Solution:

Answer: $450$

First, we count the number of distributions where each player has at least $1$ card. The possible distributions are:

- Case 1: $4 / 1 / 1$ : There are $3$ choices for who gets $4$ cards, $6$ choices for the card that one of the single-card players holds, and $5$ choices for the card the other single-card player holds, or $3 \times 6 \times 5 = 90$ choices.

- Case 2: $3 / 2 / 1$ : There are $6$ choices for the single card, $\binom{5}{2} = 10$ choices for the pair of cards, and $3! = 6$ choices for which player gets how many cards, for a total of $6 \times 10 \times 6 = 360$ choices.

- Case 3: $2 / 2 / 2$ : There are $\binom{6}{2} = 15$ choices for the cards $DBR$ gets, $\binom{4}{2} = 6$ for the cards that $RB$ gets, and $DB$ gets the remaining two cards. This gives a total of $15 \times 6 = 90$ choices.

Thus, we have a total of $90 + 360 + 90 = 540$ ways for the cards to be distributed so that each person holds at least one.

Next, we look at the number of ways that the condition cannot be satisfied if each player has at least one card. Then, $DBR$ must have no more than one card, and cannot have an even spade. We only care about cases where he has a non-zero number of cards, so he must have exactly $1$ odd spade. Then, we see that there are $2^{5} - 2 = 30$ ways to distribute the other $5$ cards among $RB$ and $DB$ so that neither has $0$ cards. Since there are $3$ odd spades, this gives $3 \times 30$ bad cases, so we have $540 - 90 = 450$ ones where all the problem conditions hold.