# 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?

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What are your thoughts? – Yuval Filmus Aug 9 '14 at 0:57
I was thinking simplistic, i thought that the expected value of this would be the sum of the probability times the value itself, but that didn't even make sense when i calculated it. Since the value was greater than 6 which is impossible since someone will obviously get it in 6 guesses. What's wrong with my thinking? – Seraphim Aug 9 '14 at 1:07
The expected number of guesses also depends on the strategy of the guesser: random guesses would result in an expected $6$ rounds rather than $3.5$ rounds with the optimal strategy of random guessing without repetition. – Yuval Filmus Aug 9 '14 at 1:26

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: $$E(X)=1\frac{1}{6}+2\frac{5}{6}\frac{1}{5}+3\frac{5}{6}\frac{4}{5}\frac{1}{4}+...+6\frac{5}{6}\frac{4}{5}\frac{3}{4}\frac{2}{3}\frac{1}{2}1=3.5$$
We can without loss of generality assume that I will use a fixed sequence of guesses, obviously non-repeating. If you choose the number to be guessed at random, then it is equally likely that I will guess it on first try, second try, and so on. Thus the expected number of trials is $\frac{1}{6}(1+2+3+4+5+6)$.