More generally: Each game, $n = 1,2,...,N$, a team has probability, $p = 0.5$, of winning. Their standing $x$ is given by $x(n) = x(n-1)\pm1$ depending on whether they win ($+1$) or lose ($-1$). Their standing starts at $x(n=0)=0$. What is the chance the team will have at least $x_w$ more wins than losses at any point during a $N$ game season? So, what is the probability $x(n) \ge x_w$ after any $n$ number of games between $0$ and $N$.
From the binomial distribution, we could find the probability of 10 more wins than loses at the end of the season. As the number of games gets much larger than 100, could we not approximate with a normal distribution as well? However, I am interested in the probability they will have at least that difference in wins vs. loses at any point in the season, whether it be game 10, game 42 or game 100.
Would I find this by
- finding the binomial distribution of wins vs losses after each game,
- calculating, from the "area under the curve," the probability of less than 10 more wins than losses,
- taking the product of these probabilities from n = 1 to n = N, and finally
- subtract this value from 1?
This question could be rephrased as, "at least 10 more heads than tails after any number of coin tosses up to 100 tosses." It could also relate to other processes described by binomial or normal distributions (noise, particle diffusion).
Sorry in advance for anything unclear, especially my notation in the first paragraph. Obviously, I am not a mathematician. Feel free to ask me to clarify. Thank you!