a game on numbers

Question

Hello, here is a little two-players game.

Players A and B choose three numbers : a, b and c for A, a', b' and c' for B. The values are numbers between 0 and 1, their sum is 1, and they are ordered: $a \geq b \geq c$ and $a' \geq b' \geq c'$.

Then, the players A and B compare their choices : $a$ vs $a'$, $b$ vs $b'$ and $c$ vs $c'$. If a player has 2 values bigger than the other, he wins (otherwise it's a tie).

I would like to study whether there is a good strategy in this game but I don't know how to start. Do you have an idea on the general way of studying this kind of game? Any reference of book/article is welcomed :)

Answer 1

It seems to me that this game is illuminated a little if one considers a huge generalization. Take a probability space $(X,\mu)$ and define on it a measurable directed graph, which we can think of as a measurable subset $A$ of $X\times X.$ The first player chooses a point $x$ and the second player chooses a point $y$. The first player wins if and only if $(x,y)\in A$. (It might also be nice to add the conditions that the measure is absolutely continuous and that if $x\ne y$ then $(x,y)\in A$ if and only if $(y,x)\notin A$, but I'm not sure that affects the discussion too much.)

Now consider a randomized strategy for the first player. This consists in choosing a different probability measure on $X$, which for convenience I'll assume is a density $f$ with respect to $\mu$ (though I may have to drop that assumption later). If the second player knows $f$, then the second player will choose $y$ such that $\int f(x)\mathbb{1}_A(x,y)d\mu(x)$ is minimized.

This produces a problem that's a continuous version of the following problem: given an $n\times n$ matrix $A$, find a non-negative vector $v$ with coordinates summing to 1 such that the smallest coordinate of $Av$ is as large as possible. If the rows of $A$ are $a_1,\dots,a_n$ then this is asking us to maximize the minimum of the inner products $\langle a_i,v\rangle$ subject to the coordinates of $v$ being positive and adding up to 1, which is similar in flavour to a linear programming problem. (Can it be turned into one? I don't see it immediately. The difference is that the objective function is a minimum of linear functions rather than a linear function. So it is a convex programming problem but with the convex function of a relatively simple form.)

I imagine that all this is either incorrect or very standard game theory. Apologies in advance if there's a Wikipedia article that says similar things more clearly and authoritatively.

Answer 2

https://arxiv.org/abs/1708.07916

This arxiv paper might be helpful. The abstract is reproduced below:

This paper explores the Nash equilibria of a variant of the Colonel Blotto game, which we call the Asymmetric Colonel Blotto game. In the Colonel Blotto game, two players simultaneously distribute forces across n battlefields. Within each battlefield, the player that allocates the higher level of force wins. The payoff of the game is the proportion of wins on the individual battlefields. In the asymmetric version, the levels of force distributed to the battlefields must be nondecreasing. In this paper, we find a family of Nash equilibria for the case with three battlefields and equal levels of force and prove the uniqueness of the marginal distributions. We also find the unique equilibrium payoff for all possible levels of force in the case with two battlefields, and obtain partial results for the unique equilibrium payoff for asymmetric levels of force in the case with three battlefields.

Theorem 3.3 on page 5 characterized all Nash Equilibrium Strategies for the game described in the question.

Answer 3

This is a rather degenerate “game”, there being only one round with no interaction between the moves of the players. But whatever.

There is no strategy that would guarantee winning or tying against any other strategy: a strategy $(a,b,c)$ can be beaten by $(a-2\varepsilon,b+\varepsilon,c+\varepsilon)$ [or $(a+\varepsilon,b-2\varepsilon,c+\varepsilon)$ if $a=b>c$, the case $a=b=c=1/3$ being left to the reader] for small enough positive $\varepsilon$.

Hence the next best result you can achieve is a strategy that would give a high probability (say, $\ge1/2$) of winning against a randomly chosen counterstrategy. Whether such a strategy exists and what it looks like will likely depend on the probability distribution on the counterstrategies, so you’d have to specify that first.

Answer 4

I think the result with proper play is always a draw as this appears to be a variant of "who can name the biggest number". I want to pick three numbers two of them are as big a possible so the obvious choice is a=1/X, b=1/2-1/X and c=1/2 where X is the biggest number I can think of. If both players do this then they tie on c and each wins one of a and b for an overall tie. And this strategy beats any other strategy that doesn't have c=1/2.