The game is as follows. Alice secretly selects three real numbers $a_{1},a_{2},a_3$ such that $1\geq a_1\geq a_2\geq a_3\geq 0$ and $a_1+a_2+a_3=1$. Bob secretly selects three real numbers $b_{1},b_{2},b_{3}$ such that $1\geq b_1\geq b_2\geq b_3\geq 0$ and $b_1+b_2+b_3=1$ They then compare their numbers. Alice gets one point if $a_i\geq b_i$, and Bob gets one point if $a_i\leq b_i$. The person with the most points wins the game.

For example, If Alice select $\frac{1}{2},\frac{3}{8},\frac{1}{8}$ and Bob selects $\frac{3}{5},\frac{3}{10},\frac{1}{10}$, then Alice wins because $a_1=\frac{1}{2}\leq\frac{3}{5}=b_1$, giving Bob one point, but $a_2=\frac{3}{8}\geq\frac{3}{10}=b_2$ and $a_3=\frac{1}{8}\geq\frac{1}{10}=b_3$, giving Alice two points.

What is the best strategy for this game?

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    $\begingroup$ no strategy wins against all the others, playing random is probably the best. This is similar to rock-paper-scissors $\endgroup$ – Denis Dec 2 '14 at 12:39
  • $\begingroup$ @Denis: "Random" doesn't mean anything much. You need to specify the distribution you are using. $\endgroup$ – user21820 Dec 2 '14 at 12:42
  • $\begingroup$ @DanielV the numbers have to be ordered, only the first one is correct $\endgroup$ – Denis Dec 2 '14 at 12:49
  • $\begingroup$ Well...the worst strategy for the game seems to be picking $1, 0, 0$...don't know if that helps... $\endgroup$ – DanielV Dec 2 '14 at 13:16
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    $\begingroup$ @Macavity: Why not? That is the hard part. $\endgroup$ – user21820 Dec 2 '14 at 14:29

The best strategy is to choose $a_1=a_2=a_3=\frac{1}{3}$, and the probability to win is of $\frac{2}{3}$.

The first thing I did was a program that calculated the probabilities of a given set $(a_1,a_2,a_3)$ to win a random $(b_1,b_2,b_3)$. In order to do this, I generated 10000 random sets for Bob using this and checked how many of them won Alice's given set.

I checked the probabilities for the 210 sets $(a_1, a_2, a_3)$ of the form $a_i = 0.05·n, \space\space n\in \mathbb{N}$ and found that the highest ones were close to the combination $a_1=a_2=a_3=\frac{1}{3}$. You can check the full table of probabilities I got here.

In order to find out why this is the optimal distribution, I came up with the following:

All the possible terns $(a_1, a_2, a_3)$ that satisfy $1≥a_1≥a_2≥a_3≥0$ can be represented as a triangle in the plane (as pointed out in the comments, we have only two independent parameters): pic1

If we now apply the restrictions $a_2\geq a_3$ and $a_2\leq a_1=1-a_2-a_3$ we obtain: enter image description here Let's see in which cases Bob will win if he plays some $(b_1, b_2, b_3)$. In order to win, he must make two of the following inequalities true: $$ 1)\space\space b_2>a_2\\ 2)\space\space b_3>a_3\\ 3)\space\space b_1>a_1\Leftrightarrow b_2 + b_3 < a_2 + a_3 $$ Let's see when this conditions are fullfilled: pic3

If you try to maximize the sum of these three areas, you find that the maximum sum is for $b_2=b_3=\frac{1}{3}$.

This is not exactly right, since not all the points in the triangle have the same probability, but it sort of points out to the right answer.


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