<s>[INST] Probability that tossing $10$ coins has $4$ consecutive heads I'd like to compute the probability of having a string of four consecutive heads after $10$ coin tosses. 
I tried to solve this problem by considering $10$ blank spaces. To have $4$ consecutive heads, we can consider $7$ blank spaces, where the $7^{\text{th}}$ space is a string consisting of $4$ heads.
Then, there are $7!$ ways to arrange the $7$ spaces, and there are $4!$ internal permutations of the $4$ heads. Also, there are $2^{6}$ choices for the $6$ non-head coins. 
This is all out of $10!$ arrangements and $2^{10}$ choices for the coins. So my probability is 
$$\frac{7! \cdot 4! \cdot 2^6}{10! \cdot 2^{10}} = 0.00208333333,$$
which is wrong. I want to know what's wrong with my reasoning. I can't figure it out. I know there's a similar post, but I don't know what's wrong with my approach.
 [/INST] The total number of possible outcomes is $2^{10}$
Using the counting method for the numerator .......
The number of outcomes with a single exact maximum $HHHH$ is 
$HHHHT----- \Rightarrow 2^5 - 3$
$THHHHT----  \Rightarrow 2^4 - 1$
$-THHHHT---  \Rightarrow 2^4$
$--THHHHT--  \Rightarrow 2^4$
$---THHHHT-  \Rightarrow 2^4$
$----THHHHT  \Rightarrow 2^4 - 1$
$-----THHHH  \Rightarrow 2^5 - 3$
$$P(4H) = \frac{2(2)^5 + 5(2)^5 - 8}{2^{10}} = .132813$$
Where two $HHHH$ are allowed......
$$P(4H+) = \frac{2(2)^5 + 5(2)^5 - 2}{2^{10}} = .138672$$
The number of outcomes with at least one run of at least $4$ consecutive heads is $2^6 + 6(2)^5 - 5$
$HHHH------ \Rightarrow  2^6$
$THHHH----- \Rightarrow  2^5$ 
$-THHHH---- \Rightarrow  2^5$
$--THHHH--- \Rightarrow  2^5$
$---THHHH-- \Rightarrow  2^5$
$----THHHH- \Rightarrow  2^5 - 2$
$-----THHHH \Rightarrow 2^5 - 3$
$$P(\ge4H) = \frac{2^6+6(2)^5-5}{2^{10}} = .245117$$
Does anyone have a more formal/elegant method to check these answers?
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