TITLE: Probability of 4 consecutive heads in 10 coin tosses
QUESTION [7 upvotes]: I am trying to compute the probability of having 4 (or more) consecutive heads in 10 coin tosses. 
I tried using recursion but it led to a complicated expression so i think i did not quite manage. 
I saw similar questions asked here that were solved with difficult approaches, but this problem looks like it could be solved in a couple of lines so I must be doing something wrong.
If anyone could help me understand it or propose a different approach for solving the problem i would be very grateful.
Have a nice day!

REPLY [7 votes]: In general, this question has been answered here several times, eg here.
For these particular small numbers, you might group the total count by the position of the first successful run
So $N = N_1 + N_2 + \cdots N_7$
$N_1 = 2^6$ [HHHH******]
$N_2 = 2^5$ [THHHH*****]
$N_3 = 2^5$ [*THHHH****]
$N_4 = 2^5$ [**THHHH***]
$N_5 = 2^5$ [***THHHH**]
$N_6 = 2^5 -2$ [****THHHH*]  minus [HHHHTHHHH*]
$N_7 = 2^5 -3$ [*****THHHH]  minus ([HHHHTTHHHH] , [THHHHTHHHH] , [HHHHHTHHHH])
Which gives $N=64 + 6 \times 32 - 5 = 251$
So the probability is $N/2^{10} = 251/1024$