Expected number of Same color run in standard 52 deck I encountered this question What is the expected number of runs of same color in a standard deck of cards?, and I understand the answer approach by @George. However, I'm unsure why my answer leads to a different answer. I cannot find my logic error, and I'm looking for some explanations or suggestions.
My approach:
There are $26$ Black cards, so we assume there are $26+1=27$ slots for the Red cards to be. For each slot i, the probability that at least one Red card is in it is $1- (26/27)^{26}$. If the slot has at least one Red card, the number of same color runs increases by 2; this holds for all $2\leq i\leq 27.$ For the first slot, if there is at least one Red, the number of run is increased by $1$, and $0$ otherwise. A illustration is below:
___ B ___ B ___ B...___ B___B___
Then I get the answer to be $(2\cdot 26+1)\cdot(1-(26/27)^{26})=33.133$ 
What caused the error?
 A: Let's try with $2$ red and $2$ black instead.
The linked question would suggest $3$ and this is indeed correct: the six equally likely possibilities are  RRBB, BBRR, RBBR, BRRB, RBRB, BRBR.
Your approach would suggest $(2\times 2+1) (1-(2/3)^{2 })=\frac{25}{9}$ which has the wrong denominator, among other issues.  Here are a couple more:

*

*you should not be multiplying your $+1$ by the second term as you have the initial black run, and in stead it should be outside the brackets

*the probability any of your gaps has at least one red is $\frac12$ as David K says (a particular black except the last is equally likely to be followed by a black or a red in a random deal, similarly the first card is equally likely to be black or red, and similarly the last card)

so the answer in this shorter example using your approach should be $(2\times 2) \times \frac12 +1 = 3$ which we already know is correct.
In the original question, using your method correctly, you should have $(2\times 26) \times \frac12 +1 = 27$, as in the original link.
