My strategy to get to the correct answer in as few a tries as possible is as follows:

This is what I know at the start: there are 6 colours, they may be duplicated.

I start by trying a set of the first 4 colours - red, orange, yellow, green.

From the score for the first attempt I know that I have two of the correct colours, but only one is in the correct position.

On the next attempt to find the secret code I use the other two colours (blue and purple) duplicated.

From the score for this attempt (attempt #2) I know that the remaining two colours are either both blue, both purple, or one blue, one purple.

On the next attempt (attempt #3) I try a red and a green, putting the red in the same position as in attempt #1, and trying green in a different position than in attempt #1 - thus being a possible combination for the score of attempt #1. At the same time I put two blues in the remaining positions being one possible solution for the score of attempt #2.

The score of attempt #3 tells me a number of things:

Two blues are not in the secret code, otherwise there would have been a 2 score for correct position. From attempt #2 we know that blue does not occupy the first two positions. There are not two blues in the secret code. This means there may be two purples - or one blue, one purple. If blue is in the code, it must be in one of the last two positions. I assume that the code has one blue and one purple - since two blues is not correct and one blue and one purple is more probable than two purples. If blue is in the code, then either red OR green is in the code, because the score of attempt#3 only allows for one more colour if the other colour is blue.

In attempt # 4, I place purple and blue in possible positions for their colour, and use two possible colours and positions for the other two positions.

The score of 2 0 for attempt #4 means that if blue and purple are two of the colours in the code, and at least one of these must be, then the other two colours red and orange are not in the code - meaning (from attempt #1) that yellow and green are in the code. I'm assuming that the correct colour in the third position is blue, which means that the green cannot be in position two otherwise there would be a 2 0 score for attempt # 3. So I guess the correct combination for attempt #5.

Those of you who like math, think about how simultaneous equations map to this type of game.