Let \( x \) be the amount of money in the original pile.
First, we examine the case where Jeri puts $50 in her left pocket and gives away \( \frac{1}{8} \) of the rest of the pile.
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After putting $50 in her left pocket, she has \( x - 50 \) left.
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She gives away \( \frac{1}{8} \) of this remaining amount, which is
\[ \frac{1}{8}(x - 50). \]
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The amount she gives away results in her keeping
\[ x - 50 - \frac{1}{8}(x - 50) = x - 50 - \frac{x - 50}{8} = x - 50 - \frac{x}{8} + \frac{50}{8}. \]
Simplifying this, we have:
\[ x - 50 - \frac{x}{8} + \frac{50}{8} = \frac{8x - 400 - x + 50}{8} = \frac{7x - 350}{8}. \]
Thus, after putting $50 in her left pocket and giving away \( \frac{1}{8} \) of the remaining money, she has:
\[ 50 + \frac{7x - 350}{8}. \]
Next, we consider the second case where she gives away $200 instead of \( \frac{1}{8} \) of the remaining money.
- She keeps \( x - 200 \).
Now, we set up the inequality since she wants the first scenario to yield more money than the second scenario:
\[ 50 + \frac{7x - 350}{8} > x - 200. \]
To simplify this inequality, we start by eliminating the fraction. Multiply everything by 8:
\[ 8 \cdot 50 + 7x - 350 > 8(x - 200). \]
This simplifies to:
\[ 400 + 7x - 350 > 8x - 1600. \]
Reorganizing gives:
\[ 400 - 350 + 1600 > 8x - 7x, \] \[ 1650 > x, \] or more conveniently,
\[ x < 1650. \]
Now we need to consider the lower limit on \( x \), which is given in the problem as \( x \geq 200 \).
Combining these inequalities, we find:
\[ 200 \leq x < 1650. \]
Thus, the possible values of \( x \) form the interval:
\[ \boxed{[200, 1650)}. \]