A true-false test contains 16 questions in how many different ways can this test be completed

I don't know if these are optimal, but they're slightly better than the other answers. The general strategy is to start by guessing "True" for all questions on the first guess, as you suggested at the end of your question. The following guesses will subdivide the test into intervals, and in the final guess we can take the best of each interval (i.e., keep our guesses or flip them if fewer than half the interval were correct).

As an aside, you didn't specify the distribution of answers, so a malicious test-setter could try to foil us by choosing the answers in such a way as to make the algorithm give the worst possible result. But we can get around this simply by guessing randomly instead of "True"; for simplicity, then, let's just assume that the answers are independent and identically distributed.

Let's use $[a]$ to denote question $a$ and $[a,b]$ to denote questions $a$ through $b$. Also let $C_k$ be the number of correct guesses on the $k$th guess. By symmetry, we can assume that at least half the guesses in any interval are correct.

2 guesses allowed: $9.571$

The resulting information from the first guess is enough to let you get an expected score of $9.571$ on your second guess. This is already better than the four-guess strategy you proposed.

3 guesses allowed: $10.367$

If $C_1 = 16$, we get full points.

If $10 \le C_1 \le 15$, then use the next guess to look at only one question, $[1]$. We're guaranteed at least $C_1$ points; if we're lucky, we'll get $C_1+1$. (Lucky means we were wrong on $[1]$ and can therefore flip it to get one more point.)

On the other hand, if $C_1 = 8$ or $9$, then use the next guess to look at $[1,7]$. By the pigeonhole principle, this guarantees at least $4 + 5 = 9$ points; in fact, if you work out the possibilities, the expected number of points is $9.713$ and $9.885$ for $C_1 = 8$ and $C_1 = 9$ respectively.

4 guesses allowed: $11.086$

If $11 \le C_1 \le 15$, then guess $[1,2]$ next. If $C_2 = 1$, then we can guarantee one more point by guessing $[1]$; otherwise guess $[3]$.

If $C_1 = 10$, guess $[1,6]$ next.

If $C_1 = 9$, guess $[1,8]$ next.

If $C_1 = 8$, guess $[1,5]$ next.

I've skipped the details for these last three cases, but I can add them if you like. In short, a rule of thumb for the subsequent guesses would be to divide a large ambiguous interval in half when possible ("ambiguous" meaning that around half the answers were correct, so we don't have as much useful information as we would like and can't gain much from flipping our guesses).

Postscript: as far as I can tell, it is always beneficial to guess the whole set of questions first, as opposed to splitting them up into thirds, as in Henning Makholm / JMoravitz's answer. That being said, that strategy still performs well and has the benefit of being straightforward.

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