No need to know how to play French tarot, I'll explain whats needed to be understood.
A deck of tarot contains 78 cards, 3 of them called oudlers play a specific and important role in the game, let's call them O1, O2 and O3. Playing with 4 players, the cards will be distributed this way :
_ Each player will be given 18 cards.
_ The remaining 6 cards will make what's called the dog.
My question is about different ways to distribute the cards and probabilities around it. Especially, what is the probability that at least one oudler is in the dog ?
Let's assume the deck is properly shuffled.
- Trivial distribution :
6 cards are given to the dog. Then each players receives one card the one after the other until there are no more cards.
- "Classic" distribution :
Cards are given to the players by 3, and after each turn of the 4 players receiving 3 cards, a card is given to the dog.
THE main question, is really, are the probability of having at least one oudler in the dog the same depending on the distribution of the cards chosen ?
Hopes it is clear. Thks
As Karl has remarked, the order in which the cards are distributed are irrelevant, so strange as it might seem, we might as well distribute $6$ cards to the "dog" first as the easiest way to compute
P(at least one oudler in the dog) = $1-$ P(no oudler in the dog) $=1-\Large\dfrac{\binom{72}{6}}{\binom{78}{6}}\;\approx 0.3917$