Two cheeky thieves counted a total of exactly 1 billion $\left(10^9\right)$ dollars in the bank vault. Now they must decide how to divide the booty. But there is one problem: the thieves have only $M$ minutes to leave the bank before the police arrives. Also, the more time they spend in the vault, the less amount they 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 $10^9 \times p^t$ dollars, where $0 \ge p \le 1$, 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 $M$th 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.
So we need to find how the distribution needs to be done.
t=2 and p=0.5
then answer would be 500000000.0 500000000.0
As usual in such games, the solution is obtained by backward induction beginning with the final round. In the final round, the proposing thief $F$ can get the entire remaining booty, $p^{M-1}$, for herself. (I'm omitting the billion.) Thus, in the preceding round, the other thief $O$ should offer that much to her, and keep the rest, $p^{M-2}-p^{M-1}$, for himself. Likewise, in the preceding round, $F$ should offer $O$ $p^{M-2}-p^{M-1}$ and keep the rest, $p^{M-3}-p^{M-2}+p^{M-1}$, for herself. Continuing like this, we find that the thief proposing at the beginning should keep
$$ 1+(-p)+(-p)^2+\cdots+(-p)^{M-1}=\frac{1-(-p)^M}{1+p}\;. $$
In the limit of infinitely slow police, the first thief keeps $1/(1+p)$, and the other one gets $p/(1+p)$.