0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 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? $\endgroup$ – lulu Feb 7 '17 at 17:13
  • $\begingroup$ 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? $\endgroup$ – Shibalicious Feb 7 '17 at 17:27
  • $\begingroup$ 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 $970-31\times 30=40$. $\endgroup$ – lulu Feb 7 '17 at 17:31
  • $\begingroup$ 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. $\endgroup$ – lulu Feb 7 '17 at 17:32
  • $\begingroup$ 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). $\endgroup$ – Shibalicious Feb 7 '17 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.