Tour de Fiddler 2025

This time around, you and a competitor are approaching the finish of a grueling stage of the Tour de Fiddler. One of you will win the stage, the other will come in second. As you approach the finish, each of you will test the feeling of your legs, which will be somewhere between 0 percent (“I can barely go on!”) and 100 percent (“I can do this all day!”). For the purposes of this puzzle, these values are chosen randomly, uniformly, and independently.

Immediately after feeling your legs, you and your opponent each have a decision to make. Do you maintain your current pace, or do you sprint to the finish? Among those who sprint for the finish, whoever’s legs are feeling the best will win the stage. But if no one sprints for the finish, everyone has an equal chance of winning the stage. In the Tour de Fiddler, you must each decide independently whether to sprint for the finish based on your legs—you don’t have time to react to your opponent’s decision.

Normally, teams at the Tour de Fiddler keep their strategy and tactics close to the vest. But earlier today, your opponent’s manager declared on international television that if your opponent’s legs were feeling 50 percent or better, they’d sprint for the finish.

As you are about to test your legs for the final sprint and see how they feel, what are your chances of winning the stage, assuming an optimal strategy?

Let's explore possible strategies that mirror our opponent. Let $U_a$ represent the strategy that if when you check your legs and sprint whenever our legs are over the level $a \in [0,1].$ Let's assume that your legs are represented by $X \sim U(0,1)$ and your opponent's legs are represented by $Y \sim U(0,1)$ and that $X$ and $Y$ are independent. Note that your win probability is $100\%$ if you sprint but your opponent does not, $0\%$ if you do not sprint but your opponent does, and $50\%$ if neither of you sprint. The only rather complicated calculation is when you both sprint. If you have chosen the strategy $U_a,$ then your win probability is given by \begin{align*}p(a) &= \mathbb{P} \{ X \geq a, Y \lt \frac{1}{2} \} + 0 \mathbb{P} \{ X \lt a, Y \geq \frac{1}{2} \} \\ &\quad\quad\quad\quad +\frac{1}{2} \mathbb{P} \{ X \lt a, Y \lt \frac{1}{2} \} + \mathbb{P} \{ X \geq a, Y \geq \frac{1}{2}, X \geq Y \}\\ &= \frac{1-a}{2} + 0 + \frac{a}{4} + \int_a^1 \int_{\frac{1}{2}}^{\max\{x,\frac{1}{2}\}} \,dy \,dx \\ &= \frac{1}{2} - \frac{a}{4} + \int_a^1 \max \{x - \frac{1}{2}, 0\} \,dx \\ &= \frac{1}{2} - \frac{a}{4} + \int_{\max\{a-\frac{1}{2},0\}}^{\frac{1}{2}} t\,dt \\ &= \frac{1}{2} - \frac{a}{4} + \frac{1}{8} - \frac{1}{2} \left( \max \{ a - \frac{1}{2}, 0 \} \right)^2 \\ &= \begin{cases} \frac{5}{8} - \frac{a}{4}, &\text{if $0 \leq a \leq \frac{1}{2}$;}\\ \frac{1}{2} + \frac{a}{4} - \frac{a^2}{2}, &\text{if $\frac{1}{2} \lt a \leq 1$.}\end{cases}\end{align*} Therefore, the optimal strategy (choice of $a \in [0,1]$) will given the maximal value of $p(a)$, which since $p$ is uniformly decreasing, occurs when $a^* = 0,$ and gives an optimal win probability of $$p^* = \max_{a \in [0,1]} p(a) = p(0) = \frac{5}{8}.$$