TITLE: What's the probability of a an outcome after N trials, if you stop trying once you're "successful"?
QUESTION [8 upvotes]: This follows on from this question about being hit by a bus.
In this question, there is a 1/1000 chance of being hit and the question was about the probability of being hit if you cross the road 1000 time.
I wondered what would happen to this probability if I stopped trying to cross the road as soon as I get hit. Does the probability change?
As far as I can figure it, the probability then just becomes the sum of the geometric series 
$$P(\text{hit by bus within 1000 crossings}) = \sum_{n=0}^{999} 1/1000 * (999/1000)^n$$
thus
$$
P(\text{hit by bus within 1000 crossings}) = 1/1000 * \frac{1-(999/1000)^{1000}}{1-999/1000}
$$
However, this is identical to
$$P(\text{hit by bus within 1000 crossings}) = 1-P(\text{not hit by bus within 1000 crossing}) = 1-(999/1000)^{1000}$$
which is the answer to the previous question. I'm curious as to why they are not different, since the first approach is specifically ignoring all instances where (for instance) I get hit by a bus on the first try and then keep trying and get hit by subsequent buses.

REPLY [4 votes]: This is the cumulative probability of the geometric distribution - http://en.wikipedia.org/wiki/Geometric_distribution

The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ...}

The CDF is $1 - (1-p)^k$l.
This is the same as Semiclassical's answer (if the probabilities are equal), but it's useful to know this has a name, as you can more easily research other properties and communicate with others about it.