# strategy for game with randomly chosen numbers

Professor $$X$$ chooses a number $$x$$ between $$a$$ and $$b$$ ($$a < b \in \mathbb{Z}$$) and asks two other students $$A_1$$ and $$A_2$$ to pick numbers between $$a$$ and $$b$$. Whichever is closer to $$x$$ gets $$90\%$$ and the other $$80\%$$, if $$a$$ and $$b$$ are distinct. Otherwise, they both get $$85\%$$. If $$X$$'s and $$A_1$$'s numbers are chosen at random and $$A_1$$ announces his number out loud, describe a strategy that leads $$A_2$$ to the highest possible mark.

I'm not sure how $$A_2$$ can get the highest possible mark. A suitable sample space for this problem would be $$\Omega = \{(a_1,a_2,a_3) : a_i\in \{a,\cdots, b\}\,\forall i\},$$ where $$a_1$$ is the Professor's choice, $$a_2$$ is $$A_1$$'s choice, and $$a_3$$ is $$A_2$$'s choice. There is a $$\frac{b-a}{2(b-a+1)}$$ probability that $$A_1$$ chooses a number below the Professor's, a $$\frac{1}{b-a+1}$$ probability that $$A_1$$ chooses a number equal to the Professors, and a $$\frac{b-a}{2(b-a+1)}$$ probability that $$A_1$$ chooses a number greater than the Professor's, but I'm not sure if this is useful. One can calculate these probabilities using the law of total probability and, defining $$A =\{(a_1,a_2,a_3) : a_i \in \{a,\cdots, b\}, a_2 < a_1\},$$ by defining $$B_i := \{(i,a_1,a_2) : a_j \in \{a,\cdots, b\}, 1\leq j\leq 2\}$$ for $$a\leq i\leq b$$, and then computing $$\sum_{i=1}^{b-a+1} P(A | B_i)P(B_i).$$ However, I'm not sure if this is useful for solving the problem. I think $$A_2$$ should choose either $$\lfloor \dfrac{a+a_2}{b-a+1}\rfloor$$ or $$\lfloor \dfrac{b+a_2}{b-a+1}\rfloor$$ (i.e. the number in between the lowest possible choice and $$A_1$$'s choice or in between the highest possible choice and $$A_1$$'s choice), though this is just from my intuition.

• Shouldn't it be whether the two guesses are distinct, not whether $a$ and $b$ are distinct? – Ross Millikan Oct 4 '20 at 18:28

Given $$A_1$$s guess, $$A_2$$ only has three reasonable strategies. He should guess the number just below or above $$A_1$$s or the same number as $$A_1$$ (if that is allowed). If he guesses a number two away from $$A_1$$ instead of next to, he gets the same result as the next number except the next number will be a tie, so it is never better and can be worse.
$$A_2$$ wants to win as many as possible, so should choose the side of $$A_1$$s guess that has more numbers in it. Finally, if $$A_1$$ picked the average of $$a$$ and $$b, A_2$$ should take the same number if that is allowed. He will get $$90$$ less than half the time if he chooses a different number but can be sure of $$85$$ by picking the same one.