What is the probability that when 5 cards are picked out of a standard, 52-card deck, they alternate in color I want to find the probability that when 5 random playing cards are picked out of the standard, 52-card deck, the cards alternate in color (Black Red Black Red Black or Red Black Red Black Red). I tried to do $1\times\frac{26}{51}\times\frac{25}{50}\times\frac{25}{49}\times\frac{24}{48}$, but it seems to large, because the order matters. How would I do this problem correctly?
 A: Let's imagine, we mark our cards with a number from $1$ to $52$, such that Red cards get numbers from $1$ to $26$ and Black cards numbers from $27$ to $52$.
Then in our probabilistic space $(\Omega, \mathcal F, \mathbb P)$, where:
$\Omega = \{ (x_1,x_2,x_3,x_4,x_5) : x_i \in \{1,...,52\}, x_i \neq x_j$ for $i \neq j$, $i,j \in \{1,...,5\} \}$
$\mathcal F = \mathcal P(\Omega)$
$\mathbb P $ is such that for any event $E \in \mathcal F$ we have $\mathbb P(E) = \frac{|E|}{|\Omega|}$
We can count all $5-$sequences in $\Omega$, clearly we have $52$ choices for first place, $51$ for second and so on... so $|\Omega| = 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 $
We are interested in probability of event $A$ - colors of our chosen cards alternate. As you noticed, either first is black or red, that means either cards at places $1,3,5$ are chosen from cards marked with numbers $1-26$ and second and fourth card is marked with number $27-52$ or otherwise. Those situations are clearly symmetric (there is easy bijection between them, just paint every black card red, and every red card black), so we'll deal with first situation and multiply our result by $2$.
So, let $A_1$ be the first case event, then $|A| = 2|A_1|$.
$A_1 = \{ (x_1,x_2,x_3,x_4,x_5): x_1,x_3,x_5 \in \{1,...,26\}, x_2,x_4 \in \{27,...,52\} , x_i \neq x_j$ for $ i \neq j \}$.
And we want to count all sequences in $A_1$. There are $26$ ways for $x_1$, then $26$ ways for $x_2$ since $x_2$ is being picked from other $26$ card-pack than $x_1$. With $x_3,x_4$ it is similar, however now they have $25$ possible ways to be picked, and lastly $24$ ways to pick $x_5$.
So we have $|A| = 2\cdot 26^2 \cdot 25^2 \cdot 24$
And our final result : $\mathbb P(A) \frac{|A|}{|\Omega|} = \frac{2 \cdot 26^2 \cdot 25^2 \cdot 24}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} = \frac{2 \cdot 26 \cdot 25}{2 \cdot 51 \cdot 2 \cdot 49 \cdot 2} = \frac{26 \cdot 25}{4\cdot 51 \cdot 49} = \frac{13 \cdot 5^2}{2\cdot 3 \cdot 17 \cdot 7^2}$
A: Your answer does correctly take order into account and is very close to $\frac 1{16}$
If you were not taking order into account you would be looking for the probability of getting 3 of one colour and 2 of the other, which is much larger
$$  P = 2 \frac { \binom {26}3 \binom {26}2  }{ \binom {52}5   }  \approx 0.65 $$
more than 10 times the probability that you calculated.
A: Your derivation considers each draw in sequence. Here is a non-sequential derivation.
There are $\binom{52}55!=52×51×50×49×48$ ways to draw an ordered hand of $5$ from the deck. Of these, the hands that alternate colours number as follows:


*

*$2$ ways to select the majority colour (as the colours alternate, there must be $3$ of one colour and $2$ of the other)

*$\binom{26}33!=26×25×24$ ways to select the cards of the majority colour

*$\binom{26}22!=26×25$ ways to select the cards of the minority colour


Thus the desired probability is $\frac{2×26×25×24×26×25}{52×51×50×49×48}$, which matches your answer.
