By introducing a specified definition of the equilibrium
value of three-person two-choice games, a multistage 
three-person game with arbitration is formulated and solved.  
Random offers $\{X_i\}_{i=1}^n$ are presented 
one-by-one sequentially, and as each offer $X_i$
comes up, each player chooses either to accept (A)
or to reject (R) it, with the aim of receiving
the most favorable partition of the offer they can get.  
When the players' choices are different,
arbitration comes in and forces the ^^ ^^ odd-man"
(the ^^ ^^ even-men") to receive $pX_i (\bar{p}
X_i/2$ each), where $0\leq p=1-\bar{p}\leq 1$ 
and the game terminates.  It is shown that, in the 
equilibrium, each player chooses R for small 
offers (A for large offers), and randomizes
between R and A for other offers, if 
arbitration favors the odd-man side, {\it i.e.}
$p\in\left(\frac13, 1\right]$ (the even-men side, 
{\it i.e.} $p\in\left[0, \frac13\right)$.)