Let's say that you have a drawer filled with 50 red socks and 50 blue socks. You want to get 10 socks total, so you randomly pick 10 socks out of the drawer one at a time without replacement. You will not accept a collection of socks with more than 7 blue socks. What will the probability distribution of the number of blue socks look like?
There are a couple of ways to interpret that last instruction. Method 1: interpret it as "draw a sock, and if it is blue and you already have 7 blue socks, put the sock you just drew back and try again." Effectively, once seven blue socks are reached, the rest are guaranteed to be red with this method.
The naive way to calculate the probability of getting $n$ blue socks is to just add up the odds gotten from multiplying the probabilities in all the possible ways. There is a $\frac{1}{2}$ chance that the first sock drawn is blue; if it is, there is a $\frac{49}{99}$ chance that the next is blue, and so on. This is hard, but it's easy for a computer, and the results of doing so are shown in plots below.
Without imposing the limit of 7, you get a distribution like this. (I didn't know it at the time, but this is a hypergeometric distribution.)
With the limit of 7, what essentially happens is that the cases of drawing 8, 9, or 10 blue socks can be treated as drawing 7 blue socks instead, because the blue socks after the 7th one are ignored. So this distribution becomes this:
Method 2: Draw socks one at a time, and if at any point you get an 8th blue sock, put them all back and start again.
I don't know how to compute the distribution using this interpretation. Were one to compute this by simulation, since the results where 8, 9, or 10 blue socks are drawn are discarded, the simulation gets run again on those instances, and the process repeats until 8, 9, or 10 blue socks are not drawn. So intuitively, it will be "like Figure 1, but cut off after 7, and bulgier."
How do you find the distribution using Method 2? And how does it compare to Method 1?