Hi I'm looking for help/advice in finding the solution to the following problem.
Suppose I have a fair $101$ sided dice labelled $0$ to $100$,
Your objective is to roll the dice any number of times to get as close as possible to $100$, without going over. The dealer then has to do the same and beat your score without busting ($\gt 100$), if the dealer busts, you win. If the dealer matches the score then it is a draw.
For example, You start by rolling a $3$, you 'hit' and roll again landing a $70$, scoring a total of $73$. You 'stand' on $73$ and hope the dealer busts i.e. sum of rolls $\gt 100$ without scoring between $73$ and $100$.
So to summarise,
- roll $0$ to $100$ with even chance of any number
- if you roll over $100$ you automatically lose
- if you roll $\lt 100$, then the dealer must beat this score without busting to prevent you winning, else if they bust, you win
- if the dealer matches your score it is a draw
Based on this I'd like to find out the optimum number on which to stand at, I'm struggling with the idea that after each roll the odds of busting increases (except for a roll of $0$).
Finally, assuming that I play the game with this figure in mind and stop rolling whenever it is achieved, what can I expect my probability of winning to be?