Expect Value of Red Cards Drawn from a Deck

Question

I'm having some trouble figuring out the right way of approaching the following problem.

You draw 2 cards from a standard deck of 52. This contains 13 of hearts, diamond, clubs and spades: clubs and spades are black, hearts and diamonds are red. What is the expected number of red cards chosen?

My current attempt is using the summation formula given in the lectures. I'm not sure how to write it in here properly though. Basically the summation of x*p(x) from x=0 to x=2. Where x is the number of red cards drawn.

This gives me 0*(1/2*25/51)+1*(1/2*26/51)+2*(1/2*25/51) = 0.745, however I feel that this is wrong as I would've thought the answer would be much closer to 1.

I feel like I'm supposed to be doing the probabilities differently, i.e. p(x=1) should be both red/black and black/red, and p(x=2) should be doubled. At this point I've messed around with the options so much I'm not really sure which is right or wrong.

If someone could check over this and give some insight as to what I'm missing it would be appreciated.

Thanks.

Answer 1

Let's define the pdf, saying that $x$ is the number of red cards:

$$p(x) = \begin{cases} \frac{26}{52}\cdot\frac{25}{51} = \frac{25}{102}& \text{if}~x=0\\ \left(\frac{26}{52}\cdot\frac{26}{51} + \frac{26}{52}\cdot\frac{26}{51}\right) = \frac{52}{102}& \text{if}~x=1\\ \frac{26}{52}\cdot\frac{25}{51} = \frac{25}{102}& \text{if}~x=2 \end{cases}$$

Then:

$$\mathbb{E}[x] = \sum_{x=0}^2 xp(x) = \\ = 0\cdot\frac{25}{102}+1\cdot\frac{52}{102} + 2\cdot\frac{25}{102} = \\ = \frac{52+50}{102} = 1,$$

which, as you expected, is exactly $1$.

Indeed, you were wrongly evaluating $p(x=1)$, because:

1. the first card is red and the second black (probability $ = \frac{26}{52}\cdot\frac{26}{51}$);

OR

2. the first card is black and the second red (probability $ = \frac{26}{52}\cdot\frac{26}{51}$);

At the end, $p(x=1) = \frac{26}{52}\cdot\frac{26}{51}+\frac{26}{52}\cdot\frac{26}{51} = \frac{26}{51}$.

Answer 2

I hope that your lecturer expects you to solve problems in the most appropriate way, which is sometimes the shortest and most elegant. Immediate considerations of symmetry tell you that the expected result for any number of cards drawn is half red and half black, likewise a quarter for each suit and a thirteenth for each denomination.

Answer 3

Expectation problems are generally much easier to deal with,
using linearity of expectation

P[get red in one draw] = E[# of reds in one draw] = $\frac12$

So by linearity of expectation, E[# of reds in 2 draws] = $2\cdot\frac12 = 1$