An odd way to write down a probability using fractions Let $k$ be a natural number and $X$ and random variable taking values
in $\left\{ 0,1,2,\ldots\right\} $. I recently saw someone claim
that 
$$
\mathbb{P}[X\geq k]=\frac{\mathbb{E}[X]}{\mathbb{E}[X|X\geq k]}.
$$
I believe I somehow must have understand something wrong, because
if I construct and easy example to test this formula, it fails (see below).
Question 1: What changes does one have to make to the formula to make it true?
Here's the counter-example: If
we toss a "two-sided" die (two-sided to make the counting
simpler), having numbers $1$ and $2$, and let $X$ count the number
of ones in a sequences of 4 tosses, and use $k=1$, then 
$$
\mathbb{P}[X\geq1]=\frac{15}{16},
$$
but $\mathbb{E}[X|X\geqslant1]=\mathbb{E}[X]$, so we would obtain
that $P[X\geq1]=1$. 
(If I use $k=2$, I also do not get equality,
rather $\mathbb{E}[X|X\geq2]=\frac{28}{16}=1.75$.)
Question 2: How to interpret "1.75"  in the last result? Given the fact that we condition on $Z\geq 2$, I would have expected to obtain a result that is greater than 2.
 A: Your counterexample is wrong. Assuming (as clarified later) that we toss a die (or flip a coin) $4$ times and want to count the number of times $1$/heads comes up, we have $\mathbb E[X]=2$, $\mathbb E[X \mid X \ge 1] = \frac{1 \cdot \frac{4}{16} + 2 \cdot \frac{6}{16} + 3 \cdot \frac{4}{16} + 4 \cdot \frac{1}{16}}{\frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16}} = \frac{32}{15}$, and $\mathbb E[X \mid X \ge 2] = \frac{2 \cdot \frac{6}{16} + 3 \cdot \frac{4}{16} + 4 \cdot \frac{1}{16}}{\frac{6}{16} + \frac{4}{16} + \frac{1}{16}} = \frac{28}{11}$. 
But the formula is also wrong: it correctly says $\mathbb P[X \ge 1] = \frac{2}{32/15} = \frac{15}{16}$, but incorrectly says $\mathbb P[X \ge 2] = \frac{2}{28/11} = \frac{11}{14}$ when $\mathbb P[X\ge2]$ is actually $\frac{11}{16}$.
Here's why. If we multiply to clear denominators, we get the incorrect formula
$$\color{red}{\mathbb E[X] = \mathbb E[X \mid X\ge k] \cdot \mathbb P[X \ge k]}$$
as opposed to the correct way to deal with conditional expectation:
$$\color{blue}{\mathbb E[X] = \mathbb E[X \mid X\ge k] \cdot \mathbb P[X \ge k] + \mathbb E[X \mid X < k] \cdot \mathbb P[X < k]}.$$
From here, if we write $\mathbb P[X < k]$ as $1 - \mathbb P[X \ge k]$ and solve for $\mathbb P[X \ge k]$, we can get $$\mathbb P[X \ge k] = \frac{\mathbb E[X] - \mathbb E[X \mid X<k]}{\mathbb E[X \mid X\ge k] - \mathbb E[X \mid X<k]}$$ which is the "correct version" of this formula.

Also, an inequality of this form holds: because $X$ is nonnegative, we have $$\mathbb E[X] \ge \mathbb E[X \mid X\ge k] \cdot \mathbb P[X \ge k]$$ by dropping the $\mathbb E[X \mid X < k] \cdot \mathbb P[X < k]$ term, and therefore $$\mathbb P[X \ge k] \le \frac{\mathbb E[X]}{\mathbb E[X \mid X \ge k]}.$$ This is tight exactly when the term we dropped was equal to $0$: either if $\mathbb P[X < k] = 0$, or else if $X=0$ whenever $X<k$.
A: $$
E[X] = \sum_{n} P(X=n)/cdot n$$$$
E[X]/E[X|X\ge k] = \frac{\sum_n P(X=n)n}{\sum_{n\ge k} P(X=n) n / P(X\ge k)}
$$
Therefore
$$
P(X\ge k) = E[X]/\sum_{n\ge k} P(X=n)n
$$
Your counterexample is wrong, the expectation without conditioning is one half.
A: $P(X\geq k) = \sum_{i=k}^{\infty}p_{X}(i)$ where $p_{X}(i)=P(X=i)$. 
Now we have that $E[X]=\sum_{i=k}^{\infty}i \cdot p_{X}(i)$ and that $E[X|X\geq k]=\sum_{i=k}^{\infty} \frac{i\cdot p_X(i)}{1-\sum_{j=0}^{k-1}p_X(j)}$. Dividing these $2$ and noting that $\sum_{i=0}^{\infty}p_{X}(i) = 1$ you get the desired result.
To elaborate on the conditional expectation bit: You can think about it as having cut a part of the probability distribution and you then have to normalise it so that the probabilities still sum up to 1
