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(+1)

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.

(1 edit)

That sounds easy enough, but I am not able to do it (I figured out that should be the process like the 5th time playing it), but I do not know if it's just that I'm not smart enough. So I request you to play and test whether it is indeed possible to win every time

(1 edit) (+1)

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.

This is the way

Huh, neat. Looks like I have a long way to go