Probability that a number chosen at random has given value

Question

There are $m$ numbers and each can take values between $1$ to $n$. Also, the mean of these numbers is given to be $x$. If someone selects one number at random out of $m$ numbers, find

• the probability that its value is $1$ ?

• the probability that its value is $2$ ?

$ \quad.\quad . \quad. $

• the probability that its value is $n$ ?

Eg. If $x=1$ then all the numbers have to be $1$. Hence probability that number will be $1$ is one and all other probabilities are zero.

Similarly if $x=n$ then all the numbers have to be $n$. Hence probability that number will be $n$ is one and all other probabilities are zero.

Answer 1

I would start with $m=2$ and $m=3$ because you can picture data points in a square or a cube.
For $m=2$, there is one formula when the sum is less than or equal to $n+1$ and another when the sum is greater than or equal to $n+1$. The turning point is the corners of the square $(n,1)$ and $(1,n)$.
The numerator of the probability is piecewise constant, and the denominator is piecewise linear.
For $m=3$, the formula changes at $n+2$ and $2n+1$ because of the cube's corners at $(n,1,1)$ and so on.
I think the numerator is piecewise linear, and the denominator is piecewise quadratic.
I think the following recursive formula is right.
I will define a sequence of functions $P_m(s)$ for the number of ways the sum of $m$ numbers can equal $s$. It is piecewise polynomial, degree $m-1$.
$$P_0(0)=1, P_0(s)=0 \text{ otherwise}\\ P_m(s)=\sum_{k=1}^n P_{m-1}(s-k)$$ Then the probability a number is $j$ is
$$\frac{P_{m-1}(s-j)}{P_m(s)}$$ The following image is for m=6, n=10

Answer 2

Your question refers to randomly selecting one of $m$ numbers (not random variables), where the mean $x$ of these $m$ numbers is given. Thus:

Let $a_I$ be the result of randomly selecting one of the fixed numbers $a_1,...,a_m$, each of which is in $[1..n]$; that is, $I$ is Uniform on $[1..m]$. Then $P(a_I=k)$ is just the proportion of these numbers that equal $k$; i.e., by the Law of Total Probability: $$\begin{align}P(a_I=k)&=\sum_{i=1}^m\,P(a_I=k\mid I=i)\cdot P(I=i)\\[2ex] &=\sum_{i=1}^m\,\mathbf{1}_{a_i=k}\cdot {1\over m}\\[2ex] &={\sum_{i=1}^m\mathbf{1}_{a_i=k}\over m } .\end{align}$$ This cannot generally be expressed as a function of the mean $x:={\sum_{i=1}^ma_i\over m},$ although certain degenerate cases can be noted (as you've already done); e.g., $x=1$ or $x=n$ imply $P(a_I=x)=1.$

---

A different question concerns randomly selecting one of m random variables (not numbers), conditional on their mean being $x$. Thus:

Let $X_I$ be the result of randomly selecting one of the random variables $X_1,...,X_m$, which are independent and identically distributed on $[1..n]$; that is, $I$ is Uniform on $[1..m]$ and independent of the $X_i$. Then, by the Law of Total Probability, letting $s=mx$: $$\begin{align}P(X_I=k\mid {X_1+...+X_m\over m} = x)&=\sum_{i=1}^m\,P(X_I=k\mid I=i, X_1+...+X_m = s)\\ &\quad\quad\quad\cdot P(I=i\mid X_1+...+X_m = s)\\[2ex] &=\sum_{i=1}^m\,P(X_i=k\mid X_1+...+X_m = s)\\ &\quad\quad\quad\cdot P(I=i)\\[2ex] &={1\over m}\sum_{i=1}^m\,P(X_i=k\mid X_1+...+X_m = s)\\[2ex] &=P(X_1=k\mid X_1+...+X_m = s)\\[2ex] &={P(X_1=k\ \cap\ k+X_2+...+X_m = s)\over P(X_1+...+X_m = s) }\\[2ex] &={P(X_1=k)\ P(k+X_2+...+X_m = s)\over P(X_1+...+X_m = s) }\\[2ex] &={1\over n}{P(X_2+...+X_m = s-k)\over P(X_1+...+X_m = s) }\\[2ex] &={1\over n}{\pi_{m-1}(s-k)\over \pi_m(s) }\\[2ex] \end{align}$$ where $\pi_{m'}(s)$ is the probability that the sum of (any particular) $m'$ of the $X_i$ equals $s$.

NB: If we make the additional assumption that the $X_i$ distribution is Uniform on $[1..n],$ then $\pi_m(s) = {P_m(s)\over n^m},$ where $P_m(s)$ is @Empy2's function that counts the number of ways that a sum of $m$ numbers, each in $[1..n],$ can equal $s$. So, in this case, $$\begin{align}P(X_I=k\mid {X_1+...+X_m\over m} = x)&={1\over n}{\pi_{m-1}(s-k)\over \pi_m(s) }\\[2ex] &={1\over n}{n^m\over P_m(s)}{P_{m-1}(s-k)\over n^{m-1}} \\[2ex] &={P_{m-1}(s-k)\over P_m(s)} \\[2ex] \end{align}$$ However, without this assumption we can still obtain a normal approximation with sufficiently large $m$, assuming only that the distribution of the $X_i$ has mean $\mu$ and variance $\sigma^2$:

$$\begin{align}P(X_I=k\mid {X_1+...+X_m\over m} = x) &\approx{1\over n}{{1\over \sqrt{(m-1)\sigma^2} }\phi\left({mx-k-(m-1)\mu\over \sqrt{(m-1)\sigma^2} }\right)\over {1\over \sqrt{m\sigma^2} }\phi\left({mx-m\mu\over \sqrt{m\sigma^2} }\right) }={1\over n}\sqrt{m\over m-1}\exp[-{1\over 2}(a^2-b^2)]\\[2ex] \end{align}$$ where $\phi()$ is the standard normal pdf and $$\begin{align}a&={mx-k-(m-1)\mu\over \sqrt{(m-1)\sigma^2} }\\[2ex] b&={mx-m\mu\over \sqrt{m\sigma^2} }. \end{align}$$

(For the uniform distribution on $[1..n],$, we have $\mu={n+1\over 2}, \sigma^2={n^2-1\over 12}.$)