In my experience, if it is difficult to calculate a probability, it is sometimes easier to calculate the inverse.
So how many 10 digit strings do not contain consecutive 0s?
We can ignore the first digit since one digit in of itself will have no impact on the probability.
From there, the probability that a single digit does not cause that string to contain consecutive 0s is inverse of the the probability that both the previous and the current digits are 0.
So if we were solving for 2 digit strings, the probability would simply be the inverse of the probability of having 0 digit followed by another 0 digit, or $1 - (\frac{1}{10\times10})$.
If we were to then apply this to a 3 digit string, the probability would be compounded with the inverse of the probability of the last digits having 0. This simply means to find the probability of not finding 0s in the first two digits and not finding 0s in the second two digits. This also covers the case of both finding digits in the first 2 digits as well as finding digits in the last 2 digits.
The probability of the first two digits not both having 0 is $1 - (\frac{1}{10\times10})$, and the probability of the last two digits not both having 0 is $1 - (\frac{1}{10\times10})$. Compounded, you would get $$1 - \frac{1}{\frac{1}{1 - (\frac{1}{10\times10})}\times\frac{1}{1 - (\frac{1}{10\times10})}}$$ Simplying a bit, we get:
$$1 - (1 -\frac{1}{10\times10})^{2}$$
Why the 2? Because we have 3 digits and 2 pairs of digits to consider. Generalizing, we would have:
$$1 - (1 - \frac{1}{10\times10})^{n-1}$$
Where $n$ is the number of digits. Take this probability and multiply times $10^n$ or in this case $10^{10}$ (total number of possible combinations) and you get the number of digits containing consecutive zeros.
It would seem that you're making the mistake of assuming the pattern is linear. When you're finding all configurations of non-0 digits in the case in which m is 1, there should be just as many configurations of 0 digits in the case in which m is 9. So:
$$xxxxxxxxx0,0xxxxxxxxx$$
If your linear calculation were accurate, then there should be 10000000000 ways to place that singular 0 digit! Hope that helps!
Best Answer
This question is very hard to solve analytically, however, I believe it is a problem on dynamic programming.
Let's say we have two arrays
XandYof sizes $x$ and $y$.X[i]will contain the number of sequences that end withi+1ones (if indexing starts with 0). Similarly,Y[i]will contain the number of sequences that end withi+1twos.Let's say $x=2$ and $y=3$. We start with arrays
X=[1,0]andY=[1,0,0]. It corresponds that we have only 1 sequence of length $n=1$ ending with 1 and 1 sequence of length $n=1$ ending with 2.For each sequence that ends with 1 we can append 2 to get proper sequence (and vice versa). However, for each sequence that ends in 1 we can append 1 only if the number of 1s is less than $x$.
Thus, the rules of transitions are:
This is outline of the code. Implementation might be more efficient.