Calculating Probability of Drawing Sword Card Last in 4 Repetitions

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The discussion revolves around calculating the probability of drawing a sword card last in a repeated card-drawing experiment involving three normal decks. The initial calculations provided include total outcomes N(Ω) and favorable outcomes N(A), but the result is deemed incorrect. Participants express confusion over the experiment's description and terminology, particularly regarding the types of cards involved. Clarification is requested to better understand the experiment's parameters and to accurately compute the probability. The conversation highlights the need for precise definitions in probability problems to ensure correct calculations.
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Repeat 4 times the following experiment:

We pull 3 cards from 3 normal decks (one sheet each), we note what some leaves pulled and repositioned every card in the deck from which we got.

What is the probability that we will see for first time any sword card exactly the last time you do the experiment?

N(Ω)= 52^12 and N(A)=39^9*(13^3+13^2*39+13*39^2) and i find this P(A)=N(A)/N(Ω)

but i the result its not correct what is the right way to solve this?
 
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ParisSpart said:
Repeat 4 times the following experiment:

We pull 3 cards from 3 normal decks (one sheet each), we note what some leaves pulled and repositioned every card in the deck from which we got.

What is the probability that we will see for first time any sword card exactly the last time you do the experiment?

N(Ω)= 52^12 and N(A)=39^9*(13^3+13^2*39+13*39^2) and i find this P(A)=N(A)/N(Ω)

but i the result its not correct what is the right way to solve this?

Your statements are unclear, particularly the underlined words.
 
it means that we note what cards we pulled from the decks and the second word means a club card (not heart , diamond or spade)
 
Could you clarify the experiment. I understand the description of what you are doing, but the question is unclear.
 
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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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