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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.

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

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