Odds of being correct X times in a row Is there a simple way to know what the chances are of being correct for a given number of opportunities?

To keep this simple: I am either right or wrong with a 50/50 chance. What are the odds that I'll be correct 7 times in a row or 20 or simply X times?

... and can the answer be put in simple terms ;)
Assuming the outcome of "answering" any question is independent of the outcome of answering other questions:

Yes, you'll be right $n$ times in a row with probability: $\left(\dfrac{1}{2}\right)^n$

If $X$ and $Y$ are independent events, then the probability that both $X$ and $Y$ occur is the product of the probabilities of $X$ and $Y$. For each question, you have a $50/50$ chance of answering correctly, which translates to a probability of $\frac 12$: for "randomly guessing" one question, you'll have a probability of $\frac 12$ that you'll be correct.

So answering correctly two questions in a row has probability: $$\frac 12 \times \frac 12 = \left(\frac 12\right)^2 =\frac 14$$

The probability of answering correctly three questions in a row: $$\frac 12 \times \frac 12 \times \frac 12= \left(\frac 12\right)^3= \frac 18$$

$$\vdots$$

The probability of answering correctly $n$ questions in a row: $$\underbrace{\frac 12 \times \frac 12 \times \cdots \frac 12}_{\large n \;\text{factors}}= \left(\frac 12\right)^n$$
You guys are making it too complicated. It goes like this:

double your last...

1st time: 1 in 2   -  (win a single coin toss)
2nd time: 1 in 4   -  (win 2 consecutive coin tosses) . . .etc.
3rd time: 1 in 8
4th time: 1 in 16
5th time: 1 in 32
6th time: 1 in 64 . . . etc . . . 1 in 2 to the nth. If you wanna get complicated with it then if 2 events have the same odds, there's no such thing as "independence" it's all in your head. Just add number of attempts together (nth)&lt;--(this is an exponent). If the odds aren't 50/50 or if there are multiple events multiply the odds together. 1 in 4 should be represented as the fraction 1/4, 2 in 17 as 2/17 and so on... An example would be: "What would the odds of winning both a 1 in 3 bet and a 1 in 6 bet?" 1/3 X 1/6 = 1/18 or just multiply the 3 and 6 to make it easy. If you wanna know the odds of winning one and losing another, you don't do anything different, just flip the losses upside-down (inverse fractions) Make sense? "What are the odds of winning a 1 in 4 bet and a 1 in 6 bet, then losing a 1 in 4 bet?" 1/4 X 1/6 X 4/1(or just 4)= 1/6. I'm not going to bother with probability vs odds syntax since both boil down to the same thing. Just make sure that a preposition is never the word that you end your sentence in.