Suppose I have an unlimited number of draws from a uniform distribution [0,1]. I want to keep on drawing until the sum of all my a values >=x (where 0<x<1).
I want to find the probability that my score ends between x and 1 when trying to do this and when it does, the expected value of my score.
Obviously the percentage chance of 'busting' is just 1 - the chance that my score is between x and 1. Intuitively, I'm tempted to say that when my score is between x and 1, on average it is halfway between x and 1, but I'm not 100% positive on that.
I looked at some other forums and found some things that seemed like they were kind of similar, but I wasn't quite sure where to take it from there.
I looked at this: Irwin Hall distribution with varying n
right now this is where I'm at, I sum up from n=1 to n=inf $$\frac{(n-z) z^{n-1}}{n!}$$
but for each n here I need to subtract the probability that the sum made it to over 1 for that n and I need to figure out how to find out that out...
for n draws out of a uniform distribution, I believe the probability that the sum is less than 1 is 1/n! but I can't just multiply what I have by 1/n! for each sum in the series because that would be 'overcounting' some of the area.
Let $f(t)$ be the probability that you succeed starting from a sum of $t$; so you want $f(0)$. A one-step analysis yields
$$ f(t)=1-x+\int_t^xf(u)\mathrm du\;. $$
Differentiating with respect to $t$ yields
$$ f'(t)=-f(t)\;, $$
and thus $f(t)=c\mathrm e^{-t}$. Since $f(x)=1-x$, we must have $c\mathrm e^{-x}=1-x$, thus $c=\mathrm e^x(1-x)$, thus $f(t)=\mathrm e^{x-t}(1-x)$ and thus
$$ f(0)=\mathrm e^x(1-x)\;. $$
You’re right that your expected score conditional on success is $\frac{1+x}2$, since the score is uniformly distributed over $[x,1]$, no matter from where you reach it.