Lottery Numbers

 

Figure 1

Figure 1 shows the Lotto results for Saturday night 21st June 2025. What's unusual about the numbers is that seven out of the eight numbers are even. I was wondering as to the probability of this happening. To begin with there are 45 numbers to choose from: 22 are even and 23 are odd. See Figure 2.

Figure 2

Looking at the draw we see that the first number drawn was a 6 and this is an even number. The probability of drawing an even number is 22/45. The next three numbers were all even as well and the probabilities are therefore 21/44, 20/43 and 19/42. Then comes the only odd number, 1, and the probability of drawing an odd number is 23/41. The final three numbers are all even with probabilities of 18/40, 17/39 and 16/38. So the probability of this configuration:$$ \text{even, even, even, even. odd, even, even, even}$$is given by:$$\frac{22}{45} \times \frac{21}{44} \times \frac{20}{43} \times \frac{19}{42} \times \frac{23}{41} \times \frac{18}{40} \times \frac{17}{39} \times \frac{16}{38} \approx 0.0023$$However, we're not concerned with that particular order of even and odd numbers. We only want to determine the probability of seven even numbers and one odd number. The odd number could be in the first to eighth position. Each of these eight configurations has the same probability, around 0.0023, so we need to multiply by eight. This gives 0.01820 to five decimal places or a probability of around 1.82% and so$$\text{probability of one odd and seven even numbers } \approx 1.82 \text{%} $$Another way to approach the probability is via combinations. Let's see if we get the same result using this approach. There are \(^{22} \text{C}_{7}\) ways of choosing seven even numbers from the 22 available and \( ^23 \text{C}_1\) ways of choosing an odd number. With no restrictions, there are \( ^45 \text{C}_8 \) ways of choosing eight numbers from the 45 available. Thus:$$ \begin{align} \text{probability } &= \frac{^{22} \text{C}_{7} \times ^{23} \text{C}_1}{^{45} \text{C}_8}\\  \\[-1.5 ex] &=\frac{170544 \times 23}{215553195} \\[-0.5 ex]  \\ &\approx 0.01820 \\[-1.5 ex]  \\ &\approx 1.82\text{%} \end{align} $$So there is agreement. The probability that all the numbers are even would be given by:$$ \begin{align} \text{probability all even } &= \frac{ ^{22} \text{C}_8}{ ^{45} \text{C}_8}\\ \\[-1.5 ex] &\approx 0.00148 \\[-1.5 ex] \\ &\approx 0.148 \text{%} \end{align}$$This is considerably more unlikely than getting seven out of eight!