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I don’t know how to handle this 3 players case, may anyone give a hand? Thanks!

When playing card game, there are 3 players playing 32 cards, each receive 10 cards, 2 remind on the table. Overall are 4 aces and 4 kings in the game. How likely is P (E | D)? D = "The table remains 2 aces" E = "I (at least) get one king".

I guess I should have steps as follows.

  1. P(E) = P(1King)+P(2K)+P(3K)+P(4K)

  2. P(D) = (4/32) * (4/32) = 1/256

  3. P (E | D) = P(E) + P(D)

Equations may be used:

Counting combinations

    The number of combinations of n objects taken r at a time is
    nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!

               Number of favorable outcomes
P(I) =     Total number of possible outcome
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  • $\begingroup$ The fact that there are three players is irrelevant. Compare your original question to the question of "What is the probability that I draw at least one red ball when drawing ten balls out of a bucket which contains $4$ red balls and $26$ blue balls?" This should be a straightforward application of the hypergeometric distribution. $\endgroup$
    – JMoravitz
    Commented May 24, 2020 at 16:01
  • $\begingroup$ It will be easier to calculate the opposite. What is the probability of getting zero red balls (and thus ten blue balls) when drawing ten balls out of a bucket which contains four red and 26 blue? $\endgroup$
    – JMoravitz
    Commented May 24, 2020 at 16:02
  • $\begingroup$ Now... to critique what work you have shown... your point #1 is fine, though tedious. It is true that the probability that getting at least one king will be the probability of getting exactly one king plus the probability of getting exactly two kings, etc... Point #2 however... the probability that both leftover cards are aces is not going to be $\frac{4}{32}\times\frac{4}{32}$. The rank of the first leftover card is not independent of the rank of the second leftover card. Instead it is $\frac{4}{32}\times\frac{\color{red}{3}}{\color{red}{31}}$ $\endgroup$
    – JMoravitz
    Commented May 24, 2020 at 16:05
  • $\begingroup$ As for your point #3... $P(E\mid D)$ is not equal to $P(E)+P(D)$...... Conditional probability satisfies $P(E\mid D) = \dfrac{P(E\cap D)}{P(D)}$ $\endgroup$
    – JMoravitz
    Commented May 24, 2020 at 16:05

1 Answer 1

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Guide:

The condition $D$ implies that a person gets $10$ cards from a stack that contains $4$ kings in total.

Find the probability $p$ that in that situation the person does not get any kings at all.

Then the answer is $1-p$.

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