I have the following problem to solve that deals with probability (something I haven't done since Grade 8 (6 years ago))
This is a one player game and it is described for $q$ sided dice. You start by rolling $m$ dice. Remove the 6's and sum the rest of the dice to get $r_1$. Now roll the remaining dice, remove the 6's and sum the rest of the dice to get $r_2$ ; continue this until all dice have shown 6 (you have none remaining). Your score is $S = r_1 + r_2 + ... + r_t$ , where $t$ was the number of rolls. If $S > 5 m$ you win, otherwise you lose.
Suppose $m = 6$ and $q = 6$. Find the chance of you winning and the average score of this game.
Any help will be appreciated. along with any websites that can potentially help with this question or probability in general
This is not a simple problem. Let us start with $m=1,q=6$. We would like to calculate the expected score, as with a lot of dice we will approach that average. Let the expected score be $e$. After one roll, $\frac 16$ of the time we have lost with a score of zero, and $\frac 56$ of the time we have scored an average of $3$ and are back at the start. So $e=\frac 16\cdot 0+\frac 56(3+e), e=15$ As this is so much larger than $5$, with a lot of dice you are almost sure to win.
To really answer it, we would have to calculate the chance that each die produces each score, then combine them. As a zero can only come from an immediate $6$, it has a chance of $\frac 16$. A one can only come from rolling $1,6$, so has a chance of $\frac 1{36}$. Two can come from $2,6$ or $1,1,6$, so has a chance of $\frac 7{216}$. It gets more complicated as we go up.