as @Bass mention I think you can lower the table to

        Pebbles left |  you have Odd number stones  | you have Even number  stones 

        0            | W                            | L

        1            | L                            | W (1)

        2            | W (2)                        | W (1)

        3            | W (2)                        | W (3)

when you count the next step as if k is odd then:

is step k-1 lose as even, color 1 stone then this move is winning

is step k-2 lose as odd, color 2 stone then this move is winning

is step k-3 lose as even, color 3 stone then this move is winning

same principle for even:

this why I think

        Pebbles left |  you have Odd number stones  | you have Even number  stones 

        4            | L                            | W (3)

        5            | L                            | W(1)

(basically searching if in the diagonal column their is a lose and you go up maximum 3 times)

(sorry cant comment not enough points)

as @Bass mention I think you can lower the table to

        Pebbles left |  you have Odd number stones  | you have Even number  stones 

        0            | W                            | L

        1            | L                            | W (1)

        2            | W (2)                        | W (1)

        3            | W (2)                        | W (3)

Let say K is the number pebble left.

when you count the next step as if k is odd then:

If your nunber of pebbles odd then you open number pebvles even.

Find losing game for k-1,k-2,k-3 For the even column If you find one you win else you lose.

same principle for even:

this why I think

        Pebbles left |  you have Odd number stones  | you have Even number  stones 

        4            | L                            | W (3)

        5            | W(1)                      | L

(sorry cant comment not enough points)