What effect does repetition or momentum have on probability? Let's say you have thirty right answers in a row, what is the probability of having another right answer?
Is there a name for this principal?
Can you give a reference?
 A: In general, either none or a lot, or somewhere in between.

Suppose someone flips a fair coin, but doesn't let you see the flip, then asks you to guess what was flipped. You can be on a serious hot streak, but the probability that your next guess is right will be $0.5,$ regardless. This shows that a "streak" need have no effect.

Let's alter the situation now, and suppose that whenever you are on a streak of $n$ correct answers in a row, the questioner flips a fair coin $n+1$ times, recording the result of each flip, and translating the result into a number (which you are to attempt to guess) as in the following example:

You're on a streak of $3$ correct guesses in a row. The questioner flips the coin $4$ times, getting heads, then tails, then heads, then heads ($HTHH$). The questioner translates $H$ to $1$ and $T$ to $0$ ($1011$), and then adds a point to the front ($.1011$). This is the binary representation of $$\frac1{2^1}+\frac0{2^2}+\frac1{2^3}+\frac1{2^4}=\frac{11}{16},$$ the number you must try to guess.

In general, after a streak of $n$ correct guesses, the correct answer will be of the form $\frac{k}{2^{n+1}},$ where $k$ is an integer with $0\le k\le 2^{n+1}-1,$ so you have a $1$ in $2^{n+1}$ chance of getting the correct answer with your guess. The correct guess will make it twice as hard to get the next guess correct, while an incorrect guess will return the difficulty to a $0.5$ chance.
We can accelerate the difficulty increase even more substantially if we were to roll a fair $m$-sided die $n+1$ times, say, and then use that to develop a fraction's base $m$ expansion. So, a streak can have a significant effect to your detriment. We could also develop some sort of contrived situation where the difficulty of a correct guess decreases with each successive correct guess, or where the difficulty changes with each successive correct guess (but not by much).

Upshot: It depends entirely on the circumstances. Ultimately, what you're wondering about is a conditional probability--namely, how does the likelihood of a right answer change, given that I've had $n$ right answers leading up to this one?
