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

  1. finding the binomial distribution of wins vs losses after each game,
  2. calculating, from the "area under the curve," the probability of less than 10 more wins than losses,
  3. taking the product of these probabilities from n = 1 to n = N, and finally
  4. 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!

  • $\begingroup$ The question is clear, and important. It is a "gambler's ruin" problem, for which there is a large literature. Your 3.) is not correct, the events are not independent. The usual approaches do not use a normal approximation. $\endgroup$ – André Nicolas Jul 14 '16 at 0:22
  • $\begingroup$ Thank you very much! From this I found exactly what I was looking for! mathpages.com/HOME/kmath683/kmath683.htm $\endgroup$ – James Jul 14 '16 at 3:52
  • $\begingroup$ You are welcome. Because in 3.) you multiplied, I interpreted your question as asking for the probability of being $10$ or more ahead at some unspecified time $\le 100$. If you make an explicit specification of the time, then the answer is easily obtained from the binomial distribution, and if $n$ is not too small you can use the normal approximation. That is not what the "gambler's ruin" calculation does. $\endgroup$ – André Nicolas Jul 14 '16 at 4:01
  • $\begingroup$ I meant, "What is the probability they will have that difference x_w at some point before the season ends." Even if they finish at 0.500 with # wins = # losses, they would still meet my criterion if at, say, game 52, they had 31 wins and 21 losses. In terms of gambler's ruin, wouldn't that be similar to the limit of the probability of losing (reach 0) or winning (reach N) within 100 iterations as N approaches infinity? They would start with $10, and in this case, I'd be interested in 10 more losses than wins. $\endgroup$ – James Jul 14 '16 at 4:14
  • $\begingroup$ Then you want gambler's ruin against a opponent with infinite wealth, and if you reach $-10$ you are broke. You do not want the limit, it is a bit more complicated than that, but you can find an answer from the recurrence. Another approach (similar, but matrix-heavy) is Markov chains. $\endgroup$ – André Nicolas Jul 14 '16 at 4:19

You can describe the number of wins after $n$ games with $X_n \sim Bin(0.5, n)$, i.e. a binomial distribution with probability of success in a single event $p = 0.5$. The number of losses is always going to be $n - X_n$ (assuming no draws). So you want to find the probability $P(X_n - (n - X_n) \ge 10) = P(2X_n - n \ge 10) = P(X_n \ge \frac{n + 10}{2})$ which just involves a standard calculation of binomial probabilities (noting that when $n$ is odd, it's equivalent to $P(X_n \ge \frac{n + 11}{2})$).

  • $\begingroup$ Thank you! Would this give, for example, the probability of 10 or more wins than losses after 23 games as P(X_23 >= 17) = 0.0173? $\endgroup$ – James Jul 14 '16 at 2:34
  • $\begingroup$ Looks good to me! $\endgroup$ – ConMan Jul 14 '16 at 3:47

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