Step 1: Find probability of winning in one throw.
Winning numbers are even numbers greater than or equal to 4.
So favourable outcomes are:
\[
4,6
\]
Hence:
\[
P(\text{win}) = \frac{2}{6} = \frac{1}{3}
\]
Step 2: Find probability of losing in one throw.
\[
P(\text{loss}) = 1-\frac{1}{3} = \frac{2}{3}
\]
Step 3: List possible cases.
The man throws until he wins or maximum of three times.
Winning on first throw:
\[
P = \frac{1}{3}
\]
Winning on second throw:
\[
P = \frac{2}{3}\cdot \frac{1}{3} = \frac{2}{9}
\]
Winning on third throw:
\[
P = \frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{3} = \frac{4}{27}
\]
Losing all three throws:
\[
P = \left(\frac{2}{3}\right)^3 = \frac{8}{27}
\]
Step 4: Find gains or losses in each case.
If he wins on first throw, gain is:
\[
1000
\]
If he wins on second throw, he loses \(200\) once and wins \(1000\):
\[
1000-200 = 800
\]
If he wins on third throw, he loses \(200\) twice and wins \(1000\):
\[
1000-400 = 600
\]
If he loses all three throws:
\[
-600
\]
Step 5: Compute expected value.
\[
E = 1000\left(\frac{1}{3}\right) + 800\left(\frac{2}{9}\right) + 600\left(\frac{4}{27}\right) - 600\left(\frac{8}{27}\right)
\]
Step 6: Simplify.
\[
E = \frac{1000}{3} + \frac{1600}{9} + \frac{2400}{27} - \frac{4800}{27}
\]
\[
E = \frac{9000}{27} + \frac{4800}{27} + \frac{2400}{27} - \frac{4800}{27}
\]
\[
E = \frac{11400}{27}
\]
\[
E = \frac{3800}{9}
\]
Step 7: Final conclusion.
Since expected value is positive, it is a gain.
\[
\boxed{\frac{3800}{9}\text{ gain}}
\]