How to divide the money based on the below conditions?

Question

Two cheeky thieves have collectively decided to rob a bank.

They have carefully counted a total of exactly $100$ dollars in the bank vault. Now they must decide how to divide the booty. But there is one problem: the thieves have only M($0\le M \le 100)$ minutes to leave the bank before the police arrives. Also, the more time they spend in the vault, the less amount could carry away from the bank. Formally speaking, they can get away with all of the billion dollars right now, but after t minutes they can carry away only $100 * p^t$ dollars, where $0 \le p \le 1$, and at t = M, they get arrested and lose all the money. They will not leave the vault until a decision on how to divide the money has been made.

The money division process proceeds in the following way: at the beginning of each minute starting from the 1st (that is, t = 0), one of them proposes his own way to divide the booty. If his colleague agrees, they leave the bank with pockets filled with the proposed amounts of dollars. If not, the other one proposes his way at the next minute etc. To escape arrest, they can only propose plans till the beginning of the Mth minute (i.e., till t = M-1). Each thief wants to maximize his earnings, but if there are two plans with the same amounts for him, he would choose the one which leads to a larger total amount of stolen dollars.

What will be the final division of money, if each thief chooses the optimal way for himself?

Answer 1

At the last minute, $A$ says he will take all the money, which by then is $\$D$.
At the second-last minute, $B$ says $A$ may take $\$D$, but $B$ will take the rest, which is $D(\frac1p-1)$.
At the third-last minute, $A$ says $B$ may take $D(\frac1p-1)$, but $A$ will take the rest....

Answer 2

Let $x(M),y(M)$ be the proportions of the original booty the first and second thief obtain given $M$ minuts to decide. Clearly $x(1)=1,y(1)=0$ as with no time left the second thief will agree to anything to avoid arrest. For general $M$, the first thief contemplates (hopefully in less than a minute):

If the other guy rejects my offer, I will get $py(M-1)$ and he $px(M-1)$ since we play the same game with less booty, less time, and roles switched. He will reject especially if I offer him less than $px(M-1)$; he will greedily accept if I offer more than $px(M-1)$; he will also accept by the tie-breaking rule if I offer exactly $px(M-1)$, which will leave $1-px(M-1)$ for me, certainly at least as much as (why?) $py(M-1)$.

We obtain the recursion $$x(1)=1,\quad x(M)=1-px(M-1)$$ (or more symmetrically $ x(M)-\frac1{1+p}=-p\cdot\left(x(M-1)-\frac{1}{1+p}\right)$) with the solution $$ x(M)=\frac{1-(-p)^M}{1+p}.$$ If we reread the thoughts of the first thieve, we notive that the first offer is accepted, thus making $$ y(M)=1-x(M)=\frac{p+(-p)^M}{1+p}.$$