# Random Numbers and Binomial Distribution

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.

• $2000$ 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 $0.9\%$. – 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 $1$.
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.