Why can't eavesdropper not make up own private key on diffie hellmann?
Eve does not have the same information. It's only with you poor choice of numbers did you get the phenomena where the number you chose for Eve happens to match. 
In general this doesn't happen, and with careful choice of parameters and large numbers the chance for this happening is negligible. 

You chose an unsafe prime $17=2^4+1$ rather than a safe prime of the form $p=2*q+1$ when q is also prime. You then chose $g=2$ which only spans a group of size 8. $2^8=1\space mod\space 17$ rather than for instance g=3 which can produce the full range. 

For Alice and Bob it is guaranteed: $(g^{P_a})^{P_b}=(g^{P_b})^{P_a} = g^{P_aP_b}$
Since Alice and Bob aren't using Eve's public key, she can't calculate the shared key with them. 
In the vast majority of cases she will end up with a different number. 

In your example the shared key came out 1, and many guesses by Eve work (but not all). Because the public key for Alice and Bob came out with low rank (2 and 4). Using a large safe prime and a generator will make this highly unlikely. 

It is considered very difficult to calculate $g^{ab}$ from $g^a$ and $g^b$ without knowing either a or b. Eve guessing a random key isn't likely to help. For random $e$ in almost all cases $g^{ae}\ne g^{ab}$