As with pretty much all the nim variants, this one can be solved by starting from the end and working backwards. With the original total number of stones being an odd number (15, as given in the title) the players will have the same parity whenever there's an odd number of pebbles left, so it's easy to work out the best strategies: they are the ones that put the opponent to a known-losing position. If no such move exists, the position is losing.
Pebbles left | Opp. parity | Odd | Even
0 | Diff | W | L
1 | Same | L | W (1)
2 | Diff | W (2) | W (1)
3 | Same | W (2) | W (3)
4 | Diff | L | W (3)
5 | Same | W (1) | L
6 | Diff | W (1) | W (2)
7 | Same | W (3) | W (2)
8 | Diff | W (3) | L
9 | Same | L | W (1)
10 | Diff | W (2) | W (1)
11 | Same | W (2) | W (3)
12 | Diff | L | W (3)
13 | Same | W (1) | L
To mechanically construct the table, first fill in row 0, and then for each position, look how many steps up you need to go to hit an "L". If "Opp. parity" is "Diff", then go up the other column instead.
As we can see, the pattern repeats after eight steps, so we don't have to count all the way to 15, but we can instead just subtract 8, and look at the strategy for 7 pebbles. Since we have an even number of pebbles, and presumably we get to start, we should colour
which leaves the opponent in a losing position that's actually included in the table.
This table works for all sensible starting positions with two players: if the starting number of pebbles were even, the only possible result would be a tie (both win, or both lose.)
Since the table may seem a bit hard to memorise, here's the strategy again in "condensed form":
If your opponent has an odd number of pebbles, leave 1 or 4 pebbles (mod 8).
If your opponent has an even number of pebbles, leave 0 or 5 pebbles (mod 8).