Move of Nature - Ultimatum game Let $N=\left \{1,2 \right \}$, $A_1=\left \{(x,y): x+y= \omega \right \}$. Player 2, upon seeing a proposal $(x,y)$ can either accept or reject the offer. Player 2 (the responder) has an outside option available which can be either: 


*

*10 with probability $p_1$

*5 with probability $p_2$

*1 with probability $p_3$


For instance, if the responder's outside option is 10, then if he rejects an offer $(x,y)$, his payoff is 10. While if he accepts $(x,y)$, it is then simply $y$. The responder knows his own outside option, but the proposer does not.
Here is a draw that how starts Move of Nature

What can be pure strategies sub-game perfect equilibrium as a function of the probabilities $p_1$,$p_2$ and $p_3$?
 A: Player 1's payoff function is given by 
$$u_1(x,y)=u_1(x,ω-x)=\begin{cases}x, &0\le x\le ω-10\\\left(p_2+p_3\right)x, &ω-10<x\le ω-5\\(p_3)x, &ω-5<x\le ω-1\\0,& ω-1<x\end{cases}$$ for $0\le x \le ω$, under the assumption that Player 2 breaks ties in favor of Player 1 (f.e. if Player 2's option is $5$ and Player 1 has proposed $5$ then Player 2 - while indifferent - accepts Player 1's offer). Note however that depending on the value of $ω$, some intervals above may be empty. 
The rest of the solution depends largely on the value $ω$ and follows an extensive case discrimination. For example if $5<ω<10$ the above becomes 
$$u_1(x,y)=u_1(x,ω-x)=\begin{cases}\left(p_2+p_3\right)x, &0\le x\le ω-5\\(p_3)x, &ω-5<x\le ω-1\\0,& ω-1<x\end{cases}$$ for $0\le x \le ω$. The first part of $u_1$ is maximized at $ω-5$ and the second part at $ω-1$ while the third part is constantly $0$ and can be ignored. So, Player 1 compares the payoffs: $(p_2+p_3)(ω-5)$ at $x_1=ω-5$ and $(p_3)(ω-1)$ at $x_2=ω-1$ and decides to play $x_1=ω-5$ iff $$(p_2+p_3)(ω-5)\ge p_3(ω-1)\iff \frac{p_3}{p_2}\le \frac{ω-5}{4}$$ If the above holds with equality then he is indifferent between $x_1$ and $x_2$.
