Probability of "winning a race"? Suppose there are three people, A, B, and C. Each person starts at $x=0$ and then randomly one person moves forward by 1, with equal probability among all three people. The winner is the first person to reach $x=3$.
By symmetry, the probability of each person winning is 1/3. But now suppose person A wins if they reach $x=2$, while person B and C still only win at $x=3$.
How can I calculate $P(\text{A wins})$?
EDIT:
Using a simulation in R I obtained these probabilities, but I'm still unsure about the analytical approach.
        A         B         C 
0.5472412 0.2263920 0.2263668 

 A: To calculate $P(\text{A wins})$, basically we need to enumerate the possible outcomes for A to win, and sum up the probabilities. First, we note that A can win after at least 2 total moves and at most 6 total moves (think about why this is true), then we consider each scenario as follows:
If A wins after 2 total moves, then the outcome can only be "AA" (i.e. A must move forward in both of the first two moves). Thus the probability is
$$\left(\frac{1}{3}\right)^2 = \frac{1}{9}.$$
If A wins after 3 total moves, then the outcome can only be "XXA" with exactly one A in the first two moves. Thus the probability is
$$\underbrace{2}_{\text{A's position}} \times \underbrace{2}_{\text{to choose B or C}} \times \left(\frac{1}{3}\right)^3 = \frac{4}{27}.$$
If A wins after 4 total moves, then the outcome can only be "XXXA" with exactly one A in the first three moves. Thus the probability is
$$\underbrace{3}_{\text{A's position}} \times \underbrace{4}_{\text{BB, CC, BC, or CB}} \times \left(\frac{1}{3}\right)^4 = \frac{4}{27}.$$
If A wins after 5 total moves, then the outcome can only be "XXXXA" with exactly one A in the first four moves and there are no three B's or three C's in the first four moves. Thus the probability is
$$\underbrace{4}_{\text{A's position}} \times \underbrace{2 \times 3}_{\text{B or C appears once and its position}} \times \left(\frac{1}{3}\right)^5 = \frac{8}{81}.$$
If A wins after 6 total moves, then the outcome can only be "XXXXXA" with exactly one A, two B, and two C in the first five moves. Thus the probability is
$$\underbrace{5}_{\text{A's position}} \times \underbrace{{4 \choose 2}}_{\text{positions for two B's and two C's}} \times \left(\frac{1}{3}\right)^6 = \frac{10}{243}.$$
Therefore, summing up above we have
$$P(\text{A wins}) = \sum_{i=2}^6 P(\text{A wins at move } i) = \frac{1}{9} + \frac{4}{27} + \frac{4}{27} + \frac{8}{81} + \frac{10}{243} = \frac{133}{243} \approx 0.5473251.$$
Note that since B and C are symmetric, we also have
$$P(\text{B wins}) = P(\text{C wins}) = \frac{1 - P(\text{A wins})}{2} = \frac{1 - \frac{133}{243}}{2} = \frac{55}{243} \approx 0.2263374.$$
