 Mahesh Godavarti
2

This is essentially dividing the deck of 52 cards into two groups - one group of all reds and one group of all blacks.

So, the probability of having an alternating red-black pattern is the same as the probability of picking one combination out of all possible ways of dividing a deck of 52 cards into two groups of 26.

The total of number of ways of dividing a deck into two equal groups is ^{52} C_{26} = \frac{52!}{(26!) (26!)} .

Therefore, the probability is given by \frac{1}{^{52} C_{26}} = \frac{(26!) (26!)}{52!} .

This is the same answer that Vivekanand got!

Interesting thing is that this answer applies to any particular pattern you can come up with. E.g. All reds followed by all blacks. Or 2 reds, followed by 2 blacks etc. Vivekanand Vellanki
1

There are 2 patterns: RBRBRB... RB and BRBRBR...BR. Both have equal probability. We will compute the probability of getting RBRBRB...RB and multiply by 2.

Probability of the first card being red = 1/2

Probability of the 2nd card being black = 26/51

Let there be 2N cards with N red and B black.

Probability of the first card being red = N/2N = 1/2

Probability of the 2nd card being black = N/(2N - 1)

The desired probability is the product of all of these for N=1 to 26.

Hence, \frac{P}{2}=\prod_{k=1}^{26}\frac{1}{2}.\left(\frac{k}{2k-1}\right)=\frac{1}{2^{26}}\prod_{k=1}^{26}\frac{k}{2k-1}

P=\frac{1}{2^{25}}\prod_{k=1}^{26}\frac{k}{2k-1}