What is the probability to win? Die game You have a die. If you get one pip at any point in the game you lose. If you get two,..., six pips you start adding the number of pips to a sum. To win the sum must get greater or equal to 100. What is the probability to win the game?
 A: An alternative approach, which still requires numerical calculation is to define a system state as the cumulative score. The system starts in state $0$ and there are a couple of absorbing states $L$ (lost) and $W$ (won). This yields 102 possible system states ${0,1,2,...,99,L,W}$. Each roll of the dice transforms the system state and it is straightforward to create the 102$\times$102 state transition matrix, $T = (p_{ij})$ where $p_{ij}$ is the probability of moving from state $i$ to state $j$. 
Multiplying $T$ by itself $n$ times, yields a matrix of the state transitions resulting from $n$ rolls of the dice and the row of this matrix corresponding to state $0$ yields the distribution of possible outcome states after these $n$ rolls. Taking $n\ge50$ ensures that for such outcomes only $L$ and $W$ have non-zero probabilities. (Note that whilst it is possible to lose from the first roll onwards, the definition of $L$ as an absorbing state with a corresponding transition $L \to L$ having a probability of 1 in $T$ means that it is legitimate to include rolls beyond the losing one. Similarly, defining $W$ as an absorbing state means we can consider rolls after the game is won.)
In my calculation procedure, I constructed $T$ and from this derived $T^2,T^4,T^8,...,T^{64}$. From the latter my results indicate that the probability of winning is 0.010197(to 6 decimal places).
A: Define $p_n$ as the probability of the sum reaching exactly $n$. Try to find a reccurence relation for $p_n$.
The starting conditions are $p_2 = \frac{1}{6}$, $p_3=\frac{1}{6}$, you will have to first work out $p_4, p_5$ and $p_6$ as they are slighly special, but from then on, it should be simpler. The probability of reaching $n>6$ is the probability of reaching $n-2$ and rolling $2$ or reacing $n-3$ and rolling $3$ and so on.
Once you have $p_n$ for general $n$, just calculate $p_{100}+p_{101}+\dots + p_{104}$
A: The probability of winning with #2=$i$, #3=$j$, #4=$k$, #5=$l$ and #6=$m$  is given by the multinomial coefficient 
$(1/6)^{i+j+k+l+m}\ \frac{(i+j+k+l+m)!}{i!\ j!\ k!\ l!\ m!}$,  
where $i$, $j$, $k$, $l$ and $m$ are constrained by
$i\leq 50$,
$j\leq \lceil (100-2i)/3\rceil=j_{max}(i)$,
$k\leq \lceil (100-2i-3j)/4\rceil = k_{max}(i,j)$
$l\leq \lceil (100-2i-3j-4k)/5\rceil = l_{max}(i,j,k)$
$m= \lceil (100-2i-3j-4j-5l)/6\rceil = m(i,j,k,l)$.
Summing over all choices gives
$\sum_{i=0}^{50}\sum_{j=0}^{j_{max}(i)}\sum_{k=0}^{k_{max}(i,j)}\sum_{l=0}^{l_{max}(i,j,k)}(1/6)^{i+j+k+l+m(i,j,k,l)} \frac{(i+j+k+l+m(i,j,k,l))!}{i!\ j!\ k!\ l!\ m(i,j,k,l)!}$.
I cannot see a way to reduce it but it would be easy to put this into a machine to solve.
