If you compare the odds for Spiel 77 with those for Glücksspirale, an immediate question arises regarding the lowest prize classes (even the numbering differs: for Glücksspirale, class 1 is the smallest prize—twice the stake—whereas for Spiel 77 class 1 is the highest prize!).
This also yields different stated odds: 1 in 10 for Glücksspirale, but 1 in 11 for Spiel 77!
Yet the lotteries are quite similar: in both cases 7‑digit numbers are drawn, etc. Logically, it’s not obvious at first glance: in both, ending digits alone can win, and there are only 10 possibilities!
So why 1 in 11?!
Right up front: the tables published by the state lottery companies are correct—this is not an error!
On closer inspection, for Glücksspirale each prize class has its own drawing, whereas for Spiel 77 there is only a single 7‑digit number drawn. “So what?” one might ask—but this small difference is crucial!
For Spiel 77 to yield 1 in 10 odds for the last digit, the preceding digits would need to be irrelevant—but they aren’t. If any preceding position already matches, a win in a higher class occurs (e.g., “two correct”), which excludes a win in the lowest class. Thus the correct last digit does not always produce a class‑7 win.
From this, the probabilities per class in Spiel 77 are:
Class 1 (highest): 1 in 10,000,000
A “round” number because there is only one winning number drawn in Spiel 77.
Class 2 (6 correct ending digits): about 1 in 1,111,111
The first digit must be wrong; otherwise it would be the class‑1 combination. There are 9 such combinations (first digit wrong × 10 choices, but the 9 that differ from the winning first digit). 9 / 10,000,000 = 0.0000009 → reciprocal ≈ 1 in 1,111,111.11… (rounded).
Class 3 (5 correct ending digits): about 1 in 111,111
The first two digits must be wrong. There are 90 such combinations. 90 / 10,000,000 = 0.000009 → reciprocal ≈ 1 in 111,111.11… (rounded).
Class 4 (4 correct ending digits): about 1 in 11,111
The first three digits must be wrong. There are 900 such combinations. 900 / 10,000,000 = 0.00009 → reciprocal ≈ 1 in 11,111.11… (rounded).
Class 5 (3 correct ending digits): about 1 in 1,111
The first four digits must be wrong. There are 9,000 such combinations. 9,000 / 10,000,000 = 0.0009 → reciprocal ≈ 1 in 1,111.11… (rounded).
Class 6 (2 correct ending digits): about 1 in 111
The first five digits must be wrong. There are 90,000 such combinations. 90,000 / 10,000,000 = 0.009 → reciprocal ≈ 1 in 111.11… (rounded).
Class 7 (lowest; 1 correct ending digit): about 1 in 11
The first six digits must be wrong. There are 900,000 such combinations. 900,000 / 10,000,000 = 0.09 → reciprocal 1 in 11.11… (rounded).
Thus the overall probability of any win—that is, the probability that a single Spiel 77 ticket 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 + p_C7
= (1 − 0.9999999) + (1 − 0.9999991) + (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!
