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First of all, my knowledge of statistics, odds and probabilities is very rusty, so pardon me if I am not making sense, but here is my problem:

  1. The odds of winning a game is 97:3 (or 97% chance of winning).
  2. Winning scores you +1 point.
  3. Losing scores you -31 points.
  4. Assume that the game is played infinite amount of times so that the chances of winning and losing converge to 97% and 3% respectively.

Now, if I take 1000 game samples (out of the infinity games played), how would one calculate the the resultant net score (if possible at all)? Am I missing something simple out?

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    Not sure what you are asking. Each game is an independent event, where success has probability $p=\frac {97}{100}$. Therefore the number of successes out of $N$ trials follows the [binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution) with parameter $p$. if there are $k$ successes out of $N$ trials then there are $N-k$ failures, hence the score will be $k-31\times \left(N-k\right)$ Does that answer your question? – 2017-02-07
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    Sorry if I had expressed myself in an unclear manner. Basically, if you know the reward and the loss for each game, and you know the chances of winning and losing, is it possible to calculate the total score after N number of trials? – 2017-02-07
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    The actual score? No. In theory you can win (or lose) every single game. Not very likely, but mathematically possible. For very large samples the probability that you are far from the mean gets smaller and smaller...is that what you meant? Thus, in your example, we expect to win $970$ times and lose $30$ times, which makes our expected score $970-31\times 30=40$. – 2017-02-07
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    The sort of question one tends to ask in this context would be along the lines of "what is the probability that I'll have a negative total balance after $1000$ trials?" Something like that. To answer those questions, it's usually a good idea to look at normal approximations to the distribution at hand. Here, it's pretty clear that you might well be negative. If instead of $970$ you only win $968$ times your score is negative. – 2017-02-07
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    Your second answer answered my question, that's what I was trying to figure out. However, I think I had missed out a crucial point in the question by assuming that the 1000 samples taken out from an infinity number of samples would follow the 97:3 ratio strictly (without any probabilistic properties). – 2017-02-07
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    Yeah I can see my mistake now. How would one try work out approximations given that the number of samples and probabilities are known? – 2017-02-07
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    The normal approximation, with $N$ samples has mean $.97\times N$ and standard deviation $\sqrt{N\times .97\times .03}$ . using those numbers (and $N=1000$) we see that the probability of getting $968$ or fewer wins is about $.3554$. So you have a very large probability of getting a negative score. – 2017-02-07
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    Could you please elaborate how you related standard deviation with the number .3554? And why is the mean 970? – 2017-02-07
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    The wikipedia article on [binomial distributions](https://en.wikipedia.org/wiki/Binomial_distribution) covers the normal approximation I used. Of course the mean is just $pN$ so here it is $.97\times 1000=970$. To get $.3554$ I computed the probability that a sample from a normal distribution with mean $970$ and $\sigma = 5.538$ (which I got from that square root) would be $968$ or less. It's just an approximation and since your total score is hovering right around $0$ I'd expect the exact answer to be slightly different. – 2017-02-07
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    With a computer, it's not hard to get the exact answer. I just ran it and got a hair over $0.38027287$. To do that computation note that the probability of getting exactly $k$ wins out of $1000$ is $\binom {1000}{k}\times .97^k\times .03^{1000-k}$. i just computed that for $k=968,967,\cdots, 0$ and added (arbitrarily cutting the sum off when the terms became negligible). – 2017-02-07
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    Thank you, that makes it very clear to me now. One last question, theoretically speaking, there is ~65% chance to have a positive outcome. Would that mean that the odds are still in favour of the player? – 2017-02-07
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    Also, is there a name for the probability that you have computed? – 2017-02-07
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    Oh, yes. On any given game, you expect to make $1\times .97-31\times .03=.04$ so, yes, the game has positive expectation. – 2017-02-07
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    The exact computation was done using the binomial distribution, as in the link I sent. Again, with large samples it's usually easier to work with the normal approximation. – 2017-02-07
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    Thank you very much for all the answers, it was really helpful! – 2017-02-07

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