Calculating Probability of Drawing Sword Card Last in 4 Repetitions

[ParisSpart]

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?

 

[mathman]

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.

 

[ParisSpart]

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)

 

[mathman]

Could you clarify the experiment. I understand the description of what you are doing, but the question is unclear.

 

[blue_raver22]

I would approach this problem by first clarifying the specific conditions of the experiment. Are we pulling cards from each deck separately and then reshuffling them back into the deck, or are we pulling cards from all three decks at once and then reshuffling them back into the deck? This distinction is important because it affects the probability calculation.

Assuming that we are pulling cards from all three decks at once and reshuffling them back into the deck, the probability of drawing a sword card last in 4 repetitions is (13/52)^4, or approximately 0.12%. This is because in each repetition, there is a 13/52 chance of drawing a sword card last. Since we are repeating the experiment 4 times, we multiply this probability by itself 4 times.

However, based on the given information, it seems that the experiment is not being repeated in this way. Instead, it appears that after each repetition, the cards are being reshuffled back into the deck. In this case, the probability calculation becomes more complex and would require knowledge of the specific order in which the cards were drawn and reshuffled. It is not possible to accurately calculate the probability without this information.

Additionally, it is important to consider the limitations of using normal decks for this experiment. If the decks are not truly random, the probability calculation will not be accurate. In order to accurately calculate the probability, a more controlled and standardized deck would need to be used.