Tour de Fiddler 2025 - Extra Credit

Instead of one opponent, now you have two—meaning three riders in total. As luck would have it, the managers for both other riders proclaimed that they’d sprint for the finish as long as their legs were feeling 50 percent or better. Note that your opponents’ feelings are independent of each other.

As the three of you near the finish, your own team manager radios you the following message: “If your legs feel <garbled> percent or better, sprint for the finish!”

You can’t make out what the garbled part of the message is, and you’re too tired to radio back for confirmation. Instead, you somehow muster the energy to randomly, uniformly pick a number between 0 and 100 to fill in the blank from your manager’s message, thereby determining your racing strategy—optimization be damned!

Right before you choose your random strategy and test your legs, what are your chances of winning the stage against both opponents?

Similar to the classic answer, we want to let your legs be $X \sim U(0,1),$ and your two opponents be $Y, Z \sim^{\text{i.i.d.}} U(0,1).$ Let your strategy be $U_a$ such that you will sprint if and only if $X \geq a.$ Here you again will win with $100\%$ probability if you are the only one sprinting. You will never win if you are not sprinting and at least one of opponents is sprinting. Here if no one is sprinting then your win probability drops to $\frac{1}{3}$ since you each have equal probability of winning. It gets a little more complicated here if multiple people are sprinting. Obviously, if only one opponent is sprinting, then your win probability is equivalent to case where both you and your single opponent from the Classic question are sprinting, that is, $\frac{1}{2} \left(p(a) - \frac{1}{2} - \frac{a}{4}\right).$ Leaving only the more complicated problem of what happens when all of the racers are sprinting. So to summarize we have \begin{align*}\tilde{p}(a) &= \mathbb{P} \{ X \geq a, \max\{Y,Z\} \lt \frac{1}{2} \} + 0 \mathbb{P} \{ X \lt a, \max \{Y,Z\} \geq \frac{1}{2} \} \\ &\quad\quad\quad +\frac{1}{3} \mathbb{P} \{ X \lt a, \max \{Y,Z\} \lt \frac{1}{2} \} + 2 \mathbb{P} \{ X \lt a, Y \geq \frac{1}{2}, X \geq Y, Z \lt \frac{1}{2} \} \\ &\quad\quad\quad\quad +\mathbb{P} \{ X \geq a, \min \{Y, Z \} \geq \frac{1}{2}, X \geq \max\{Y, Z\} \} \\ &= \frac{1-a}{4} + 0 + \frac{a}{12} + \frac{1}{8} - \frac{1}{2}\left( \max\{a-\frac{1}{2},0\}\right)^2 + \int_a^1 \left( \int_{\frac{1}{2}}^{\max\{\frac{1}{2},x\}} \,dy \right) \left( \int_{\frac{1}{2}}^{\max\{\frac{1}{2},x\}} \,dz \right) \,dx \\ &= \frac{3}{8} - \frac{a}{6} - \frac{1}{2} \left(\max \{a-\frac{1}{2},0\}\right)^2 + \int_a^1 \left(\max{x-\frac{1}{2},0} \right)^2 \,dx \\ &= \frac{5}{12} - \frac{a}{6} - \frac{1}{2} \left( \max \{ a - \frac{1}{2}, 0 \} \right)^2 + \frac{1}{24} - \frac{1}{3} \left( \max\{a - \frac{1}{2}, 0\}\right)^3 \\ &= \begin{cases} \frac{5}{12} - \frac{a}{6}, &\text{if $0 \leq a \leq \frac{1}{2}$;}\\ \frac{1}{3} + \frac{a}{12} - \frac{a^3}{3}, &\text{if $\frac{1}{2} \lt a \leq 1$.} \end{cases}\end{align*}

Therefore, the probability of winning just before choosing the value of $a$ for your strategy is \begin{align*}\Pi = \int_0^a \tilde{p}(a) \,da &= \int_0^{\frac{1}{2}} \left(\frac{5}{12} - \frac{a}{6}\right) \,da + \int_{\frac{1}{2}}^1 \left(\frac{1}{3} + \frac{a}{12} - \frac{a^3}{3}\right) \,da \\ &= \left[ \frac{5a}{12} - \frac{a^2}{12} \right]_{a=0}^{a= \frac{1}{2}} + \left[ \frac{a}{3} + \frac{a^2}{24} - \frac{a^4}{12} \right]^{a=1}_{a = \frac{1}{2}} \\ &= \left( \frac{5}{24} - \frac{1}{48} \right) - 0 + \left( \frac{1}{3} + \frac{1}{24} - \frac{1}{12} \right) - \left( \frac{1}{6} + \frac{1}{96} - \frac{1}{192} \right) \\ &= \frac{3}{16} + \frac{7}{24} - \frac{33}{192} = \frac{36 + 56 - 33}{192} = \frac{59}{192} = 0.3072916666\dots \end{align*}