Here's a problem I'm having.
I have 7 dice. I want to rank the difficulty of achieving a particular 5 dice combination after all 7 dice have been rolled.
A straight is when at least 5 of the seven dice are in sequential order.
A full house is when you have at least 3 dice of one number and at least 2 dice of another number.
And a 3 of a kind is when you have 3 dice of one number but NOT a full house.
So the first thing that makes sense to me is that 3-of-a-kind is actually more difficult than a full house after all dice are out.
But the problem I'm having is whether 3-of-a-kind is more difficult or a straight is more difficult. Because if a straight is more difficult, then I want to award the roll as a straight and that changes the probability of seeing a 3-of-a-kind. And if a 3-of-a-kind is more difficult then I want to award the roll as a 3-of-a-kind, not check for straights, then straights become more difficult and the odds switch places.
Should I be checking for these independently from each other? For instance, when you ask how often do I get 3-of-a-kind in poker does it include 4-of-a-kind?
Am I thinking about this correctly?
I am not sure whether I understand your question. Is it about the overlap of straight and 3 of a kind? The following scheme might be helpful anyway.
$\begin{array}{cccccccc} \text{split-up} & & \text{full house} & \text{straight} & \text{3 of a kind} & \text{orders} & \text{selections}\\ 7 & & & & & 1 & 6 & 6\\ 6|1 & & & & & 7 & 30 & 210\\ 5|2 & & \text{yes} & & & 21 & 30 & 630\\ 5|1|1 & & & & & 42 & 60 & 2520\\ 4|3 & & \text{yes} & & & 35 & 30 & 1050\\ 4|2|1 & & \text{yes} & & & 105 & 120 & 12600\\ 4|1|1|1 & & & & & 210 & 60 & 12600\\ 3|3|1 & & \text{yes} & & & 140 & 60 & 8400\\ 3|2|2 & & \text{yes} & & & 210 & 60 & 12600\\ 3|2|1|1 & & \text{yes} & & & 420 & 180 & 75600\\ 3|1|1|1|1 & \text{consecutive} & & \text{yes} & \text{yes} & 840 & 10 & 8400\\ 3|1|1|1|1 & \text{not consecutive} & & & \text{yes} & 840 & 20 & 16800\\ 2|2|2|1 & & & & & 630 & 60 & 37800\\ 2|2|1|1|1 & \text{consecutive} & & \text{yes} & & 1260 & 20 & 25200\\ 2|2|1|1|1 & \text{not consecutive} & & & & 1260 & 40 & 50400\\ 2|1|1|1|1|1 & & & \text{yes} & & 2520 & 6 & 15120\\ \text{sum} & & & & & & & 279936\end{array}$
Giving the dice an order there are $6^{7}=279936$ possible outcomes.
To explain: split-up $4|2|1$ stands for the possibilities a number is thrown $4$ times, another number is thrown $2$ times and again another number is thrown $1$ time. For instance: $1211616$.
Then the numbers $1,2,6$ are selected but of course it can be done with other numbers. There are $6\times5\times4=120$ selections for split-up $4|2|1$.
Also there are several orders. Next to the mentioned $1211616$ we can have $6111261$ or $1621116$ et cetera. There are $\frac{7!}{4!2!1!}=105$ orders for split-up $4|2|1$.
The product gives you all possibilities for split-up $4|2|1$ and the probability that it occurs is $\frac{105\times120}{279936}=\frac{12600}{279936}$.
In two cases I discerned consecutive and not consecutive. For instance at result $4164423$ we are dealing with a split-up $3|1|1|1|1$ but the numbers are not consecutive (hence we do not have a straight, if I understood you well). At result $4564423$ we are also dealing with a split-up $3|1|1|1|1$ and at this result the numbers are consecutive (hence we have a straight).