This is beyond tough, if you ever get repeated numbers, there is no hope to figure out the right number in the given number of tries
Logical it's the opposite. If you do it right you should never lose a round by the amount of guesses and only by time.
Use the first two tries to just to get through the nummbers 1-8. Dont repeat anything. If by then you still did not get atleast 4 yellow nummbers. Use the thrid guess to look in to 9 and 0 and already start to try figure out repeatet nummbers. So that in every field was at least a yellow/green number. After that all, should fall in to place.
As a example by the first two guesses i only got a yellow 2 and a yellow 8. For the third guess i put the 2980. So the unsure numbers are under already yellow/green blocks and in the rest i already look for possible repeats.
Because there are only a maximum of three nummbers when they repeat, you should figure in two guesses, per process of elimination, out where the numbers belong.
So to show of a little i put up the high score of 10100. I forcefully finished there. Only thing i can add now is if you got only two colored numbers under each other you must spread the 9 and 0 out over the third and fourth guess. For the fourth guess you want to be sure to always have at least a colored number in every column thats not repeating in the column.