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March 17th, 2011, 12:29 PM   #1
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Fair deck of 52 Cards

Arrangement of 5. (5 cards are dealt out of a 52 card deck)
# of arrangements: 52! / (52-5)! 5! = 2, 598, 960 Correct?

Order does not matter.
Random.
Dependent.

I need to know the probability of getting exactly no Jacks, 1 Jack, 2 Jacks, 3 Jacks, 4 Jacks --- In the 5 cards that are dealt to us.
We do not know any card before knowing the next one. They are all revealed at once.
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March 17th, 2011, 09:05 PM   #2
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Re: Fair deck of 52 Cards

The key to this is that the C(52,5) = 2598960 possible hands are all equally likely.
Therefore, all you have to do is to find the number of ways to choose the cards in question, and divide by 2598960.

No jacks
How many arrangements are there without jacks? Suppose you were to take the jacks out of the pack of 52, and then deal. There are 48 cards left, and C(48,5) possible hands. Therefore the probability this would happen if the jacks had stayed in the pack is C(48,5)/C(52,5).

One jack
Again, separate the jacks from the rest of the cards. You select one jack from the pile of 4, and then four cards from the pile of 48. There are C(4,1)*C(48,4) ways you could do this, so the probability is C(4,1)*C(48,4) / C(52,5).

Two jacks
Separate the jacks, then select two jacks from the pile of 4, and three cards from the pile of 48. There are C(4,2)*C(48,3) ways... etc, etc.

Finally, check your answer by adding the five different probabilities. They should sum to 1.
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