I'm using the ternary numeral system, i.e. numbers in base 3. There's a catch: The numbers I'm representing will never have a 2 as a digit. For instance, I use 0,1,10,11,100,101,110,111,... I.e. they look like binary digits, but they're really ternary numbers.
Now I'd like to know, for a random ternary number, in this form, of $m$ digits, what is the probability that the first (least significant) $n$ bits are all zero?
For $m=3$, we'd have the following numbers written out in base 3 (remember they only look like they're bits): 000,001,010,011,100,101,110,111. These correspond to the decimal values 0,1,3,4,9,10,12,13 respectively. So representing these in binary, they are 0,1,11,100,1001,1010,1100,1101 respectively. If I take $n=2$, I'd be finding the probability that the 2 least significant bits are all zero. Of the numbers in binary, 0,100, and 1100 have the 2 least significant bits as zero. So out of the 8 possible numbers that I have, 3 have this property. So the probability that I'm looking for is 3/8.
Now, I'm asking, for generalized $m$ and $n$ using this system, what are the probabilities?