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I have a problem on my hands and I am a little confused on how to solve it.

Say you have $50$ people and each chooses a number from $0$ to $100$. Now you take the average of all the numbers and denote it $X$, then calculate $\frac{2}{3}(X+9)$ and the person who chose the number closest to it wins the game.

The question is, which numbers are such, that if you choose them, then there is some other number that yields a closer number for all other peoples choices/results. The first thing to note is that $0\leq X \leq 100$ and so $6 \leq \frac{2}{3}(X+9)\leq 72.6$. Now I am not sure what to do at this point. It would seem like a bad move to put in anything outside of this bound (except possibly $73$). But I am not sure how to formalize this problem mathematically. I tried making it that you have the function

$$f(x_1,x_2,x_2.,x_{50}) = |2/3(X+9)-x_{50}|$$

and try to minimize this or something. This eventually turns into $|2/3Y-\frac{74}{75}x_{50}+6|$ where

$$0 \leq Y=\frac{x_1+x_2...x_{49}}{50}\leq 98$$

I think what confuses me is that what number you pick influences $X$ hence i tried to divide it up. Any help would be appreciated!

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  • $\begingroup$ Is repetition allowed i.e. can two persons pick the same number or does each person have to choose a unique number? $\endgroup$ – Nilotpal Kanti Sinha Mar 6 at 4:50
  • $\begingroup$ @NilotpalKantiSinha Yes, repetition is allowed $\endgroup$ – Sorfosh Mar 6 at 4:53
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Once you show that nobody should choose a number outside the range $[6,72]$ you can repeat the argument that led to that conclusion. If all the numbers are in that range, $\frac 23(X+9)$ is in the range $[10,54]$ so nobody should choose a number outside that range. This process will converge where $X=\frac 23(X+9), X=18$ so everybody should choose $18$.

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  • $\begingroup$ I understand that, but how do we prove nobody should choose a number outside the range $[6,72]$? I think I got a solution just by inspecting the radical cases of 5,6 and 72,73. IS there a slicker solution? $\endgroup$ – Sorfosh Mar 6 at 5:09
  • $\begingroup$ You say that if you were going to choose a number outside the range, you would be strictly better off to choose $6$ or $72$ because you will be closer to $\frac 23(X+9)$ If you were going to win with the number outside, you will still with with the one on the edge and you have extra chances. The next step presumes that your opponents are as smart as you are. If you know they all love $100$ you should choose $72$ (or anything in the range $[72,99]$ ) and win. $\endgroup$ – Ross Millikan Mar 6 at 5:13

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