$$P(N) = \sum_{i = 1}^{3N} \left(\binom{3N - 1}{3N - i} \left(\frac{N - 1}{N} \right)^{3N - i} \left(\frac{1}{N} \right)^{i - 1} \left(\frac{1}{i} \right) \right)$$
$0 \leq 3N - i \leq 3N - 1$ racers dedicate all fuel to $N - 1$ races (that you are not in) shown by $\left(\frac{N - 1}{N} \right)^{3N - i}$.
The remaining $1 \leq i \leq 3N$ racers (including you) dedicate all fuel to last remaining race (that you are already assumed to be in) shown by $\left(\frac{1}{N} \right)^{i - 1}$.
The probability of winning in this case is then $\frac{1}{i}$, since $i - 1$ racers were added to your race, and ties are broken randomly.
Sum over all values of $i$ in your race. $i \geq 1$ since this is your race.
$$probFilledForSpecificArrangement(numOtherRacers, p, numRacesFilled = 1) = p^{numOtherRacers}$$
$$probFilledForSpecificArrangement(numOtherRacers, p, numRacesFilled > 1) = \sum_{i = numRacesFilled - 1}^{numOtherRacers - 1} \binom{numOtherRacers}{i} \cdot probFilledForSpecificArrangement(i, p, numRacesFilled - 1) \cdot p^{numOtherRacers - i}$$
$$probExactlyNumRacesFilled(N, numRacesFilled) = \binom{N}{numRacesFilled} \cdot probFilledForSpecificArrangement(3N - 1, \frac{1}{N}, numRacesFilled)$$
$$P(N) = 1 - probExactlyNumRacesFilled(N, N)$$
The new strategy is to put a nonzero amount of fuel in each race, so you will only win if there is a race that no one else puts any fuel in.
Your fuel will be less than 1 in each race. Everyone else will put all of their fuel (1) in one race.
You will win when there is at least one race of all zeros (not counting yourself).
So $P(\text{Win}) = P(\text{at least one race with all 0s})$
$ = 1 - P(\text{no race with all 0s})$
$ = 1 - P(\text{all races with nonzeros})$
$ = 1 - P(\text{all races filled})$
where a race is filled if at least one racer (not counting yourself) dedicates all energy to this race.
$probExactlyNumRacesFilled(N, numRacesFilled)$ is the probability that exactly $numRacesFilled$ races out of $N$ races are filled.
$probFilledForSpecificArrangement(numOtherRacers, p, numRacesFilled)$ is a helper function for $probExactlyNumRacesFilled$ inspired by the binomial distribution.
Place $i$ racers in $numRacesFilled - 1$ races and the remaining $numOtherRacers - i$ racers in the last remaining race such that each of the $numRacesFilled$ races are filled.
$numRacesFilled - 1 \leq i \leq numOtherRacers - 1$ since we need at least one racer in each race.
Thus $numOtherRacers - i \geq 1$.
$p$ is the probability of a racer dedicating all fuel to a certain race.
We could increase the efficiency of $probFilledForSpecificArrangement(numOtherRacers, p, numRacesFilled)$ by removing the multiplication $p^{numOtherRacers}$ and $p^{numOtherRacers - i}$, and instead just multiply once by $p^{numOtherRacers}$ in $probExactlyNumRacesFilled$. This is because in the recurrence above we basically get something like $p^{numOtherRacers - i} \cdot p^i = p^{numOtherRacers}$.
If done this way then we can more easily build up a solution using dynamic programming instead of a cache.
$$dp[i, j] = \text{the number of ways to fill $i$ races using $j$ unique racers}$$
$$dp[1, j] = 1$$
$$dp[i, j] = \sum_{k = i - 1}^{j - 1} \binom{j}{k} \cdot dp[i - 1, k]$$
The recurrence is correct because the number of ways to fill $i$ races using $j$ unique racers is the sum of the number of ways you can choose $k$ of these racers to fill $i - 1$ races (multiplied by the number of ways you can fill these races) and have the rest of the $j - k$ racers to fill the last remaining race. We must have that $k \geq i - 1$ since $i - 1$ races would not be able to be filled otherwise. Similarly, we must have that $j - k \geq 1$ to fill the last race. We can rearrange this to be $k \leq j - 1$. Thus $i - 1 \leq k \leq j - 1$.
Run $i$ from $2 \leq i \leq N$ since $i = 1$ is the base case and more than $N$ races cannot be filled.
Run $j$ from $i \leq j \leq 3N - 1$ since we need at least $i$ racers to fill $i$ races, and have at most $3N - 1$ racers that we care about.
$O(N^3)$ (if you precompute factorials)
Then we can just replace $probFilledForSpecificArrangement(numOtherRacers, p, numRacesFilled) = p^{numOtherRacers} \cdot dp[numRacesFilled, numOtherRacers]$
since this is just the probability of getting to this arrangement multiplied by the number of ways to fill numRacesFilled races using numOtherRacers racers.
We can rewrite $probExactlyNumRacesFilled$ to be non-recursive based on the first set of equations above, but this function is much slower to run this way compared to the recursive way (due to caching) and the DP way.
Let $S$ be the set of all ordered sets whose $numRacesFilled$ integer elements are all greater than or equal to 1 and sum to $3N - 1$ (given that order matters).
$$S = {S_i : S_i = {s_{ij} : 1 \leq j \leq numRacesFilled, s_{ij} \geq 1, \sum_{s \in S_i} s = 3N - 1}}$$
$S$ is equivalent to all of the ways that all of the other $3N - 1$ racers could fill $numRacesFilled$ races (this part does use recursion to find).
$$probExactlyNumRacesFilled(N, numRacesFilled) = \binom{N}{numRacesFilled} \sum_{S_i \in S} \left(\frac{numRacesFilled!}{\prod_{s \in S_i} s!} \cdot \left(\frac{1}{N} \right)^{3N - 1} \right)$$
$$ = \binom{N}{numRacesFilled} \cdot numRacesFilled! \cdot \left(\frac{1}{N} \right)^{3N - 1} \sum_{S_i \in S} \frac{1}{\prod_{s \in S_i} s!}$$
Note in the first equations we have $p = \frac{1}{N}$ and $numOtherRacers = 3N - 1$.
The first equation can reduce to this equation, since the first equation is multiplying many combinations together, and simplifying these multiplications just results to this equation.
Finally, the official solution uses a 3D grid instead of a 2D grid like shown above.