I am trying to figure out the odds of a certain situation in the card game French Tarot, specifically what are the odds of the 'attacker' being out of the same suit of the 'R' that a defender has played on round 1. This is a bad situation since the attacker would win 10 points on the first round.
As a start to this, I want to calculate the odds of any player who takes the dog ending up out of a suit of cards (i.e. having 0 cards of that suit). The dog is a set of 6 cards that the attacker can pick from and must put back 6 cards before starting play.
To calculate this, I want to see the probability that the person has at most 6 of one suit after adding the 6 cars from the dog to their 18 cards (assuming a 4 player game). With 6 or less cards for one suite they can all be placed in the dog, ending up with none of that suit for the final 18 cards before the playing begins.
I wrote the following formula to capture this:
C(78-14,24-6)*C(78-(24-6), 6) / C(78,24)
The first term [C(78-14,24-6)] represents choosing 18 cards that are not from that suite, and the second term [C(78-(24-6), 6)] represents picking 6 more cars (of any suite) from the remaining cards, which is 78 total minus the first 18. The last term in the division [C(78,24)] represents the total number of hands of 24.
However, when I plug this into Wolfram Alpha I get ~2280 which means my equation is fundamentally wrong since it should be less than 1.0.
If someone can tell me where I am going wrong I'd appreciate it.
Update: I realize that in a real game of French Tarot the attacker who wins the bet is likely to have a better hand than the others, but I am not factoring that in to my calculations at this point, and am assuming an even distribution. Mainly I want to see what is wrong with my equation and then I can build off that.
Update 2: corrected a few typos where I had said 20 and 26 instead of 18 and 24.
I'll assume that by "at most $6$ of one suit" you mean "at most $6$ of one particular suit" (namely the one whose 'R' the defender plays).
I'll also ignore, like you did, the fact that a defender is known to have played the 'R', so this card is known not to be in the attacker's hand.
Your calculation overcounts hands that have more than $18$ cards not from the given suit (equivalently, hands with less than $6$ cards from that suit). For instance, a hand with $19$ cards not from the given suit is being counted $\binom{19}1=19$ times, since any of the $19$ cards can be taken to be the one in the group of $6$. What you need is
$$ \sum_{k=0}^6\binom{78-14}{24-k}\binom{14}k\;. $$