1
$\begingroup$

What would be the correct answer for expected number of diamonds in a 5-card hands? I am thinking as follow:

Let $X_i = 1$ if the i-th card is diamond, or $0$ otherwise.

$X_i = (1 \times P($i-th card is diamond$)) + (0 \times P($i-th card is NOT diamond$)) = 13/52 = 1/4$

Then, expected number of diamonds in a hand is $X_1 + X_2 + X_3 + X_4 + X_5 = 5/4$

Is this correct? Do I need to consider if first card is diamond, then probability of second card is diamond will change to $12/51$ ?

Many thanks for the help.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

You're absolutely right. Another way to think about this problem is that the sum of the expected numbers for the different suits should be $5$, and there is no reason why any suit is more likely than another. This means that each suit has an expected number of $5/4$.

Notice the value of $X_1$ does not affect that of $X_2$ because these are expected values and not outcomes of an event. Before any cards are dealt to you, there is no reason for you to expect that the second card will be a diamond more or less than you expect that the first card will be a diamond.

$\endgroup$
4
  • $\begingroup$ Thanks, Jared. So, do you mean in expected values, the event is always independent? $\endgroup$
    – user350954
    Commented Aug 6, 2013 at 7:29
  • $\begingroup$ @user350954 What he means is that the probability that the second card is a diamond (without any conditionals) is $\frac{1}{4}$. That is what is asked for when computing expected value. You can calculate $E(X_2)$ and take the dependency on $X_1$ into account (they are not independent) via $P(X_1 = 0)\cdot E(X_2|X_1 = 0) + P(X_1 = 1)\cdot E(X_2|X_1 = 1)$, but you still get $\frac{1}{4}$. $\endgroup$
    – Arthur
    Commented Aug 6, 2013 at 7:37
  • $\begingroup$ The first paragraph uses linearity of expectation, which is valid even for random variables that are not independent. $\endgroup$
    – Tony Huynh
    Commented Aug 6, 2013 at 8:34
  • $\begingroup$ I guess, I understand now. Thanks a lot guys. $\endgroup$
    – user350954
    Commented Aug 6, 2013 at 16:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .