A game where you roll a fair die, repeatedly, adding up the faces that show up, until the face 6 appears. What is the expected sum (including the 6)?
All I can think of is that the expected value of each roll is $7/2$.
I'd appreciate your help!
[Math] Rolling a die until obtaining the face 6. Whats the expected amout of the sum
probability
Related Solutions
Well, you can make numbers from 1 to 110, that's 110 possible states you can be in. That's a large transition matrix, but not intractable. Define transition matrix $M$ as:
$$\begin{align} M_{i,j} = \begin{cases} \frac{1}{10} & \text{ if } 1 \le j - i \le 10 \text{ and } 1 \le i \le 100 \\ 1 & \text{ if } i = j\text{ and } 100 < i \le 110 \\ 0 & \text{ otherwise} \end{cases} \end{align}$$
Define the initial state vector: $$\begin{align} V_{j} = \begin{cases} \frac{1}{10} & \text{ if } 1 \le j \le 10 \\ 0 & \text{ otherwise} \end{cases} \end{align}$$
This matrix compute the probability that you end up in state $j$. I won't wrote out the whole thing here. The steady state $S$ is given by:
$$S = VM^{\infty}$$
I don't recommend doing this by hand.
Anyway, you get the following probabilities:
$$\begin{array} {c|c} \text{End State} & \text{Probability} \\ \hline 101 & \frac{{ 14545454540916238222253308031039403263876427137099 \atop 728738149747953197899302063661139633020606426446001 }}{10^{101}} \\ \hline 102 & \frac{{ 18181818143103207886299794518678653455112915813572 \atop 471568730433172695421560362344285718841548526446001 }}{10^{101}} \\ \hline 103 & \frac{{ 16363636337149401877562367260240328253764897737948 \atop 745622241485421850822156839220501392260147526446001 }}{10^{101}} \\ \hline 104 & \frac{{ 12727272741576429962125352735631301525861758140731 \atop 984148880861015973102098820410322797857111216446001 }}{10^{101}} \\ \hline 105 & \frac{{ 10909090931874858870242133363925460156752233410373 \atop 601637391699278967219484863685353489177266485446001 }}{10^{101}} \\ \hline 106 & \frac{{ 90909091116377120481295620461022602720063194921283 \atop 52571281564919296780386873223909380629437281346001 }}{10^{101}} \\ \hline 107 & \frac{{ 72727272852713068490158133017227152668140086810036 \atop 91839439446721479480174650845945205326825156836001 }}{10^{101}} \\ \hline 108 & \frac{{ 54545454582842347527531906998645390126303113111522 \atop 24370063982379262253640845972771391003951819875001 }}{10^{101}} \\ \hline 109 & \frac{{ 36363636346255163502142715869656509893927302281916 \atop 05910755643757924951410231819125651609791149217901 }}{10^{101}} \\ \hline 110 & \frac{{ 18181818155610931814042064558296878037883980477975 \atop 93593065135379352069784448816645340273314411495091 }}{10^{101}} \\ \hline \end{array}$$
Expected state is calculated as always, and the final answer is :
$$\sum_{k=100}^{109} k\cdot S_{j,k} = \frac{{ 2080000000053214123545556126144601328836935442611255 \atop 847240374822309959920561393884518831211143790906011 }}{2\cdot10^{100}}$$
which is almost exactly $104$ (to eight decimal places).
EDIT-- Corrected post, originally answered the question "at least 100" rather than "more than 100".
The expected value of a sum of random number $N$ of iid random variables $X_i$ is $$E\left[\sum_{i=1}^N X_i\right]=E[N]E[X_i]$$ In your case you add $E[N]$, so the answer is $$E[N]E[X_i]+E[N]=3.5\cdot 3.5+3.5 =4.5\cdot 3.5 = 15.75$$
Best Answer
Solution 1: We roll the die and add the value. With $5/6$ probability we start again. Hence,
$$E(X) = \frac{7}{2} + \frac{5}{6}E(X)$$
and this gives $E(X) = \boxed{21}$.
$E(X) = \infty$ is also a solution, so we should show that the expectation is finite. This is equivalent to showing that the expected number of rolls is finite, since the expected value is bounded between the number of rolls and six times the number of rolls. By standard methods, the expected number of rolls is
$$\sum_{n=0}^{\infty} (n+1) \frac{1}{6} \left(\frac{5}{6}\right)^n$$
which converges to $6$. We can use this to motivate a secondary method:
Solution 2: On average we roll $6$ times. The average on a roll is $\frac{7}{2}$. Hence the answer is $21$.
You might ask, but doesn't it matter that we can't roll a six or else we stop? Each roll still has the same expectation of $\frac{7}{2}$, it is just that we stop arbitrarily when we happen to roll a six.
(This perhaps unintuitive logic is similar to that in the problem involving a town where couples birth babies until their first boy and then stop. The proportion of boys to girls in this town is still $1:1$ since each birth is still just an independent event; they just choose to stop arbitrarily when the first boy comes)