Suppose there is a fare die with $6$ sides and we roll it again and again until we meet the pattern $123123$ for the first time. Let $\mathscr{F}_n$ be the filtration generated by the die. I want to know the expectation of the rolling times $\tau$.
Without using the law of total probability, suppose there is an infinite sequence of gamblers who join the game on each round, paying for $1$ dollar first and get the reward according to the following laws:
-
Only the gamblers seeing $1$ are possible to get paid.
-
Starting from the round when some gambler seeing $1$, his reward is $6$ dollars when seeing $1$, and if the next round is $2$, his reward becomes $6^2$, and if the following round is $3$, his reward becomes $6^3$, and if the following round is $1$ again, his reward becomes $6^4$, etc. His reward becomes largest only when he sees $123123$, then his reward does not change any more.
- If the next $6$ round is not $123123$, once he discovers, his reward immediately becomes $0$ and will not change. For example, a gambler sees that the next $6$ rounds follows $123141$, his temporary reward follows $6$, $6^2$, $6^3$, $6^4$, $0$, $0$.
Let $X_n$ be the total temporary reward of the gamblers from number $1$ to number $n$ up to $n$th round. Then the total wealth after each gambler paying $1$ dollar entering the game, is $X_n – n$. If we can prove $X_n – n$ is a martingale, and if the condition for stopping theorem is met, we can conclude $E[\tau]=E[X_\tau]=X_\tau = 6^6+6^3$.
My problem is how to justify $X_n-n$ is a martingale and the usage of stopping theorem.
Let $Y_n = X_n – n$. Surely $Y_n \in \mathscr{F}_n$ and $E|Y_n| < \infty$ for any $n$. Let $\xi_i(n)$ be the temporary reward of the $i$th gambler at the round $n$. Then $X_n=\sum_{i=1}^{n}\xi_i(n)$. We only need to prove: $$
E[\sum_{i=1}^{n-1}(\xi_i(n)-\xi_i(n-1))+\xi_n(n)|\mathscr{F}_{n-1}] = 1
$$
namely
$$
E[\sum_{i=1}^{n-1}(\xi_i(n)-\xi_i(n-1))|\mathscr{F}_{n-1}] = 0
$$
We only need to solve the case when $n=2,3,4,5,6$, since $\xi_1(7)-\xi_1(6)\equiv0$. Actually $E[\xi_i(n)-\xi_i(n-1)|\mathscr{F}_{n-1}]=0$. This can be justified by brutal calculation if I did it right. Please tell me if there is any easier way or my calculation is wrong.
My main problem is how to justify the usage of stopping theorem. I guess we should consider to prove $E[\tau]<\infty$. My attempt is the following:
Let $d_i$ be the result of the dice at $i$th round, and $D_i=(d_i, d_{i-1},…,d_{i-5})$. Certainly $D_i$ determines if the game stops. Then we have: $P(\tau > 6) = 1-\frac{1}{6^6}$. I am stuck from now. I try to get something like $P(\tau>n\alpha)<\beta^n$ and then we can estimate the upper bound of $E[\tau]$. Could anyone help me with this? Thank you so much for any help!
Edit:
I know there is a way using law of total probability. I am trying to avoid that. Also if $E[\tau]<\infty$ could be justified using law of total probability, I am still interested in how to prove it. Anyway I prefer the approach not using that law. Thank you!
Best Answer
For an upper bound on $E[\tau]$ you can divide your sequence into blocks of 6 dice. Let $D_i = (d_{6i - 5}, d_{6i - 4}, \ldots, d_{6i})$. Then, on event $D_i = (1, 2, 3, 1, 2, 3)$ you know for sure that $\tau \le 6i$. Let $T = \min \{i; D_i = (1, 2, 3, 1, 2, 3)\}$. It is important to separate the blocks to get independence. Now, $T$ is a geometric with parameter $6^{-6}$ and $\tau \le 6T$.