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I've got a logical problem with my mathematics skills. so far, I have to calculate a prizing system to say "you won" or "you lose".

here my way until now:

  • u = max. users
  • p = prizes
  • w = winchance in percent: p*100/u

to say, if you won or not I use this:

$win = (rand(0,10000) > 100*(100-w)) ? TRUE : FALSE;

is this the correct way to calculate it? or is there still a better solution? thanks a lot, ~frank

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  • $\begingroup$ Does max.users u=10000 ? $\endgroup$ – CAGT Jan 22 '14 at 8:31
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    $\begingroup$ Minor problem for rand: if $w=100$, there is still slight chance for the code to give a FALSE. I suggest changing the rand part to rand(1,10000). Bigger problem, but maybe still minor: rand might be giving integers only, so if the right hand side of inequality $100\times(100-w)$ is not an integer, the result may not be precise. $\endgroup$ – peterwhy Jan 22 '14 at 9:21
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Apart from the problem of the random function I wrote above, the logical problem here is that the result from every run of the code is supposed to be independent. And so, if $w$ is fixed and the code is run for $u$ times, the number of wins follows binomial distribution: $$W\sim B\left(u,\frac pu\right)$$

And while the expected number of wins is $p$, there is quite some chance that the resultant number of wins is different from $p$. Actually, $$P[W\neq p] = 1 - \binom{u}{p}\left(\frac pu\right)^p\left(1-\frac pu\right)^{u-p}$$

If $p$ is fixed and $u$ approaches infinity, the distribution converges to Poisson distribution with expected number of wins $p$. The probability

$$P[W\neq p]\approx 1-\frac{p^p}{p!}e^{-p}$$

For $p=1$ and $u=10000$, there is over $63\%$ chance that the number of wins is not equal to your $p$.

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  • $\begingroup$ could you please explain this in "my words"? like I did with $win = .... cause my skills in mathematic syntax is to low to understand you. I've got the % of the winchance, so how can I say: you win or lose, based on the % winchance? $\endgroup$ – frank allen Jan 22 '14 at 9:52
  • $\begingroup$ In short, you can almost keep your code $win=(rand(1,10000) > 100*(100-w)) ? TRUE : FALSE;, but the problem of this logic is that the number of wins after running the code for $u$ times is not guaranteed to equal to $p$. $\endgroup$ – peterwhy Jan 22 '14 at 9:57
  • $\begingroup$ in my example, I startet with 4 prizes and 10 users, after a few runnings, I had 3 prizes left for 3 users, so it's a 100% chance to win? $\endgroup$ – frank allen Jan 22 '14 at 10:03
  • $\begingroup$ OK, if you modify $u$ to "number of remaining users" and $p$ to "number of remaining prices", that would be better. $\endgroup$ – peterwhy Jan 22 '14 at 10:08

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