Expert: Paul Klarreich - 12/11/2006

Question
5 cards are drawn without replacement from a standard 52 card deck. Find the probability that a. the 1st card drawn is an ace or a spade. b. the cards draen are 2 aces, 2 kings , and 1 other. c. the 3rd drawn is a ten given the 1st card was kingand the second was the ten of diamonds.


Answer
Questioner:  Ozzie
Category:  Advanced Math
Subject:  Probability
Question:  5 cards are drawn without replacement from a standard 52 card deck. Find the probability that a. the 1st card drawn is an ace or a spade. b. the cards drawn are 2 aces, 2 kings , and 1 other. c. the 3rd drawn is a ten given the 1st card was king and the second was the ten of diamonds.
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Hi, Ozzie,

a. The probability that the 1st card drawn is an ace or a spade.

Since you don't know anything about the first four cards, they are irrelevant. There are four aces and 12 'other' spades, which gives 16 cards out of 52, so that is the probability:  16/52 = 4/13
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b. the cards drawn are 2 aces, 2 kings , and 1 other.

Assuming that:

-- 'other' means NOT an ace and NOT a king,
-- these cards are drawn in any order. (the two aces don't have to be first)

then count the possible 5-member subsets satisfying these conditions.

There are C(4,2) pairs of Aces.  C(4,2) = 4*3/2*1 = 6
There are C(4,2) pairs of Kings. C(4,2) = 4*3/2*1 = 6
There are C(44,1) = 44 other cards.

The total number of 'good' 5-card hands -- I mean subsets -- is the product:

6*6*44

The total number of 5-card hands is C(52,5).
         52 51 50 49 48
C(52,5) = --------------
            5 4 3 2 1
Divide:
   6 * 6 * 44   6 6 44 5 4 3 2 1
p = ---------- = ------------------
     C(52,5)      52 51 50 49 48

Do some reducing and work that out:

     6 44   4 3 2 1
p = ------------------
    52 51 10 49  8

     6 11   4  2 1
p = ------------------
    52 17 10 49  2

       3 11    
p = -----------
    13 17 5 49

That's fairly small.  Now you know why Aces over Kings can win a lot of pots.
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c. the 3rd drawn is a ten given the 1st card was a king and the second was the ten of diamonds.

After those two are drawn, the remaining 50 cards contain 3 remaining 10's.  The probability of drawing one of them is 3/50.