Question

If 2 cards are drawn from a deck of 52 cards, what is the expected number of spades?

Answer 1

First do 52/4

Answer 2

Next, you will take the 1/4 of 52 and divide it until it = 2. How many times did you divide it? You have to solve from there.

Answer 3

For this problem, it would appear at first that a critical point is whether the draw is done "without replacement." I will assume that it is done this way. The sample space is as follows:

A. SP SP
B. ~SP SP
C. SP ~SP
D. ~SP ~SP

(where ~SP means the card was not a spade). The probability of the outcome at the second draw changes depending on what happens in the first draw. I get:
\[P(A) = \frac {13}{52} \frac {12}{51} = \frac{156}{2652} = 0.0588 \]
\[P(B) = \frac {39}{52} \frac {13}{51} = \frac{507}{2652} = 0.1912 \]
\[P(C) = \frac {13}{52} \frac {39}{51} = \frac{507}{2652} = 0.1912 \]
\[P(D) = \frac {39}{52} \frac {38}{51} = \frac{1482}{2652} = 0.5588 \]
Notice that the sum of P's equals 1.0
\[156 + 2(507) + 1482 = 2652\]
The expected number is just the average. It's the sum of the number of spades times P for each case:
\[2 P(A) + 1 (P(B) + 1 (P(C)) = \frac{}{} = 2(0.0588) + 2(0.1912) = 0.5\]
So, after doing such a fancy calculation, I get a very simple answer! The reason is easy to see: after two draws I have two cards. There is symmetry with respect to the four suits---no reason why any one should be different than the others. I have:
\[\frac{2}{4} = 0.5\]