>>16805528 (OP)
let's say that lottery numbers are compromised of [math]n[/math] digits (between 0 and 9, including 0 and 9). we will have a total of [math]10^n[/math] lottery numbers.
what does "picking random" means? it can pick randomly with equal probability two different numbers ignoring others (0 probability for others). let's us say that every number's probability equal to [math]\frac{1}{10^n}[/math], for simplicity we will denote it as [math]p(n)[/math]. so now the question is: if we pick numbers for infinite ammount of time, will we get every number?
what is the probability of picking a number [math]x[/math] for our first attempt? well, obviously [math]p(n)[/math], not a big number for a big [math]n[/math]. maybe we could pick [math]x[/math] in two attempts? we draw two numbers from the machine, what is the probability that [math]x[/math] is one of them? we have three variants: pulled [math]x[/math] in first attempt, pulled [math]x[/math] in second attempt, pulled [math]x[/math] twice. the total probability will be [math]p(n)(1-p(n)) + (1-p(n))p(n) + p(n)^2[/math]. bigger than [math]p(n)[/math] but still a small number for sufficiently large [math]n[/math].
okay, what about the general case? quite easy: it will be complementary probability of not getting [math]x[/math]. so the probability of getting at least 1 time the number [math]x[/math] in [math]k[/math] equals to [math]1 - (1-p(n))^k[/math]. as you can see, because [math]0<1-p(n)<1[/math] as [math]k[/math] grows we will have that [math](1-p(n))^k \rightarrow 0[/math]. in this sense, in infinite ammount of time we will get [math]x[/math] no matter what.
concrete example: let's say [math]n[/math] equals to 6, then we have [math]1-p(6) = 0.999999[/math]. let [math]k[/math] be [math]10 000 000[/math] (ten million), then the probability of [math]x[/math] being picked at least once in ten million is [math]1 - 0.999999^10000000 \approx 0.9999546[/math]. yep, a safe bet that it will be picked.
