If you compare the odds for the Super 6 lottery with those for Glücksspirale, the immediate question is why the stated odds in the lowest prize classes differ: 1 in 10 for Glücksspirale, but 1 in 11 for Spiel 77!
At first glance this isn’t obvious: in both cases, the ending digits alone can win, and there are only 10 possibilities!
So why 1 in 11?!
Right away: the tables published by the state lottery companies are correct—this is not an error!
On closer inspection, for Glücksspirale the winning numbers for each prize class are drawn independently of each other (so you get different winning numbers per class). In the supplementary lottery Super 6, however, only a single six-digit number is drawn. “So what?” one might ask—but this small, subtle difference is exactly the point!
For Super 6 to yield exactly 1-in-10 odds for the last digit, the preceding digits would have to be irrelevant—but they aren’t. If any earlier position already matches, a win in a higher prize class occurs—namely two correct—so a correct last digit in that case does not produce a win in the lowest class.
This gives us the probabilities for each prize class in the Super 6 lottery as follows:
Prize class 1 (highest): 1 in 1,000,000
A “round” number, since only a single winning number is drawn in Super 6.
Prize class 2 (5 correct ending digits): about 1 in 111,111
The first digit must be wrong; otherwise it would be the class‑1 combination. There are 9 such combinations in which the 5 ending digits are correct and the first digit is wrong. 9 / 1,000,000 = 0.000009 → reciprocal ≈ 1 in 111,111.11… (rounded).
Prize class 3 (4 correct ending digits): about 1 in 11,111
The first two digits must be wrong. There are 90 such combinations. 90 / 1,000,000 = 0.00009 → ≈ 1 in 11,111.11… (rounded).
Prize class 4 (3 correct ending digits): about 1 in 1,111
The first three digits must be wrong. There are 900 such combinations. 900 / 1,000,000 = 0.0009 → ≈ 1 in 1,111.11… (rounded).
Prize class 5 (2 correct ending digits): about 1 in 111
The first four digits must be wrong. There are 9,000 such combinations. 9,000 / 1,000,000 = 0.009 → ≈ 1 in 111.11… (rounded).
Prize class 6 (1 correct ending digit): about 1 in 11
The first five digits must be wrong. There are 90,000 such combinations. 90,000 / 1,000,000 = 0.09 → ≈ 1 in 11.11… (rounded).
Thus the overall probability of any win—that is, the probability that a single ticket in the Super 6 lottery wins something at all—is the sum of the class probabilities (they are mutually exclusive):
p = p_C1 + p_C2 + p_C3 + p_C4 + p_C5 + p_C6 = (1 − 0.999999) + (1 − 0.999991) + (1 − 0.99991) + (1 − 0.9991) + (1 − 0.991) + (1 − 0.91) = 0.1, i.e., 10%.
And finally, a direct link to the latest lotto numbers! Good luck!
