Is My Card Game Probability Calculation Correct?

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SUMMARY

The probability calculation for the card game involving 6 type A cards, 9 type B cards, and 6 type C cards is accurately derived. The probability that the leftover counter matches the hidden type A card is 1/6. The distribution of cards results in three players receiving 4 cards and two players receiving 3 cards, leading to a total probability of (1/6)*(4/18)^3 *(3/18)^2 for all players having the matching type A card. The initial assumption that the answer could be simplified to (1/6)*(1/5)*(1/4)*(1/3) is incorrect due to the nature of card distribution.

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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)
 

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