I guess that I am not understanding the question.
In a 3 digit game there are 1000 combinations of numbers. That means that roughly once every 100 drawings the same last two digits will appears in a drawing. The first digit then has a 1 in 10 chance of appearing every time that pair shows.
So if you want to know the most likely pairs to appear in the next 1-25 drawings, look at the last few 100-150 drawings until there are 25 remaining pair that have not yet appeared. There is a reasonably good chance that one of them will appear.
Once you have isolated that pair look at the last 5 drawings where that pair has appeared and since which 1st digit that it appeared with. You will probably wind up with 5-7 digits for you to choose from that have not yet appeared with that pair. And you can probably use some other methodology based on how many times that number has appeared with other pairs in the last X drawings to reduce your choice down to 2-3 numbers.
If you want to factor the number of times that digit was odd versus even as a means of narrowing own your choices you can certainly do that, but in the end all you are doing is widening the number of drawings that it will take for that pattern to reappear.
The more conditions used to define the strategy, the larger the window between drawings when that pattern has a chance of winning. There is no way to circumvent that which is what keeps the lotteries in business.
If you play all odd or all even numbers there are 250 combinations where the numbers are all odd or all even. There are only 50 pairs where the last two digits are all odds or all even. This means roughly once every four drawings you will see that kind of result, no system is going to alter that fact although in the short term there may be times when it occurs once every two drawing or once every ten.