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For a game I generate various block types given certain odds. Say, there's a $0.001$ chance for the karma block. If a typical game has $600$ blocks, what's the distribution of games that have $0$ karma blocks, $1$ karma blocks, $2$ karma blocks, etc.

I've tried reading the wikipedia page on probability distributions, but I can't figure out how it applies to this.

Maazul
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HappyEngineer
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2 Answers2

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You want the binomial distribution because the probability of each block being karma or not-karma (whatever that means), that is, the probability of a success or a failure in each trial is constant and independent of the other trials.

In general, the probability that there are k successes out of n trials in a binomial distribution is ${n\choose k} p^k(1-p)^{n-k}$ (where p is the probability of a success). In your case the probability of 'k' "karma blocks" would be ${600 \choose k} (0.001)^k(0.999)^{600-k}$.

What's actually happening is you determine there must be exactly $k$ karma-blocks, and that probability is $0.001^k$. The rest of the 600 have to be failures, hence $0.999^{600-k}$ (because if it's not a success, it's a failure, 1-0.001=0.999). And you don't care about order so you multiply by the number of different orders there could be for k karma blocks in 600 blocks, which is $600 \choose k$

Zen
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  • Adding to this answer, the Normal Distribution is often used when $n$ is large. There are some other conditions which should be satisfied as they are mentioned on the wiki. – Jerry May 27 '13 at 18:41
  • @Jerry Oh please! The Poisson distribution is a far better candidate than the normal distribution in this case. – Dilip Sarwate May 27 '13 at 19:13
  • @DilipSarwate sure. I never did Poisson. My class stopped at Normal... – Jerry May 27 '13 at 19:15
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A very good approximation to the desired probabilities is $$P\{X = n\} = e^{-0.6}\frac{(0.6)^n}{n!}, ~~ n = 0, 1, 2, 3, \ldots$$ This is called the Poisson approximation to the binomial distribution (cf. @Zen's answer) and applies whenever $n$ is large ($600$ in your case) and $p$ is small ($0.001$ in your case) while $np = 0.6 \ll n$.

Dilip Sarwate
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