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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!

James
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  • 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. – André Nicolas Jul 14 '16 at 00:22
  • Thank you very much! From this I found exactly what I was looking for! http://www.mathpages.com/HOME/kmath683/kmath683.htm – James Jul 14 '16 at 03:52
  • 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. – André Nicolas Jul 14 '16 at 04:01
  • 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. – James Jul 14 '16 at 04:14
  • 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. – André Nicolas Jul 14 '16 at 04:19
  • Alright, I really appreciate your help! At the risk of seeming even more obtuse: Say you have a particle on a line that moves either one step right or one step left every second with equal probability. It’s 10 steps away from a wall to its right. There is nothing that will ever stop it on its left. What’s the probability of it hitting the wall within 100 seconds? Would I not use the limit but instead use an opponent with infinite wealth (or Markov chains) for this case? – James Jul 14 '16 at 04:33
  • That's the same problem as the one you posted. Opponent of infinite wealth is not important with $N=100$. And you find the probability of not having hit the wall, which can be obtained from the recurrence. It is somewhat messy. The limit is not relevant, with probability $1$ we will hit the wall, but probably well after $100$. – André Nicolas Jul 14 '16 at 04:46
  • Got it. Thank you very much! I knew they were the same. Sorry to be redundant. Just wanted to be clear. Also, in my second comment, N is not the max number of iterations. N is some number of wins that would end the game. My mistake. I'm not able to edit that comment anymore. Also, I would have moved this to a chat if my reputation were high enough. Sorry. – James Jul 14 '16 at 05:07
  • I finally had time to check the article you referred to. It does discuss, among others, your question, giving the recurrence and generating function. From the recurrence or the generating function one can compute an exact answer (by machine). I know there has been work on estimates but a short search has not yielded results. – André Nicolas Jul 14 '16 at 14:38

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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 $P(X_n \ge \frac{n + 11}{2})$).

ConMan
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