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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?

Mazzer
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3 Answers3

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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).

DMM
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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 $p_{100}+p_{101}+\dots + p_{104}$

5xum
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  • The probability of the sum reaching 2 is only $1/6$? Oh, maybe you mean reaching exactly 2. – Gerry Myerson Feb 11 '14 at 08:30
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    That's what I meant, I fixed it. – 5xum Feb 11 '14 at 08:31
  • OK, so you are going to get the recurrence $$p_n=(p_{n-2}+p_{n-3}+p_{n-4}+p_{n-5}+p_{n-6})/6$$ Solving that doesn't look too easy. – Gerry Myerson Feb 11 '14 at 08:34
  • Depends. It's easy to write a program to calculate $p_n$ for all $n<105$. It's not easy to calculate $p_n$ in a closed form, that is true. But I see no better way of solving this. – 5xum Feb 11 '14 at 08:42
  • @Arthur The question says "If you get one at any point in the game you lose" – 5xum Feb 11 '14 at 10:13
  • In my opinion, this is the "best" method to solve the problem. –  Feb 11 '14 at 20:17
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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.

alphanzo
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