Cards with Expected Value

So I have been working on a really hard problem: The cards of a standard 52-card deck are dealt out in a circle. What is the expected number of pairs of adjacent cards which are both black? Express your answer as a common fraction.

I have no idea how to solve it. Any full solution would be really helpful!

#Combinatorics

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

For each pair of cards, the probability that they are both black is $\frac{26}{52} * \frac{25}{51} = \frac{25}{102}$ , so since there are 52 pairs in the circle, the expected number of pairs which are both black is $\frac{25}{102}*52 = \frac{650}{51}$ .

But wait, you say. The probabilities for each pair aren't independent of each other, so shouldn't that mess with the expected values? The answer is no, due to one of my favorite theorems, Linearity of Expectation. Take a look at it here, here, or (if you feel like dealing with the usual issues that math wikipedia articles have) here.

Brian Reinhart - 3 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$