Probability of picking 3 numbers in sequence, with k random picks I am struggling a bit with this problem, I think I am somehow close but I miss something.
Let's assume we have numbers from $1$ to $30$, we pick $5$ random numbers (with repetition) and we want to know the probability to pick at least $3$ numbers in sequence.
Denominator:
The total number of combinations without repetition should be $$30!/(5!*(30-5)!)$$
and with repetition: ( 5 5 6 7 8 ) is a valid sequence, this is what i am actually interested in
$$(30+5-1)!/(5!*(30-1)!)$$
Numerator:
I should get the ways to pick $5$ numbers where at least $3$ are in a sequence.
There are $28$ ways to pick $3$ numbers in a row, if we just would pick $3$, but with $5$ I am having the problem.
My approach is $28$ $+$ the way to pick the other $2$ numbers, that is $C(30+2-1,2)$
The flaw here I think is that I think I am counting twice some combinations.
How can I get the number of combination of picking $3$ consecutive numbers with $5$ picks?
Edit: To clarify
repetition is allowed, so $5\; 5\; 6\; 7\; 8$ is a valid pick.
Order is not important, so $5$ $5$ $6$ $7$ $8$ is the same as $6$ $7$ $8$ $5$ $5$. 
 A: This answer looks at the situation where we draw with replacement and with order.
We draw $(x_1,\ldots,x_5)\in\{1,\ldots,30\}^5$ and are interested in 
$$
N = \#\left\{(x_1,\ldots,x_5)\in\{1,\ldots,30\}^5 \mid \exists i\in\{1,2,3\}\quad\text{st}\quad x_{i+2}=x_{i+1}+1 = x_{i} + 2\right\}
$$
As you said we have 28 ways to pick 3 consecutive numbers from $\{1,\ldots,30\}$. 
Given that we have 3 consecutive numbers at a specific place we are free to choose the other two numbers as we please. So for each $i\in\{1,2,3\}$ we have $28\cdot30^2$ possible combinations that have 3 consecutive numbers on $x_i,x_{i+1},x_{i+2}$. 
Now note that the total amount of combinations is less than $3\cdot28\cdot30^2$. For example $(1,2,3,4,30)$ is counted twice, since it has three consecutive numbers for $i=1$ and $i=2$. In general: we count all sequences that contain 4 consective numbers twice and all sequences that contain 5 consecutive numbers thrice. We conclude
$$
N = 3\cdot28\cdot30^2 - \#\{\text{4 consecutive}\} - 2\#\{\text{5 consecutive}\}
$$
By similar calculations we get 
\begin{align}
\#\{\text{4 consecutive}\} &= 2\cdot27\cdot30 - \#\{\text{5 consecutive}\} \\
\#\{\text{5 consecutive}\} &= 26
\end{align}
Therefore we get that
$$
N = 3\cdot28\cdot30^2 - 2\cdot27\cdot30 + 26.
$$
A: This is combination with repetition, so total number of possibilities is $$ \frac{(N + r - 1)!}{r!(N-1)!} = \frac{(34)!}{5!(29)!} = 278256$$
There are 28 sequences of 3 in 30, so to pick 3 consecutive out of 5, I have 3 numbers in the sequence, plus 2 that needs to be picked out of 29 (so as to not pick the next one in the sequence) with repetition where order is not important (combination with repetition) i.e. $$ 28 * \frac{(30)!}{2!(28)!} = 28 * 435 = 12180  $$
Similarly, the number of ways of picking a sequence of 4 numbers is then $$ 27 * \frac{(29)!}{1!(28)!} = 27 * 29 = 812 $$
And finally, the number of ways of picking a sequence of 5 numbers is $$ 26 * \frac{(28)!}{0!(28)!} = 26 * 1 = 26 $$
The sets containing sequences of 3,4 and 5 consecutive numbers do not overlap, they are all different numbers, so the probability of selecting a number with  at least a sequence of 3 consecutive numbers out of 5 from a set of 30 with repetition where order is not important is then $$ \frac{ 12180 + 812 + 26}{ 278256} = 0.0468 $$
A: Let $E_{i,k}$ denote the event that the consecutive numbers $i,\dots,i+k-1$
are picked.
Then you are interested in $P\left(\bigcup_{i=1}^{28}E_{i,3}\right)$
and applying inclusion/exclusion we find:
$$P\left(\bigcup_{i=1}^{28}E_{i,3}\right)=\sum_{1\leq i\leq28}P\left(E_{i,3}\right)-\sum_{1\leq i<j\leq28}P\left(E_{i,3}\cap E_{j,3}\right)+\sum_{1\leq i<j<k\leq28}P\left(E_{i,3}\cap E_{j,3}\cap E_{k,3}\right)$$
For the terms here that do not equal $0$ we find:
$P\left(E_{i,3}\right)=P\left(E_{1,3}\right)$ for $i=1,\dots,28$
$P\left(E_{i,3}\cap E_{i+1,3}\right)=P\left(E_{1,4}\right)$ for $i=1,\dots,27$
$P\left(E_{i,3}\cap E_{i+2,3}\right)=P\left(E_{i,3}\cap E_{i+1,3}\cap E_{i+2,3}\right)=P\left(E_{1,5}\right)$
for $i=1,\dots,26$
So it's enough to calculate $P\left(E_{1,k}\right)$ for $k=3,4,5$ and to do some counting.
