Probability of Picking the Correct Choice. My friend and I play this game where we must guess a number between 1 and 10. We are given 3 chances and the person who is thinking of the number tells the other person whether their guess is right or wrong; no other information. I was wondering if there is an optimal strategy to this, such as eliminating 3 numbers and pick from the remaining 7. So my question is:
Given the set $S = \{1,2,3,4,5,6,7,8,9,10\}$, what is the probability of picking the wining number $w$ with $m$ amount of guesses after eliminating $n$ elements from $S$?
 A: We will consider first the case that you do not eliminate any numbers from the set $S$. 
Then, the probability of picking the winning number with the first try is $\frac{1}{10}$. The event of picking the winning number with a second try implies that the winning number was not picked in the first try, thus due to the independence between two tries the probability is $\frac{9}{10}\cdot\frac{1}{9}=\frac{1}{10}$. Then the probability to guess the number with two tries is $\frac{2}{10}$. In general, the probability of picking the right number with $m$ guesses is $\frac{m}{10}$.
Now assume $n\in\{1,2,\dots,9\}$ numbers were randomly eliminated from $S$. Let $A:=\{\text{picking the right number\}}$ and $B:=\{\text{the right number was not eliminated}\}$. Clearly, $\{B, B^{c}\}$ is a partition and we can write
$P(A)=P(A|B)+P(A|B^{c})$. 
Finally, since $P(A|B^{c})=0$, we obtain $P(A)=\frac{P(A\cap B)}{P(B)}=\frac{m}{10-n}\cdot({1-\frac{n}{10}})=\frac{m}{10}.$ Hence, the second strategy does not give you any advantage.
