Is My Card Game Probability Calculation Correct?

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The discussion revolves around calculating the probability that all five players in a card game receive the type A card matching their counter. The game involves 6 type A cards, 9 type B cards, and 6 type C cards, with one card of each type hidden. The proposed solution calculates the probability as (1/6)*(4/18)^3*(3/18)^2, considering the distribution of cards among players. There is confusion regarding whether the calculation is correct, especially in light of the fact that one player has two type A cards. The thread seeks validation of the probability calculation method used.
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A card game involves dealing 3 types of cards to players. There are 6 type A cards, 9 type B cards and 6 type C cards. Each type A card has one matching counter.

In a game with 5 players, one card of each type is selected at random and hidden without knowledge. Each player then chooses a counter with one left over. All the remaining three types of card are combined, shuffled and dealt as evenly as possible to the players.

What is the probability that all of the 5 players have the type A card that matches their counter?




My solution:
Probability that the left over counter matches the hidden type A card = 1/6

18 of the three card types are dealt with three players receiving 4 cards and two players receiving 3 cards.

Probability of each player receiving the correct type A card = 4/18 and 3/18 respectively so the total probability that all players have the type A card matching their counter is
(1/6)*(4/18)^3 *(3/18)^2

Dues this seem correct to anybody? I was expecting this problem to be a lot harder so not sure if I'm missing something/done something wrong. Thanks in advance
 
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I would have thought that since one player has two cards of type A then the answer would simply be (1/6)*(1/5)*(1/4)*(1/3)
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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