The Mathematical Game

In the whole solution i've assumed that A will not get the correct answer without guessing the required number of times. Let's assume that A's first guess is ${ p }_{ 1 }$ . Then the chances of ${ p }_{ 1 }$ being less than the answer is $\frac { 100000-{ p }_{ 1 }}{ 100000 }$ and the chances of p being greater than the answer is $\frac { { p }_{ 1 }-1 }{ 100000 }$ . Let these two fractions be ${ x }_{ 1 }$ and ${ x }_{ 2 }$ where ${ x }_{ 1 }<{ x }_{ 2 }$ . We need two get minimum possible value of ${ x }_{ 1 }$ to reduce the number of turns required. By solving we will get, ${ p }_{ 1 }$ =50000. So the first guess should be 50000. Now, there will be 2 cases,

1. Secret Number<50000
2. Secret Number>50000

In case 1, A's second guess would be ${ p }_{ 2 }$ , on solving, gives us 25000. In case 2, A's second guess would be ${ p }_{ 2 }$ + ${ p }_{ 1 }$

We will repeat this n times, till ${ p }_{ n }$ becomes 1 (actually less than 1). We are halving p on every step. By halving 100000 n times, we get a value less than 1. So we get the equation:-

$\frac { 100000 }{ { 2 }^{ n } } <1\\ 100000<{ 2 }^{ n }\\ By\quad putting\quad values\quad we\quad get\\ \min { (n)=\boxed { 17 } }$

The mechanism behind the solving of this problem is similar to a searching method used in programming called Binary Search.

The maximum number of guesses for a certain answer would be if you were to start from the middle ( $50000$ ) and divide by 2 (or multiply by 1.5) each step through according to the result.provided (Greater/Less than). The minimum would be when the secret number is $50000$ and the greatest would be when the secret number is $1$ or $100000$ . The minimum would require only one guess.

For the maximum, the number of guesses required would be $\large{\lceil\log_{2}{100000}\rceil}$ $17$