q^{\frac{\alpha -1}{\alpha -1}{n}_1}+q^{\frac{{\alpha }^2-1}{\alpha -1}{n}_1}+q^{\frac{{\alpha }^3-1}{\alpha -1}{n}_1}+q^{\frac{{\alpha }^4-1}{\alpha -1}{n}_1}+\cdots \\ =q^{-\frac{1}{\alpha -1}{n}_1}\left(q^{\frac{n_1}{\alpha -1}}+{\left(q^{\frac{n_1}{\alpha -1}}\right)}^{\alpha }+{\left(q^{\frac{n_1}{\alpha -1}}\right)}^{{\alpha }^2}+{\left(q^{\frac{n_1}{\alpha -1}}\right)}^{{\alpha }^3}+\cdots \right)\end{array}$ $■$

Example Here are some values of $E\left[X\right]$ for various $n_1$ and $\alpha$ in Corollary 3: ( Table 3 )

The winless lottery streak, while disheartening for those experiencing it, is a mathematical phenomenon rooted in the principles of probability, randomness, and statistical patterns. Understanding the mathematics behind lotteries helps us appreciate the rarity of winning and the statistical inevitability of winless streaks. While participants may yearn for a breakthrough, it is crucial to remember that lotteries are primarily games of chance, where each drawing offers an independent opportunity for success. The mathematics of the winless lottery streak reminds us of the complexities of probability and the unpredictable nature of random processes, inviting us to approach lotteries with both hope and a realistic understanding of the odds.