Solution

Answer: 0.5276662759

Let

$X$ = Louise’s last number (the one that first pushes the total above $1$),
$Y$ = Julie’s last number (the one that first pushes the total above $2$).

We want

$$\mathbb P(Y>X).$$

1. Mathematical analysis

Step 1 — Distribution of Louise’s stopping state

Suppose before Louise’s final draw, her accumulated sum is $r\le 1$, and then she draws $x$ such that

$$r+x>1.$$

Thus:

$$0\le r\le 1,\qquad 1-r < x \le 1.$$

Step 2 — Density of the pre-final sum $r$

Assume Louise had already drawn $n$ numbers before the final draw.

The density that those $n$ numbers sum to $r$ is the simplex volume

$$\frac{r^{n-1}}{(n-1)!}.$$

Summing over all possible $n\ge1$,

$$\sum_{n=1}^\infty \frac{r^{n-1}}{(n-1)!} = e^r.$$

Hence the joint density of $(r,x)$ is simply

$$f_{R,X}(r,x)=e^r,$$

on the region

$$0\le r\le1,\qquad x>1-r.$$

Step 3 — Transform to the remaining gap for Julie

After Louise finishes, the total is

$$T=r+x.$$

Julie must reach $2$, so define the remaining gap

$$a = 2-T = 2-r-x.$$

Then $0<a<1$.

Since $r=2-a-x$, the joint density becomes

$$f_{A,X}(a,x) = e^{2-a-x},$$

with domain

$$1-x \le a \le 1,\qquad 0\le x\le1.$$

(Indeed this integrates to $1$.)

2. Probability Julie eventually wins

Define

$$H(a,x)$$

as the probability Julie eventually wins (i.e. her final draw exceeds $x$) when she still needs an amount $a$ to reach $2$.

Step 4 — Integral equation for $H(a,x)$

Julie draws a uniform $u\in[0,1]$.

Case A: $u>a$

She stops immediately, and wins iff $u>x$.

Contribution:

$$1-\max(a,x).$$

Case B: $u\le a$

She continues with remaining gap $a-u$.

Contribution:

$$\int_0^a H(a-u,x),du.$$

Therefore,

$$H(a,x) = \int_0^a H(v,x),dv + 1-\max(a,x).$$

Differentiate with respect to $a$.

Step 5 — Solve the differential equations

Region 1: $a<x$

Then $1-\max(a,x)=1-x$, so

$$H'(a,x)=H(a,x).$$

Boundary condition:

$$H(0,x)=1-x.$$

Thus

$$H(a,x)=e^a(1-x), \qquad a<x.$$

Region 2: $a>x$

Now $1-\max(a,x)=1-a$, giving

$$H'(a,x)=H(a,x)-1.$$

Matching continuously at $a=x$ yields

$$H(a,x) = 1+\bigl(1-x-e^{-x}\bigr)e^a, \qquad a>x.$$

3. Final integration

We now average $H(a,x)$ against the joint density:

$$P = \iint H(a,x)e^{2-a-x},da,dx.$$

The domain naturally splits at $x=\tfrac12$.

After carrying out the integrations,

$$P = -\frac{5e^2}{4} +\frac14 +\frac{7e}{2}.$$

Numerically,

$$P \approx 0.5276662759433455\ldots$$

So Julie (the second player) actually has the advantage.

4. Python implementation

from mpmath import mp

mp.dps = 50  # high precision

e = mp.e

# Closed-form probability
P = -5 * e**2 / 4 + mp.mpf(1)/4 + 7 * e / 2

print(P)
print(mp.nstr(P, 12))

5. Code walkthrough

mp.dps = 50

Use high precision arithmetic.

e = mp.e

The constant $e$.

P = -5 * e**2 / 4 + mp.mpf(1)/4 + 7 * e / 2

Implements the exact closed-form formula:

$$P = -\frac{5e^2}{4} +\frac14 +\frac{7e}{2}.$$

print(P)

Prints the full high-precision value.

print(mp.nstr(P, 12))

Prints enough digits to round correctly to 10 decimal places.

6. Final verification

The probability is slightly larger than $1/2$:

$$0.527666\ldots$$

This makes intuitive sense because Julie benefits from starting with Louise already near $1$, making large finishing jumps more likely.

Rounded to 10 decimal places:

$$0.5276662759$$

Answer: 0.5276662759

Project Euler Problem 436