I think the analysis of the strategy&#x27;s chances of winning is faulty. We can assume that the adversary chooses A and B such that P(A &lt; B) = ½, since otherwise the optimal strategy is to always guess A &lt; B or B &gt; A depending on if P(A &lt; B) &gt; ½ or P(A &lt; B) &lt; ½.<p>Now, analyze the game using the standard approach of breaking it down into simple, non-overlapping cases:<p><pre><code>    R &lt; A &lt; B: lose
    A &lt; R &lt; B: win
    A &lt; B &lt; R: win

    R &lt; B &lt; A: win
    B &lt; R &lt; A: win
    B &lt; A &lt; R: lose
</code></pre>
Thus the probability of losing is<p><pre><code>    P(R &lt; A &lt; B) + P(B &lt; A &lt; R)
    = P(R &lt; A ∧ A &lt; B) + P(B &lt; A ∧ A &lt; R)
    = P(R &lt; A)*P(A &lt; B) + P(B &lt; A)*P(A &lt; R)
    = P(R &lt; A)&#x2F;2 + P(A &lt; R)&#x2F;2
    = 1&#x2F;2
</code></pre>
There is no paradox: no matter how you choose R, you have exactly a ½ chance of winning.