Probability of a 75% freethrow shooter making at least 5 shots in a row out of 10. What is the probability that a 75% free throw shooter, given the assumptions listed below, can make at least $5$ in a row of $10$ shots?  So in effect he must make $5$, $6$, $7$, $8$, $9$, or all $10$ in a row.  He only gets $10$ shots total.
The main difference of this question vs. others I've seen on this site is that mine asks for "in a row" which changes the math needed to solve it.  I can solve this using placeholders using a divide and conquer method but I was told a recurrence relation can be used so I would like to see that solution method please.
Some assumptions:
A) The 75% free throw (f.t.) percentage remains constant during the 10 shots.  So things like fatigue, distractions... are not of any concern for this question. The shooter is a professional and the 75% is his average over a long period of time as a professional so it can be considered a very accurate prediction of his future f.t. performance.
B) The shooter makes an honest attempt to make as many shots as he can (in other words, he doesn't intentionally miss any shots).
C) When I say at least $5$ in a row, I am talking about the longest streak only of the made shots so for example, using $M$ for make and $x$ for miss, this is NOT considered $5$ in a row $MMMxxxMMxx$ because the longest streak of makes is only $3$ in a row. However, $xxMMMMMxxx$ is $5$ in a row.  Note that $MMMMxMMMMx$, even though it is $8$ out of $10$ made shots, is only considered $4$ in a row for this question (kinda like in bowling, $2$ strikes in a row, then an open frame, then a single strike is not considered a "turkey" ($3$ strikes in a row)).
D) Note that when I ask for the "longest" streak of makes, it could actually be the only streak (such as all $10$ in a row) or it could be the longer streak (if there are only $2$ streaks of makes) so I wanted to clarify that.  Also cases such as $MMxMMxMMxx$ has no longest make streak (they are all of length $2$), but 2 is considered the "longest" in this case.
A bonus question related to this main question is which is the shooter more likely to get, $8$ in a row or all $10$ in a row?
Thank you.
 A: We can get an explicit expression by considering the cases where the consecutive successful free throws occur between two immediately flanking failures, or they do not (as in the case where the shooter begins or ends the streak at the beginning or end of the 10 trials).
In the first case, we see that we must calculate the sum $$\sum_{m=0}^3 (4-m) p^{5+m} (1-p)^2 = p^5(4-5p+p^5).$$  In the second case, we must calculate $$p^{10} + \sum_{m=0}^4 2p^{5+m}(1-p) = p^5 (2-p^5).$$  Thus our desired probability for a general $p \in [0,1]$ is given by $$p^5(4-5p+p^5)+p^5(2-p^5) = p^5(6-5p),$$ which for $p = 0.75$ is $2187/4096$.

In case one is interested, the following Mathematica code calculates the above polynomial via direct enumeration:
Total[Times @@ # & /@ Cases[Tuples[{p, 1-p}, 10],{___,p,p,p,p,p,___}]] // Simplify

And the subsequent command
% /. p -> 3/4

gives the probability for $p = 0.75$.  Note that the speed of this command scales poorly with the number of trials, since for each additional trial there is a doubling of the length of the list generated by Tuples[].  However, generalizing the above sums is easy.
A: The probability of the shooter getting at least five in a row is the sum of the probabilities of getting 5, 6, 7, 8, 9, or 10 in a row. These probabilities assume that any shot that is not part of the make sequence is missed.  This is not what the original author asked so please be aware of that. 
\begin{align*}
p_5 &= 6* (0.75)^5 * (0.25)^5 \\
p_6 &= 5* (0.75)^6 * (0.25)^4\\
p_7 &= 4* (0.75)^7 * (0.25)^3\\
p_8 &= 3* (0.75)^8 * (0.25)^2\\
p_9 &= 2* (0.75)^9 * 0.25\\
p_{10} &= (0.75)^{10}
\end{align*}
$P= p_5+p_6+p_7+p_8+p_9+p_{10} = 0.126$ or 12.6%.
Bonus: $p_8= 3*0.75^8 * 0.25^2= 0.018$ and
$p_{10}= 0.75^{10} = 0.056$.
So the probability of shooting 10 straight is higher
A: $P(5)=$Prob(hits 5 but misses the previous and following shots)
$=0.75^5*0.25+4*0.25*0.75^5*0.25+0.25*0.75^5\\=(\frac34)^5(\frac14+\frac4{16}+\frac14)=729/4096$
$P(6)=(\frac34)^6(\frac14+\frac3{16}+\frac14)$
and so on.
A: Let $p$ be the probability of scoring and let $P_{m,n}$ be the probability of making at least $m$ in a row from $n$ shots. Conditioning on the first throw,
$$
P_{m,n} = (1-p)P_{m,n-1} + p[(1-p)P_{m,n-2}+p(1-p)P_{m,n-3}+\ldots+p^{m-2}(1-p)P_{m,n-m}+p^{m-1}]
$$
with initial conditions
$$
P_{m,m} = p^m \qquad P_{m,m+1} = p^m +(1-p)p^{m}
$$
Hence when $p=0.75$, $m=5$, and $n=10$,
$$
P_{5,5} \approx 23.7\% \qquad 
P_{5,6} \approx 29.7\% \qquad 
P_{5,7} \approx 35.6\% \qquad 
P_{5,8} \approx 41.5\% \qquad 
P_{5,9} \approx 47.5\% \qquad 
P_{5,10} \approx 53.4\%
$$
For bonus: 
$$
P_{10,10} \approx 5.6\% \\
P_{8,8} \approx 10.0\% \qquad P_{8,9} \approx 12.5\% \qquad P_{8,10} \approx 15.0\% 
$$
