Representing a real number as the value of a countably infinite game Is it true that for any real number $p$ between 0 and 1, there exist finite or infinite sequences $x_m$ and $y_n$ of positive real numbers, and a finite or infinite matrix of numbers $\varphi_{mn}$ each of which is either 0 or 1, such that 
1) $\sum x_m=1$
2) $\sum y_n=1$
3) $\forall n\;\sum x_m\varphi_{mn}=p$
4) $\forall m\;\sum y_n\varphi_{mn}=p$
 A: Generalization of my previous answer.  Is this right?  $p$ irrational...
For $0\lt p \lt 1/2$ let $x_1(p)=y_1(p)=p$, $\varphi_{1,1}(p) = 1$,
$\varphi_{1,j}(p)=\varphi_{j,1}(p)=0$ for $j>1$,
$x_n(p)=(1-p)x_{n-1}(p/(1-p))$ for $n>1$. And
$\varphi_{n,m}(p) = \varphi_{n-1,m-1}(p/(1-p))$.
For $1/2 \lt p \lt 1$ let $x_n(p)=x_n(1-p)$ and $\varphi_{n,m}(p) = 1-\varphi_{n,m}(1-p)$.
A: I will show the following facts:


*

*If $p\in\mathbb Q$, then your question has positive answer. (see Edit 2)

*For general $p$ your question has positive answer up to $\varepsilon$, for all $\varepsilon$. (see Edit 2)

*For general $p$ your question has positive exact answer if you allow the use of finitely additive probability measures. (see Proposition below). This is particularly motivated by the fact that, for countable games, people usually allow the use of finitely additive probability measures and not only countable additive ones. 


Proposition. Let $p\in[0,1]$. There are finitely additive measures $\mu,\nu$ on the power set of $\mathbb Z$ and a countably infinite matrix $\phi_{mn}$ with entries either $0$ or $1$ such that


*

*$\int_{\mathbb Z}d\mu(m)=1$

*$\int_{\mathbb Z}d\nu(n)=1$

*For all $m\in\mathbb Z$, $\int_{\mathbb Z}\phi_{mn}d\mu(n)=p$

*For all $n\in\mathbb Z$,, $\int_{\mathbb Z}\phi_{mn}d\nu(m)=p$


Proof. Consider the following game. Choose $W\subseteq\mathbb Z$ having upper mean value[1] equal to $p$ and consider the following two-person game: two players choose simultaneously $m,n\in\mathbb Z$ and player 1 wins if $m+n\in\mathbb W$. Player 1's payoff function is exactly the countable infinite matrix $\phi_{mn}=\chi_W(m+n)$. Now consider the mixed extension of this game obtained by allowing the players to play all finitely additive strategies with Player 1's mixed extension payoff 
$$
\int\int \chi(m+n)d\mu(m)d\nu(n)
$$
where $\mu,\nu$ are finitely additive strategies. By Theorem 4.1 in [2] this game has a Nash equilibrium with value exactly $p$ and obtained with translation invariant probability measures $\mu,\nu$. These two measures verify your properties. The first two because they are probability measures; the third and the fourth because they are translation invariant.
[1] The upper mean value of $W$ is the supremum of $\int\chi_W(x)d\lambda(x)$, where $\lambda$ runs over the set of translation invariant finitely additive probability measure on $G$. This set of measures is weak* compact and then the supremum is attained.
[2] Capraro V., Scarsini M., Existence of equilibria in countable games: an algebraic approach, http://arxiv.org/pdf/1203.2301.pdf
Edit. I have realized that in the proof of the proposition it is not necessary to construct the game. Just take $W$ with with upper mean value $p$, construct $\phi_{mn}=\chi_W(m+n)$ and take $\mu=\nu$ be an invariant measure that gives measure $W$ to $p$. I leave that proof since the Original Question ask explicitly for a game theoretical interpretation of the result.
Edit 2. One can get an approximative solution in countable finitely additive measures as follows. Fix $\varepsilon >0$ and take $n$ big enough such that $\mathbb Z/(n\mathbb Z)$ contains set of normalized counting measure equal to $p$ up to $\varepsilon$. Apply the construction above to get the desired result. Notice that if $p\in\mathbb Q$, then this construction can be done with $\varepsilon=0$, giving an exact solution in countably additive measures.
A: Golden Section  The golden section works.  
Write $u = (\sqrt{5}+1)/2$, I don't call it $\varphi$, since that
symbol is already in the problem.  So of course $u^m+u^{m+1}=u^{m+2}$.
Let $p=u^{-2} \approx 0.3819$.  Define:
$$
x_1 = u^{-2}, x_2=u^{-3}, x_3=u^{-4},\dots,x_m=u^{-m-1},\dots
$$
and $y_m=x_m$.  Then compute
$$
\sum_{m=1}^\infty x_m = \sum_{j=2}^\infty u^{-j} =
\frac{u^{-2}}{1-u^{-1}} = \frac{u^{-2}}{u^0-u^{-1}}=
\frac{u^{-2}}{u^{-2}} = 1.
$$
Define:
$$\begin{align*}
&\varphi_{1,1}=1, \qquad\varphi_{1,j}=\varphi_{j,1}=0, j>1.
\cr
&\varphi_{2,2} = 0,\qquad\varphi_{2,j}=\varphi_{j,2}=1, j>2.
\cr
&\varphi_{m,n} = \varphi_{m-2,n-2}, m,n>2.
\end{align*}$$
Now by induction on $n$ we will show
$\sum_{m=1}^\infty x_m\varphi_{m,n} = u^{-2} = p$
for all $n$.
For $n=1$, compute
$$
\sum_{m=1}^\infty x_m \varphi_{m,1} = x_1\cdot 1 +
\sum_{m=2}^\infty x_m \cdot 0 = u^{-2}.
$$
For $n=2$, compute
$$
\sum_{m=1}^\infty x_m \varphi_{m,2} = x_1\cdot 0 + x_2\cdot 0
+ \sum_{m=3}^\infty x_m\cdot 1 =
\sum_{j=4}^\infty u^{-j} = u^{-2}.
$$
For $n>2$ apply the inductive hypothesis:
$$\begin{align*}
\sum_{m=1}^\infty x_m\varphi_{m,n} &=
x_1\cdot 0 + x_2\cdot 1 + \sum_{m=3}^\infty x_m \varphi_{m-2,n-2}
= u^{-3}+u^{-2}\sum_{j=1}^\infty x_j \varphi_{j,n-2}
\cr &=u^{-3}+u^{-2}u^{-2}=u^{-3}+u^{-4}=u^{-2}.
\end{align*}$$
