0

I have a question and I will be happy for help. there is a normal deck of cards (ace = 1...king=13, 52 cards, 4 cards for each value) someone takes out without return two cards. what is the expectation of the sum of the cards? I tried to seperate it to odd sums and even sums but it is very difficult and I think there is an easier way. does someone have an Idea how to solve it?

Lidor Hacham
  • 21
  • Try it with a smaller number of cards.. You will see that the expectations for the second card equal the expectations of the first card. – Doug M Dec 26 '20 at 07:01

2 Answers2

0

Hint

See Linearity of Expectation.

user2661923
  • 35,619
  • 3
  • 17
  • 39
0

Suppose $X$ is random number which is drawn from the full deck, and $Y$ is random number drawn at the second step. We want to find $\mathbb{E}[X+Y] = \mathbb{E}[X] + \mathbb{E}[Y]$. Clearly, $\mathbb{E}[X] = \frac{13 \cdot 14}{2} \cdot \frac{4}{52} = 7$. It remains to find $\mathbb{E}[Y]$. Let's use the law of total expectations $$\mathbb{E}[Y] = \sum_{x=1}^{13} \mathbb{E}[Y|X=x]P(X=x)$$ $$\mathbb{E}[Y|X=x] = 1\cdot \frac{4}{51}+2\cdot \frac{4}{51}+\dots+(x-1)\cdot\frac{4}{51}+x \cdot \frac{3}{51}+(x+1)\cdot\frac{4}{51}+\dots+13\cdot\frac{4}{51} = \frac{(x-1)x}{2}\frac{4}{51}+\frac{3x}{51}+\left( \frac{(13-x)(13-x+1)}{2} +x(13-x)\right)\frac{4}{51} = \frac{364-x}{51}$$ So, $$\mathbb{E}[Y] = \sum_{x=1}^{13} \frac{364-x}{51} \frac{4}{52} = 7$$ Thus, $\mathbb{E}[X+Y] = 7+7 = 14$

D F
  • 1,342
  • Way too complicated. $E(X+Y)=E(X)+E(Y)$ – saulspatz Dec 26 '20 at 08:02
  • @saulspatz Ok, but how do you find $E[Y]$? – D F Dec 26 '20 at 08:03
  • It's the same as $E(X)$. – saulspatz Dec 26 '20 at 08:03
  • @saulspatz why? we draw cards without replacement – D F Dec 26 '20 at 08:04
  • 1
    Draw the first card. Don't look at it yet. Draw the second card. Don't look at it yet. Why is the expected value of the first card different from the expected value of the second? Each is equally likely to be any one of the $52$ cards. – saulspatz Dec 26 '20 at 08:07
  • @saulspatz You draw the second card out of the deck of 51 cards...for me it is not at all obvious that these expectations should be equal. Although in the end they are... – D F Dec 26 '20 at 08:10
  • @saulspatz +1 : nicely phrased – user2661923 Dec 26 '20 at 08:36
  • thank you very much! – Lidor Hacham Dec 26 '20 at 11:55
  • do you have any idea how to calculate the variance? – Lidor Hacham Dec 26 '20 at 11:56
  • @LidorHacham You may show that $X$ and $Y$ are identically distributed i.e. $P(Y = k) = P(X=k)$ (you can show this using the law of total probability). That means that $Var[X]=Var[Y]$. And then $Var(X+Y) = 2Var(X) + 2*cov(X, Y)$. $Cov(X, Y) = E[XY] - (E[X])^2 = E[X(364-X)/51] - (E[X])^2$. So, it remains to find $E[X^2]$ which you may find by definition. – D F Dec 26 '20 at 12:09
  • isn't there a "trick" to solve it by symmetry as the there is in the expectation? – Lidor Hacham Dec 27 '20 at 05:21
  • @LidorHacham You use it in showing that variances of $X$ and $Y$ are equal. But you still need to calculate a covariance to find the variance of sum. And I think there is no other way as calculate it directly by definition. – D F Dec 27 '20 at 06:24
  • ok thank you very much! – Lidor Hacham Dec 27 '20 at 16:22