After many hours on the court I think I have the answer.
Starting with a hits and b misses the chance of going to a+1 misses is a/(a+b) and the chance of going to b+1hits is b/(a+b).
If the target is A misses and B hits then before each shot a will run from 1 to A-1 and b will run from 1 to B-1. a+b will run from 2 to A+B-1.
So the probability of any particular sequence which works will be (A-1)!(B-1)!/(A+B-1)!
Surpisingly the probability is the same whichever sequence we choose.
The number of sequences which work is the number of ways of ordering A-1 misses and B-1 hits. The length of the sequence is A+B-2 so the number of sequences is the number of ways of choosing A-1 steps to be misses from A+B-1 steps. This is (A+B-2)!/(A-1)!(B-1)!.
The probability of getting A misses and B hits is the product of the number of sequences times the probability of each sequence, which is
(A-1)!(B-1)!/(A+B-1)! x (A+B-2)!/(A-1)!(B-1)! = 1/(A+B-1)
The total number of shots taken is A+B.
So the probability of any particular score from n shots is 1/(n-1), regardless of the score.
In the example this is 1/9.
This is a very neat result, it is a flat distribution which is quite different to the bell curve you mentioned for coin tosses. I guess we are always favouring the more extreme result which compensates exactly for the fact there are less ways of reaching the extreme results.
target is 5-each which I call p(5,5)
now p(5,5) = (4/9)p(4,5) + (1-5/9)p(5,4)
by continuously substituting I do indeed get a result with 9! in the denominator but my answer reduces to 8/105. Perhaps I made a little mistake in there. I’ll check
That’s totally right! But there is something amiss with the rest of the comment— perhaps the starting out values were incorrect. Recall that p(0,n) = p(n,0) = 0 (since one shot was missed, and one made, at the beginning); p(1,2) = p(2,1) = 1/2….