Finding an Infinite Binary Sequence with Average Frequency of 1's = p

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SUMMARY

The discussion centers on the existence of an infinite binary sequence $(x_1, x_2, x_3, \ldots)$ where each element is either 0 or 1, such that the limit of the frequency of 1's approaches a specified probability $p$ in the interval $[0,1]$. The sequence is defined recursively, with $f(n)$ representing the cumulative count of 1's up to the nth term. The proposed method simulates a biased coin flip, ensuring that the average frequency of "heads" matches the probability $p$. This approach effectively constructs a machine that outputs either "H" or "T" based on the defined probability.

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caffeinemachine
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The following question came up when me and a friend of mine were discussing some basic things about probability:Let $p$ be a real number in $[0,1]$.

Does there exist a sequence $(x_1, x_2, x_3, \ldots)$ with each $x_i$ being either $0$ or $1$, such that

$$
\lim_{n\to \infty} \frac{f(n)}{n} =p
$$

where $f(n)= x_1+x_2+\cdots+x_n$, that is, $f(n)$ is the number of times $1$ has appeared in the first $n$ slots. Motivation: Consider a coin which may or may not be fair, and say the probability of "heads" showing up is $p$.
Suppose we want to have a machine which simulates this coin.
That is, we want to have a machine which shows either "H" or "T" every second ('second' here is a unit of time) on its screen.
If the machine properly simulates the coin, then we must have the average frequency of "H" occurring is $p$.
 
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Consider the sequence:
$$
x_1=1,\\
x_{n+1}=\begin{cases} 1, \mbox{ if } f(n)/n<p \\
0, \mbox{ otherwise}
\end{cases}
$$,
 

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