homebrew – Is it still a binomial distribution when rolling dice with different probabilities that must be met to count a success?

How do you calculate the probability of success when success is two different criteria / probabilities?

I theorize a resolution system for a homebrewed mechanic that involves launching a 6d10 pool and triggering abilities based on certain outcome criteria (all in pairs of 2 dice). The remaining value pool is always important because it is allocated to something else, but for the calculation of the probability, the dice can be any result. All dice are rolled at once and the order does not matter.

A player rolls 6d10 and tries to use a single dice or a combination of two dice to get a final value. If you use two dice, combinations can be made based on the facial values ​​of the pair to trigger increases in the sum. The conditions for increasing the value are defined as abilities and have varying success criteria.

How to calculate the probabilities of different abilities defined as 2 dice on 6d10 with disparate success conditions such as:

9 (1/10); any odd number less than 9 (4/10)

any even number X (5/10); X / 2 (1/10)

1 or 2 (2/10); ten[if 1] 9[if 2] (1/10)

ect.

I had initially assumed that being combinations of pairs would produce an equal probability for each discrete option and that the total probability would be the combination of all possible pairs that trigger success. Although I have studied how to proceed, I have not learned all the calculation, I still have trouble knowing if it is fundamentally correct or to correctly combine each discrete probability to generate the final result. I do not know how to test my results without an incredible sample size.

If all else fails, I suppose I could produce a table of all 10 ^ 6 combinations and find a way to query the rows for the corresponding criteria. Although as a dyslexic artist, I tend to avoid programming and I have found no function in Excel allowing me to do what I need without getting me started in the SQL language.

Using the first example (9 and any odd number <9), would the probability be calculated as such?

$$
left[ frac 1 {10} times frac 4 {10} times frac {10} {10} times frac {10} {10} times frac {10} {10} times frac {10} {10} right] times 15 = 0.6
$$

15 being the result of 6choose2

I am a frequent user of Anydice for which I have examined the distribution of the 2 highest dice, namely 6d10. Looking at the result for a perfect 20 × 2×10 in 6d10, I'd assumed it would be the probability for any pair of 1/10. When I calculate it that way, I get 15% instead of the 11.43% of AnyDice. What am I doing wrong?

(Also for my continuing education, how to generate mathematical notation problems and produce them in this text field?)

Thanks for your consideration! Sorry, this question is more related to my weaknesses in mathematics than to a singular problem ….