If I put a copper and add 4 more random cards into my hand from my starting deck, does it change the odds of getting the 5:2 split?
Short Answer:

No.

Long Answer:

The odds of getting a 5:2 split over the more common 4:3 split is unchanged, even if the first card is intentionally chosen to be a copper.

Mathematically, this can be proven using hypergeometric distribution.  This can be calculated using the following function, where N is the size of the deck, n is the number of cards drawn, and m is the number of coppers available:

P(X=k) = (m choose k) * ((N-m) choose (n-k)) / (N choose n)


With a normal starting deck, you would use values of N=10, n=5 and m=7.  Without boring you with the math proper, these would give you a result breakdown as follows:

2 copper: 1/12
3 copper: 5/12
4 copper: 5/12
5 copper: 1/12


So you would normally have a 1/12 + 1/12 = 1/6 chance of getting the 5:2 split.

Now, if the first card is preemptively selected to be a copper, this changes our starting numbers.  Since the first card is known, your starting draw would effectively be a four card draw from a nine card deck containing only six coppers.  Punching in the values N=9, n=4 and m=6 into our function, we end up with the following table:

1 copper: 1/21
2 copper: 5/14
3 copper: 10/21
4 copper: 5/42


Even though the spread of numbers is significantly different, your odds of a 5:2 split work out to 1/21 + 5/42 = 7/42 = 1/6, which is exactly the same as before.

However, note that although the odds of the split themselves are unchanged, by selecting out the first copper you are now over twice as likely to get the 5:2 draw as you are to get the 2:5 draw.