You add independent Uniform(0,1) draws until the running total first exceeds 2. What is the expected number of draws?

You add independent Uniform(0,1) draws until the running total first exceeds 2. What is the expected number of draws?

Approach: Let m(x) be the expected count needed to pass x, condition on the first draw to obtain an integral equation, solve it on [0,1] and then continue the solution onto [1,2].

4.6708. Conditioning on the first draw gives the integral equation m(x) = 1 + ∫_0^x m(x-u) du. On [0,1] its solution is m(x) = e^x, which recovers the familiar m(1) = e. Substituting the known piece and solving on [1,2] gives m(x) = e^x - (x-1)e^{x-1}. A direct check uses the identity P(N > n) = P(sum of n uniforms <= 2), which is (2^n - n)/n! for n at least 2, and summing that series returns the same value. So the expected number of draws is m(2) = e^2 - e = 4.6708. The second unit of target costs only 1.95 draws against the first unit's 2.72, because the overshoot past 1 already carries part of the way.

Follow-up: What is the expected size of the overshoot above 2 at the stopping draw?

Key concepts: integral equation, conditioning on the first draw, overshoot, expected number of draws.