I guess I’m revealing my intellectual ineptitude for all to see, but I actually wrote a short routine to explore that.
First, the probability of 000111 is 1/64. That is obvious.
Allowing for repeated occurrences, so that 0000 would count as two occurrences for instance, this codelet surprised me with the result (length(z)):
for k1 = 1:64
q(k1,:) = dec2bin(k1-1,6);
z3{k1} = strfind(q(k1,:),'000');
end
z = find(cell2mat(z3));
I wouldn’t mind an analytical proof of this. It’s not obvious to me.
