The devil’s logic

The strategy is as follows. First, the logician chooses one of the two envelopes with probability $1/2$. Let $V$ be the number he sees.

He then chooses a real number $\omega \in \mathbb{R}$ at random (according to a probability distribution that is strictly positive on $\mathbb{R}$ -- for example, a Gaussian distribution).

Finally, he compares $V$ and $\omega$: if $V \geq \omega$, he says that his number is larger; otherwise, he says that it is smaller.

Why does this work? Let $a<b$ be the numbers chosen by the devil. Thus, $V=a$ or $V=b$, each with probability $1/2$.

Three cases may occur:

• Suppose that $\omega<a$. In this case, we always have $\omega<V$, so the logician will always say that his number is larger. If $V=a$, he loses, and if $V=b$, he wins. Therefore, he goes to heaven with probability exactly $1/2$.
• Suppose that $\omega\geq b$. By the same reasoning, he goes to heaven with probability $1/2$.
• Suppose that $a\leq\omega<b$. In this case, if $V=a$, then since $V\leq\omega$, the logician will say that his number is smaller, and he will be correct. If $V=b$, then $V>\omega$, so the logician will say that his number is larger, and once again he will be correct. In both cases, the logician wins.

Thus, if we denote by $p>0$ the probability that the logician chooses $\omega$ in the interval $[a,b]$, then the logician goes to heaven with probability

$$(1-p)\frac{1}{2}+p = \frac{1}{2}+\frac{p}{2} > \frac{1}{2}.$$