In the Central station there is a long train composed of two sorts of cars, green cars and blue cars.
Thus, the adjacent cars of the trains form blocks of the same color. As always, Alexander calculates beforehand the expected length of a block and obtains almost 2 (since the train is long but finite)
So, Alexander boards the train over night being convinced that the block his car belongs has a normal average length of 2 cars.
Later in the morning, by summing the average number of the backward cars in his block and the average number of forward cars and adding 1 for its own car, Alexander obtains a nice average length of 1+1+1 = 3 cars for his block.
By simply boarding one car in some block, Alexander increased the average length of his block from 2 to 3.
Why the Alexander's block is longer than a normal block ?
NOTE: Originally I was trying to evaluate the length of colored sequences in the Zuma game. I got, by two approaches, say "global" and "local", two different figures. Nor today I do not know exactly which is the right one. Since this is not the only context that provides different solutions depending on approach, I brought Alexander in.
This is a very basic probability question. I do not think it should be downvoted, because it can be interesting for the beginner.
As was pointed out in the comments, Alexander is more likely to see a longer block because a longer block is more likely to see Alexander. It's interesting to derive the answer analytically, assuming an infinite train. The expected number of blocks of size $n+1$ or longer equals that of blocks of size exactly $n$ (because the $(n+1)^{th}$ wagon has equal chances of terminating the block or continuing it). Then, the probability that a block chosen uniformly at random has size exactly $1$ is $1/2$, the probability that it has size exactly $2$ is $1/2 \cdot 1/2$, and in general the probability that a block has size exactly $n$ equals $2^{-n}$.
Then, the probability $p(n)$ that a specific wagon belongs to a block of length $n$ is proportional to $n\cdot 2^{-n}$ - because the number of wagons in all blocks of size $n$ is proportional to the frequency of such blocks times the number of wagons in such blocks. The expected number of wagons Alexander will see in his block is then:
$\left(\sum_{i=1}^\infty n \cdot p(n)\right)/ \left(\sum_{i=1}^\infty p(n)\right) = \left(\sum_{i=1}^\infty n \cdot (n2^{-n})\right)/ \left(\sum_{i=1}^\infty n2^{-n}\right)=3$
(can you see why it adds up to $3$?)