Casino, last turn between Alice and Bob: Alice has $300$ chips and Bob has $700$. They can only play once and they bet simultaneously.
Our roulette gives $5$ options: double your money with probability $\frac{1}{2}$, triple with probability $\frac{1}{3}$, and so on with$\frac{1}{9}$, $\frac{1}{18}$ and $\frac{1}{36}$. Each player can do simultaneous bets at the same time and we assume that each bet does not affect the probability of the others. For example, he can bet some chips at $\frac{1}{2}$ and some at $\frac{1}{9}$ at the same time. (This is a fair roulette!!)
Alice decided to play all her money with probability $\frac{1}{3}$, so that, if she wins, she will have $900$ chips.
Bob has already seen Alice’s bet and is trying to figure out the optimum way to play, so as to maximize the probability to win.
My attempt was as follows: Since he wants to win and knowing that Alice may also win, in which case she will have $900$ chips, Bob must play in such a way to ensure he will have $>900$ chips and at the same time ensure the maximum probability.
So he must keep some number of chips and bet the rest with probability $\frac{1}{2}$ so as to have a total of $901$ chips. This means he must keep $499$ chips and play $201$ which if doubled will be $402$ so in total $499+422=901$.
The probability in this case will be $$\frac{(499 \times 100 \%) \ +\ (201 \times 50\%)}{100}$$ What do you think?
I'm not entirely sure of what you mean by "Bob has already seen Alice's bet..."
If Bob knows the details of Alice's bet, he can just make the exact same bet (on the same numbers) for 67 chips. If the bet loses, they both lose and Alice has nothing; Bob has 633 chips. If the bet wins, they both win and Alice has 900 chips, and Bob has 901 chips.