I'm trying to calculate the possible ways a set of 6-sided dice d can fall so an amount of bonus points b can be distributed among them to result in at least t amounts of 6es, with the assumption that no die will roll as a 6 naturally.
Example: we have 4 dice and 4 points, need to achieve at least 2 times a 6. This can be achieved with 2 dice rolling 5-5, 4-5, 5-4, 4-4, 3-5, or 5-3, 3 dice rolling 5-5-5, 5-5-4, 5-4-5, 4-5-5, or 4 dice rolling 5-5-5-5.
Basically this can be calculated by dividing the points over the dice with repetition, or k multichoose n where n = t -> b and k = t -> d. However, for each of these steps I need to multiply the amount of combinations with the possible states of the uncounted dice. My reasoning first was that each of the uncounted dice could have a value of 1-5 and thus (n multichoose k) * 5^(d-k).
However in that case I will be counting some cases double, such as when I say 2 dice roll 5-5, the other two dice can both be 5 as well which equals the case of 4 dice rolling 5-5-5-5. If I say the excess dice can be 1-4, then it still matches the case of 5-5-4 for the 3 dice. But if I say the excess dice can be 1-3, I will miss cases with the 3 dice rolling: 5-5-4-4 is also valid, but won't be reached if the last die can only be 1-3.
Lastly I will also need to keep into account when not using all the dice in the pool, that it's also possible to make the "winning" combination several ways among the dice, basically multiplying the result with the amount of combinations without repetition or d choose k. This seems to open up the way for me to do even more double-counting...
This is becoming quite a rabbit hole I'm finding difficult to wrap my head around, and I'm wondering if there's a simpler way of calculating this?