So I am trying to solve the following question:
X and Y are dividing a pie according to the following procedure. First, X proposes a division. If Y agrees then the division is implemented. Otherwise, they wait for an hour and Y proposes a division. If X agrees then the division is implemented, otherwise, they wait for an hour and X proposes a division and so on. A fraction p of the pie is worth $p × (\frac{1}{2})^t$ after t hours of waiting if t ≤ 5 and then the cake becomes totally useless. Find a subgame perfect Nash equilibrium. How long will they wait?
I know that I have to use backward induction to find the nash equilibrium but I am confused about how to draw the subgame tree. Could you please help me with the problem? Thank you in advance.
You do not need to draw a tree to solve this. Just think the problem through. What happens in hour 5? In hour 5 $X$ makes a proposal. The Pie is worth $(1/2)^5$. She offers $Y$ a zero share of the pie. $Y$ accepts as $Y$ is indifferent between getting nothing now and nothing in the next round as the pie is not worth anything then. So what happens in hour 4? The pie is worth $(1/2)^4$. $Y$ has to make $X$ weakly better off by accepting her proposal now rather than waiting for an hour and getting $(1/2)^5$. Thus, $Y$ will propose half of the cake to the $X$. Same logic applies for all previous hours. Thus, in the first offer $X$ will propose half the cake to $Y$ and the game is over after the first proposal.