# Expectation of number of different "blocks" [duplicate]

Suppose we have 52 decks with 26 red and 26 black cards. We shuffle them at a random order. Then we define a "block" as cards with same colors, for example, BRRB has 3 blocks and BRRRBBRRRR has 4 blocks. So how many blocks we would have, in the sense of expectation?

My though is that, we set indicator function $$I_i$$ representing the i-th card has different colors with the card on the left of it. So $$I_i=1$$ if it is different, and is zero if it's same. So the final number of blocks would be equal to $$\sum_{i=1}^{52}I_i$$, and $$E\sum_{i=1}^{52}I_i = \sum_{i=1}^{52} P(i)$$, where P(i) is the probability that i-th card has different color comparing to his left one. But I got stuck at this step cuz I don't know how to calculate the probability of $$i-th$$ term. Is there anybody who can help me?

• yes it is @User1865345! Thanks for showing that. But I am still wondering, why Pi = P(i-1) = ... P(1)=26/51? this is the point where I dont understand... Apr 18, 2023 at 10:33
• @User1865345 That answer was essentially copied from math.stackexchange.com/questions/2763/… where other responses go into more detail Apr 18, 2023 at 10:54
• I see @Henry. Just noticed Whuber's comment. It's unethical but at least this could be used to mark the duplicate. Any further detail in the query has to be specified by OP. Apr 18, 2023 at 10:56
• @User1865345 May I ask which comment? I suspect you might be referring to stats.stackexchange.com/questions/1865/…. My concern is that your remark "it's unethical" appears to apply to that comment, but I hope you intended it instead to apply to the behavior I was commenting about!
– whuber
Apr 21, 2023 at 18:11
• Of course, I was pointing at OP not explicitly citing the original post - that is unethical to me. I would even taken a copied one too if the original source was mentioned. I upvoted your comment too. I hope there is no misunderstanding. @whuber. Apr 21, 2023 at 19:44