Let's say we have generated n=100000 numbers from a set: {1, 2, ..., 50}.

Let X a discrete random variable that expresses the number that 1 appears. Then X~Binomial(n,p), where n=100.000 and p=1/50 (uniform distribution of numbers). The mean is n*p=2000.

Why the probability: P(X=2000) does not equal to 1, or something close to 1? Isn't this the expected value, so why the probability doesn't equal 1?

This is just a theoretical question, I know that all the real random generators are not that random but pseudo-random.

  • $\begingroup$ $2000$ is the most common and most likely result, but $1999$ isn't much more uncommon, and neither is $2001$. Basically, any result between $1950$ and $2050$ are more or less equally common, with $1950$ and $2050$ having probability around $0.5\%$, and $2000$ having probability around $0.9\%$. $\endgroup$ – Arthur Jun 14 '15 at 19:30

For a six-sided die, the expected value of a roll is $3.5$, but the probability of rolling $3.5$ is zero.

Even if the expected value is attainable by the random variable, it still doesn't hold: for a $99$-sided die, the expected value of a roll is $50$, but the probability of rolling $50$ is $1/99$, nowhere near $1$.

For a single trial, there is no guarantee that the value of a random variable is near its expected value. However, in some cases you can quantify how close it should be; if you consider over many trials, you can talk about some sort of convergence.

In fact, you can actually apply the Law of Large Numbers here to say something that your intuition seems to be grasping for. If you let $X_1,\ldots, X_n$ be coin flips with probability $p=1/50$, then the law of large numbers states that $$\mathbb{P}\left(\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n X_i = p\right)=1.$$ Note that $\sum_{i=1}^n X_i$ is the binomial random variable $X$ that you considered above.


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