6

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?

Seraphim
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    What are your thoughts? – Yuval Filmus Aug 09 '14 at 00:57
  • 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 09 '14 at 01:07
  • 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 09 '14 at 01:26

2 Answers2

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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: $$E(X)=1\frac{1}{6}+2\frac{5}{6}\frac{1}{5}+3\frac{5}{6}\frac{4}{5}\frac{1}{4}+...+6\frac{5}{6}\frac{4}{5}\frac{3}{4}\frac{2}{3}\frac{1}{2}1=3.5$$

ingenbryrsig
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7

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 $\frac{1}{6}(1+2+3+4+5+6)$.

André Nicolas
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