One of the numbers 1 through 10 is randomly chosen. You are to try to guess the number chosen by asking questions with "yes-no" answers. Compute the expected number of questions you will need to ask if your $i$th question is to be "Is it $i$?", $i = 1, 2, \ldots , 10$.
The solution almost everywhere states that the answer is $5.5$ because the number of questions needed to accurately determine the unknown number would be $x$ only if the unknown number is $x$ itself (where $1\leq x\leq 10$). By this logic the answer is given by:
$$E[X]=1\bigg(\frac1{10}\bigg)+2\bigg(\frac1{10}\bigg)+\ldots+10\bigg(\frac1{10}\bigg)=5.5$$
My problem with this method is that we don't need to ask $10$ questions in any case. If we already asked $9$ questions, then based on the answer to the $9$th question we could deduce whether the unknown number is $9$ or $10$. Therefore, I believe the actual answer should be $5.4$ instead, which we attain by setting the $P[X=9]=\frac{2}{10}$ and $P[X=10]=0$. Is my thought process correct or am I missing something?
By the description of the problem:
Your $ i ^{th}$ question has to be:
"Is it $i$?"
You stop asking questions when the answer is finally yes.
So your last question is, if you have heard No's so far,
"Is it 10?"
even if you already know that you wiil hear
"Yesss."
The probability that this happens is again 1/10.
To be honest, this problem of Ross is not the best possible one. The biggest mistake is that the author forgot to add that
"You stop asking questions when the answer is yes."
With this, the answer is 5.5. It is Sheldon who missed something--not you.