What is the probability that in 7 drawn numbers in lottery (out of 39) there are 4 consecutives?
I have been chasing my tail here. So, we have $39-3=36$ sequences of 4 consecutive numbers. If we pick first or last one of those sequences, remaining 3 numbers we can choose in $\binom {39-4-1 } 3=11968$ ways. On the other hand, if we pick any other sequence (34 of them),we would have $34\cdot {39-4-2 \choose 3}=185504$ possible ways.
so, $p(\text{4 consecutive numbers})=\frac{11968+185504}{{39 \choose 7}}=0.01284$ right??
I perceive a more complex logic .
To take the more difficult original problem of exactly $4$ consecutive "chosens" in a $39/7$ lottery,
denoting "non-chosens" by bullets, and "chosens" by circles, there can be $3$ "chosens" patterns:
$\bullet \bullet \circ\circ\circ\circ\bullet\circ\circ\circ\bullet\;,\;\bullet\circ\circ\circ\circ\bullet\bullet\circ\circ\bullet\circ\bullet\; and\;\bullet\circ\circ\circ\circ\bullet\bullet\circ \bullet\circ\bullet\circ\bullet\bullet $
The number of bullets shown is only illustrative, there will be a total of $32$ of them with $33$ gaps between them including the two ends for placing "chosens"
$Thus\; number\;of\;favorable\;ways\;of\;placing\;chosens\;=\binom{33}22!+\binom{33}33! + \binom{33}3\frac{4!}{3!}$
Divide by $\binom{39}7$ to get the probability.
You should now be able to easily work out the simpler problem posed in your comments.