Given a unbiased dice how to find the average number of rolls required to get $n$ consecutive sixes given already $m$ number of sixes occurred where $m\leq n$. I know how to solve using n consecutive sixes with out any occurrences from this link How many times to roll a die before getting two consecutive sixes?, can anyone help me finding out no of rolls if any occurrences occured before?
[Math] How many times to roll a die before getting $n$ consecutive sixes given $m$ occurences
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Related Solutions
You can calculate the average this way also.
The probability of rolling your first $6$ on the $n$-th roll is $$\left[1-\left(\frac{5}{6}\right)^n\right]-\left[1-\left(\frac{5}{6}\right)^{n-1}\right]=\left(\frac{5}{6}\right)^{n-1}-\left(\frac{5}{6}\right)^{n}$$
So the weighted average on the number of rolls would be $$\sum_{n=1}^\infty \left(n\left[\left(\frac{5}{6}\right)^{n-1}-\left(\frac{5}{6}\right)^{n}\right]\right)=6$$
Again, as noted already, the difference between mean and median comes in to play. The distribution has a long tail way out right pulling the mean to $6$.
For those asking about this graph, it is the expression above, without the Summation. It is not cumulative. (The cumulative graph would level off at $y=6$). This graph is just $y=x\left[\left(\frac{5}{6}\right)^{x-1}-(\left(\frac{5}{6}\right)^{x}\right]$
It's not a great graph, honestly, as it is kind of abstract in what it represents. But let's take $x=4$ as an example. There is about a $0.0965$ chance of getting the first roll of a $6$ on the $4$th roll. And since we're after a weighted average, that is multiplied by $4$ to get the value at $x=4$. It doesn't mean much except to illustrate why the mean number of throws to get the first $6$ is higher than around $3$ or $4.$
You can imagine an experiment with $100$ trials. About $17$ times it will only take $1$ throw($17$ throws). About $14$ times it will take $2$ throws ($28$ throws). About $11$ times it will take $3$ throws($33$ throws). About $9$ times it will take $4$ throws($36$ throws) etc. Then you would add up ALL of those throws and divide by $100$ and get $\approx 6.$
For $n\in \{0,1,2\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.
We note $$E[2]=\frac 16\times 1+\frac 56\times \left(E[1]+1\right)$$ $$E[1]=\frac 16\times \left(E[2]+1\right)+\frac 56\times \left(E[1]+1\right)$$
$$E=E[0]=E[1]+1$$
this system is easily solved and, barring error (always possible), yields $$\boxed {E=43}$$
Best Answer
For $m\in\left\{ 0,\cdots,n\right\} $ let $\mu_{m}$ denote the expectation of number of rolls required to get $n$ consecutive sixes on the moment that the throwing of exactly $m$ consecutive sixes has just occurred. Then $\mu_{n}=0$ and for $m<n$ we find the recursion formula: $$\mu_{m}=1+\frac{1}{6}\mu_{m+1}+\frac{5}{6}\mu_{0}$$ It is based on the observation that with probability $\frac{1}{6}$ by throwing a six we land in the situation of having $m+1$ consecutive sixes and with probability $\frac{5}{6}$ we will have to start all over again. Substituting $m=0$ leads to $\mu_{0}=6+\mu_{1}$ and substituting this in the recursion formula gives: $\mu_{m}=6+\frac{1}{6}\mu_{m+1}+\frac{5}{6}\mu_{1}$.
Substituting $m=1$ in this new formula leads to $\mu_{1}=6^{2}+\mu_{2}$ and... Now wait a minute...
This starts to 'smell' like: $$\mu_{m}=6^{m+1}+\mu_{m+1}$$ doesn't it? Presume this to be true. Then we find:
This enables us to actually prove that the presumption is correct. For this it is enough to check that indeed $\mu_{n}=0$ and $\mu_{m}=1+\frac{1}{6}\mu_{m+1}+\frac{5}{6}\mu_{0}$ for $m<n$, which is just a matter of routine.
Note that for $n=2$ we find $\mu_{0}=\frac{1}{5}\left(6^{3}-6\right)=42$ corresponding with the answers on the question that you referred to.