There are $N$ dollars in a box. You have a probability $P$ (fixed) to win everything in the box and $1-P$ to pay a fee of $200$ dollars ($100$ go in the box, the other half is lost). You can try it as much as you want.
I've calculated mathematical expectation $E$ if one tries one time, two times or three times and solved the equation saying that $E\gt0$ to know when it is worth it to steal the box (depending on $N$).
I'd like to solve the problem for $n$ times trying to steal the box (stopping on a first success).
Here are the first iterations :
- Trying one time : $E = P\times N-200\times (1-P)$ (it is worth to steal the box when $E\gt0 \Leftrightarrow N>128$)
- Trying two times (stopping if succeed on first attempt) : $E = P\times N-P\times(1-P)\times(N-100)-400\times(1-P)^2$ (it is $N-100$ and not $N-200$ because 100 dollars go in the box if you fail the first time). It is worth it if $E\gt0 \Leftrightarrow N>99$.
- Third iteration : Becomes really complicated, I've solved it using Soulver (numerical method)
Is it possible to generalize the problem to $n$ tries ?