Robot Racing!

Since finishing times are unique to each gear shift and the order in which the shifts are made is of utmost import, the race will only end in a tie if the robots make the same shift as each other for all four shifts.

Since the gearshift mechanism is in a 3-by-3 grid and both robots start in the center, each has four gears to randomly shift to. The probability that they choose the same gear is $\frac{1}{4}$ .

At this point, the robots are positioned in a edge gear. The options they have to shift to are corner gears and the center gear. There are two corner gears and one center gear that the robots can shift to, making a total of three possible gears. The probability that they choose the same gear is $\frac{1}{3}$ .

From here, the robots have either shifted to a corner gear or the center gear. Since at the last shift, they had two corner gears and one center gear to shift to, the probability of the robots being in a corner is $\frac{2}{3}$ , and the probability of them being in the center is $\frac{1}{3}$ . If they are in a corner, the probability that they make the same choice of gear is $\frac{1}{2}$ , and if they are in the center, that probability that they make the same choice of gear is $\frac{1}{4}$ . We take the weighted average of these two options to get a $\frac{5}{12}$ chance that the robots chose the same option.

After this, the robots are once again both in an edge gear. The chances that they choose to make the same shift are once again $\frac{1}{3}$ .

Multiplying all of the probabilities together gives us the total chance that the race will end in a tie, which is $\frac{1}{4} \times \frac{1}{3} \times \frac{5}{12} \times \frac{1}{3} = \frac{5}{432}$ . Converting this to the requested format, we get the decimal equivalent of $\frac{5}{432}$ as approximately $0.01157$ . Multiplying by $10000$ gives us about $\boxed{116}$ .

Edit: At the time of posting, the answer posted by the author was incorrect, being cited as 104.