What's more likely to show up first when rolling a $6$-sided die, $556$ or $234$?
The expected number of rolls of getting $556$ is solving for $a$ in the following equations: $$a = {5\over6}(a + 1) + {1\over6}(b + 1)$$ $$b = {5\over6}(a + 1) + {1\over6}(c + 1)$$ $$c = {2\over3}(a + 1) + {1\over6}(c + 1) + {1\over6}$$ Solving, we get $a = 216$, $b = 210$, $c = 174$, so the expected number of rolls it takes to get $556$ is $216$.
The expected number of rolls is getting $234$ is solving for $x$ in the following equations: $$x = {5\over6}(x + 1) + {1\over6}(y + 1)$$ $$y = {2\over3}(x + 1) + {1\over6}(y + 1) + {1\over6}(z + 1)$$ $$z = {2\over3}(x + 1) + {1\over6}(y + 1) + {1\over6}$$ Solving, we get $x = 216$, $y = 210$, $z = 180$, so the expected number of rolls it takes to get $234$ is also $216$.
So I have two questions:
- Given that the expected number of rolls for getting $556$ and $234$ is $216$ for both, does it follow that they're equally likely to show up first when rolling a $6$-sided die? If not, which one is more likely to show up first? And what's the calculation that would demonstrate this?
- Without writing out the Markov chain and calculating, on an intuitive level, how do we see which one is more likely to show up first (or that both of them are equally likely to show up first)?
In general, for two random variables $X$ and $Y$ with $\mathbb{P}(X=Y)=0$, $\mathbb{E} X = \mathbb{E} Y$ is not sufficient for $\mathbb{P}(X \leq Y) = 1/2$. Even more, $X$ and $Y$ having the same distribution is not sufficient. But this situation with $556$ and $234$ is not a coincidence.
We want to show that the number of strings (of length $n$) in which $556$ appears first (at position $i$) is equal to the number of strings in which $234$ appears first (at position $i$). The bijection suggested in the comments should work: given any string of die-rolls, replace the first occurrence of $556$ or $234$ with the other, if it exists. This bijection is an involution, since, e.g., replacing $234$ with $556$ does not create any new instances of of $234$ or $556$, because $234$ and $556$ are non-self-overlapping and do not overlap each other.
To summarize from the comments the limits of this approach: note that $565$ overlaps with itself ($56565$), so the above bijection does not work ($56234 \to 56565 \to 23465$). Additionally, $234$ and $456$ overlap with each other ($23456$), so the above bijection again does not work ($23234 \to 23456 \to 45656$).