An urn contains 20 white balls and 10 black balls. If you draw a white ball, you get 1 dollar, but if you draw a black ball, you lose 2 dollars.
How many balls should you draw in order to maximize the probability of winning? Hint: Use a computer.
Really have no idea how to do this.
Best Answer
HINT
The advanced version of this question is that you get to decide whether to keep drawing after seeing past results. Do you think this (homework?) problem is about that? Anyway, instead I assume this is asking about the basic version: You must pick a number $N \in [0, 30]$ in advance and then do exactly $N$ draws.
Since the expected value per draw is ${20\over 30} \times 1 + {10 \over 30} \times -2 = 0$, the expected value after $N$ draws is also $0$, so the question must be asking about the probability of ending up positive.
So for each $N$, you are supposed to calculate the probability $p_N$ of being positive after $N$ draws. Then you pick the $N$ that maximizes $p_N$.
Now after $N$ draws, if you get $k$ whites and $N-k$ blacks, your result is $R=k - 2(N-k)$. You need to (use a computer to) calculate $p_N$ by summing all the cases where $R>0$.
BTW the question didn't say whether the drawing is with or without replacements. The answers may be different. (With the wording as is, I would presume without replacements.)
Can you finish from here?