This week the #puzzle is: Can You Sprint to the Finish? #probabilities #strategy #optimal
| … 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 (and only 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? |
And for 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 if (and only if) 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? |

Highlight to reveal (possibly incorrect) solution:
Formulating what we already know:
- There are 2 competitors, A (me) and B.
- There will be 1 winner.
- legs(X) describes the state of X’s legs, somewhere between 0 and 100 percent.
- For each competitor these 2 steps will be done in order:
- Determine legs(X).
- Decide whether to sprint.
- If both sprint, winner will have max(legs(X)).
- If 1 sprint, they will win.
- If 0 sprint, p(A wins) = p(B wins) = 50%
- If legs(B) >= 50%, they will sprint.
- A strategy is a list of values, that will lead to a sprint. For B the strategy is 50-100.
Now lets look at the different situations. Let a = legs(A) and b = legs(B).
Situation 1: a < 50.
| A sprints? | b < 50 | b => 50 |
| Yes | p(A wins) = 1 | p(A wins) = 0 |
| No | p(A wins) = 50% | p(A wins) = 0 |
In this situation, the strategy of always sprinting dominates. Situation 2, a = 50.
| A sprints? | b < 50 | b => 50 |
| Yes | p(A wins) = 1 | p(A wins) = p(a > b) = 0 |
| No | p(A wins) = p(a > b) = 1 | p(A wins) = 0 |
In this situation, all strategies lead to the same results. Finally situation 3, a >= 50.
| A sprints? | b < 50 | b => 50 |
| Yes | p(A wins) = 1 | p(A wins) = p(a > b) = 50% |
| No | p(A wins) = 50% | p(A wins) = 0 |
The dominant strategy in this situation is that A sprints. This leads to an overall strategy of always sprinting. However, that wasn’t the question.
- p(A wins)
- = p(a < 50) * (p(b < 50) * 1 + p(b > 50) * 0) + p(a > 50) * (p(b < 50) * 1 + p(b > 50) * 1/2)
- = 1/2 * (1/2 + 0) + 1/2 * (1/2 + 1/4)
- = 1/4 + 3/8
- = 5/8
My result: 5/8 = 0.625 = 62.5%.
And for extra credit:
I couldn’t quite get the math right in my head, so I monte carloed the problem. My result: 0.307 / 30.7%.