Losing at Lotto

Figure 1

Each week, I play four games in the Saturday Night Gold Lotto and, though I seldom win a prize, I always manage to choose at least one winning number. For each game in this particular lottery, you must choose six numbers out of a total of 45 (numbered from 1 to 45) and, during the draw, six main numbers are chosen as well as two supplementary numbers (so there are eight "winning" numbers). If you select all six main numbers, you share in the first division prize of $4,000,000. The amount is divided among the total numbers of winners which on average is between four and five. Combinations of main numbers and supplementary numbers win lesser prizes. Tonight, I was surprised to see that I'd managed to select none of the eight winning numbers. I've attached a screenshot of my results (see Figure 1). This got me wondering what was the probability of doing this? It had to be fairly low because I couldn't recall it ever happening to me before.

To tackle the problem, let's consider a single game in which there are 37 "losing" numbers along with the eight "winning" numbers. How many ways can these 37 numbers be chosen if I am selecting six numbers at a time, disregarding the order in which they are chosen. There are \( ^{37}C_6 \)= 2516 ways to do this out of a total of \(^{45}C_6\) = 8815 possibilities. So the probability of choosing six losing numbers in a single game is 2516/8815 which is almost 0.29 or 29%. Thus for every 100 games I play, around 29 will result in my missing all of the winning numbers. However, it is highly unlikely that I'll "succeed" in doing this four times in a row because each game is independent of the other and so the probabilities are multiplied. Thus the fraction 2516/8815 is raised to the fourth power and this gives a probability of about 0.0066 or 0.66%. In other words, if I play 10,000 sets of four games each, I'll only miss all of the winning numbers about 66 times. Tonight was one of those times.

Below is shown the WolframAlpha calculation:

For an earlier post on the probability of winning at Oz Lotto, follow this link.

UPDATE: on Saturday March 6th 2021 I also managed to select no numbers out of four games. Thus it is almost three years since this last happened. Figure 2 shows a screenshot of the results.

Figure 2