You flip a coin having a 5 on one side and a 10 on the other. At the same time, you throw a normal 6-sided die. So there are 12 possible combinations. You add the two numbers. This gives 10 possible unique results and two possibilities to get 11 (5+6 and 10+1).
If you perform the experiment twice, what is the probability to get 11 at least once?
I thought that this has to be calculated by using the counter-probability. The probability to get 11 is $\frac{2}{12}$ or $\frac{1}{6}$. So the counter-probability when doing the experiment twice is $\frac{5}{6}^2 = \frac{25}{36}$. Therefore the probability to get 11 at least once should be $\frac{11}{36}$ (30.55 %).
But this seems to be incorrect. What am I doing wrong?
Your calculation is correct, the probability is $\frac{11}{36}$. Your code (given in comments) is not giving the right answer because it outputs #successes / #failures instead of #successes / #attempts. Just change the last line to
print(t/1000000).