Odds of obtaining "the" point

Discounting the come-out completely, What would be the odds of getting one point? Generally, any number?
How about 2 in a row? The question here is to not calculate the odds of a roll, again and again, but the odds of a point scoring a win for the shooter, once, or again and again, outside of the come-out roll.
What are the odds that the shooter will continue to get points (ignoring the come-out)?

1-point odds=
2-points in a row, same shooter=
3-points in a row, same shooter=
4-points in a row, same shooter=
5-points in a row, same shooter=
6-points in a row, same shooter=
7-points in a row, same shooter=
8-points in a row, same shooter=

So what I think I see is a continually degrading odd for success, a maturity of chances (not sure the terminology). I believe this is also referred to as the Gambling fallacy but not quite... I call it the Monte Carlo Mystery because I cant find a name if it exists. I think that this arrangement is a little different than the fallacy. It also seems like this calculation is going to be a lot harder than normal which is why I'm reaching out for help.

What do you see, or more importantly, what is REALLY there that I am looking at?
It can be difficult for me to make sense in this area where I have little education, and no mentor. Please bear with me, I do have a point and am not trying to waste your time. Any help is appreciated and thank you. Even if I'm really really way off in left field, Id sure like to see the math!

"One does not become enlightened by imagining figures of light but by making the darkness conscious"-Carl Jung

Given than an unknown point has been established, the probability of hitting it is 40/99 = 40.4%. The probability of doing that n times in a row, or more, is (40/99)^n.

Thank you very much for the reply I know you are extremely busy.

To me this is very helpful and thanks again, if there's ever anything I can do...