Probability and Expected Value of a guessing game Let's say I think of a number between one and six. I will tell you to guess the number and tell you when it's wrong until you guess the correct number. What is the expected number of guesses before the correct number?

The probability of the guesser guessing the on the first guess correctly is $\frac{1}{6}$. Given failure on the first attempt (probability $\frac{5}{6}$), the probability of success on the second attempt is $\frac{1}{5}$. Similarly, given failure on the first AND second attempts, the probability of success on the third attempt is $\frac{1}{4}$. This continues until the sixth attempt, when the probability of success is 1. Thus, the expected value of the random variable $X$, where $X$ is the number of guesses before the correct number is: $$E(X)=1\frac{1}{6}+2\frac{5}{6}\frac{1}{5}+3\frac{5}{6}\frac{4}{5}\frac{1}{4}+...+6\frac{5}{6}\frac{4}{5}\frac{3}{4}\frac{2}{3}\frac{1}{2}1=3.5$$