Thanks :)

Math was always my favorite subject, I was just too chicken to go for a phd in it 

Not sure if the proof I gave is the proof you had in mind


south_korea_ln

science

[Daily puzzle] Probability of ending at 1

Let $$f(x)$$ be the probability that the game ends on 1 when the current position is $$x$$.

We can easily verify that $$f(0)=0$$, $$f(0.5)=0.5$$ and $$f(1)=1$$.

We also know that for $$0 < x < 0.5$$, 

$$
f(x) = 0.5 f(2x)
$$

And for $$0.5 < x < 1$$,

$$
f(x) = 0.5 + 0.5f(2x-1)
$$

We'll assume that $$f$$ is continuous and differentiable (and verify later).

We can therefore write that when $$0 < x < 0.5$$,

$$
f^\prime(x) = f^\prime(2x) 
$$

And when $$0.5 < x < 1$$,

$$
f^\prime(x) = f^\prime(2x-1)
$$

These conditions can only be satisfied if $$f^\prime(x)$$ is a constant. And with the boundary conditions of $$f(0)=0$$, $$f(1)=1$$, we obtain

$$
f(x) = x
$$

(which is continuous and differentiable.)
