$50$% lifetime basketball free throw shooter with $5$% confidence boost if he makes a shot. This is a probability question but instead of a fixed $50$% such as a fair coin flip, the $50$% basketball free throw shooter gets a confidence boost of $5$% (from his $50$% lifetime rating) each time he makes a shot, but then returns to (or stays at) $50$% confidence level if he misses a shot or shots.  For example, for the 1st shot he has a $50$% confidence level he will make it.  If he misses it, he stays at $50$% confidence level for the 2nd shot, however if he makes the first shot, his confidence level climbs to $52.5$% for the 2nd shot.  Similarly if he makes the 2nd shot after already having made the 1st shot, his confidence level climbs another $5$% to $55$%....  If he misses multiple shots in a row, his confidence level "bottoms out" at $50$%, it never goes lower.
So for this question, I am asking what is the probability that the shooter will make exactly $5$ shots out of $10$ sequential attempts, that is, any $5$ out of $10$.
Assume the shooter makes $10$ honest attempts to make all $10$ shots.
If it helps, I calculated the probability of making $5$ out of $10$ if the shooting % was fixed at $50$% to be $10 \choose 5$ * $(\frac 1 2)^5$ * $(\frac 1 2)^5$ = $63/256$ = $24.609$%.  I am wondering how his "confidence boost" will change this "simplified" answer to a similar question.
The answer to my original question may be difficult to compute but if someone really wants a challenge, a variation of it would be as follows:
If the shooting % was fixed at $50$% confidence level, it seems the most likely outcome would be $5$ out of $10$ shots made which would be around $25$% probability.  How much of a confidence boost would be needed so that the most likely made number of shots would change to $6$ out of $10$?  Let x = the confidence boost after each made shot as a multiplier such that, (for example), if x = $4$%, then if he makes the first shot, his confidence level on the second shot would be $1.04 * 50$% = $52$% ....  Obviously x has to be a small percentage like $8$% or less otherwise his confidence level on the last shot if he makes all $9$ previous shots would exceed $100$% since $1.08 ^ 9$ = $1.999$.
Note that $10 \choose 6$ * $.57^6$ * $.43^4$ = $24.623$% so it "edges out" the $24.609$% mentioned previously for exactly $5$ out of $10$ made at a fixed $50$% confidence level so it looks like the average confidence level needs to be around $57$% to make $6$ out of $10$ to be the most likely but this is not the answer to my question it is just a hint/approximation/simplification.
Another interesting "tidbit" is $10 \choose 6$ * $.5874^6$ * $.4126^4$ = $24.99999$% 
My motivation for this question and harder variation is to show that many math problems are simplified to have fixed probabilities such as fair coin flips but when a variable is introduced such as a confidence boost, the problem becomes significantly harder and very few people even attempt to answer it.  I created this question and variation and I am not even a math person but I suspect it will "stump" many "mathheads" as either unsolvable or too difficult and thus not worth it but whoever solves it correctly will likely get respected for it here on this site as well as "pave the way" for similar variable probability questions that may come "down the pike".
Math to me is like a tool such as a screwdriver but the tool should not have the requirement that the screw be perfectly straight with a perfect head cuz that is not real world.  Many screws are slightly bent and partially stripped so a good screwdriver would still work on those so similarly, math should not always (only) be applied to "perfect" scenarios like fair coin flips.
 A: Here's a Python/Sage program that will compute what you want. This is for the version where the increase is linear (e.g., the probability of making shot number 3 is exactly $55\%$, not $50\% * (1.05)^2=55.125\%$), but it wouldn't be too hard to modify it to do the other version instead.
num_shots = 10

bit_lists = [list(ZZ(n).binary().zfill(num_shots)) for n in range(0, 2 ** num_shots)]
makes_misses_lists = [[string == '1' for string in bit_list] for bit_list in bit_lists]

t = var('t')

def monomial(makes_misses_list):
    to_return = 1
    run_length = 0
    for bool in makes_misses_list:
        if bool:
            to_return *= (QQ('1/2') + t * run_length)
            run_length += 1
        else:
            to_return *= (QQ('1/2') - t * run_length)
            run_length = 0
    return to_return

def makes_n_poly(n):
    return sum(monomial(makes_misses_list) for makes_misses_list in makes_misses_lists
        if makes_misses_list.count(True) == n)

Now you can call makes_n_poly(n)(t) to find the probability that you'll make exactly $n$ shots, if making a shot increases that probability by $t$. So
makes_n_poly(5)(t=0)

returns $63/256$, the probability of making $5$ shots if there's no confidence boost.
You're interested in
makes_n_poly(5)(t=0.025) # 0.025 is the same as 2.5%

which evaluates to $0.227617657165527$: that is, the chance of making exactly $5$ shots decreases from about $24.6\%$ to about $22.7\%$ if your chances of making a shot increase by $2.5\%$ after each success.
What about your second question, how much of a confidence boost you need for $6$ shots to be more likely than $5$? You can plot the likelihood of both $5$ and $6$ shots:
P = plot([makes_n_poly(5), makes_n_poly(6)])
show(P, xmin = 0, xmax = QQ(0.1), ymin = 0, ymax = 0.3)

We stop at a confidence boost of $0.1 = 10\%$ because after that the $6$-shot version will involve probabilities larger than $1$.
The result of this plot is:
As you can see, the probability of making $6$ shots never exceeds the probability of making $5$ (though both decrease, in favor of the probability of making more than $6$ shots).
If you plot the probability of making each number of shots from $5$ to $10$ over the range where they all make sense (up to a little over $5\%$), you'll see that none of them ever actually intersect...
Q = plot([makes_n_poly(5), makes_n_poly(6), makes_n_poly(7), makes_n_poly(8), makes_n_poly(9), makes_n_poly(10)])
show(Q, xmin = 0, xmax = QQ('1/18'), ymin = 0, ymax = 0.3)



It's worth noting that statisticians have found very little evidence that athletes' winning streaks are meaningful, in most situations. Free throws are actually one of the few exceptions — it really does seem to be true that making a free throw makes you more likely to make subsequent free throws. But given that people in general see streaks everywhere that probably don't have any real meaning, the size of the effect for free throws is almost certainly a lot smaller than you think it is...
