We roll 3 fair dice. In each round of rolling, we remove the dice with the value '1'. The game ends when all the dice are removed.
Let:
$N$ - the number of dice showing '1' at the first round.
$Y$ - the total number of rounds.
Prove: $\mathbb{E}(Y)>\frac{1}{1-\mathbb{P}(N=0)}$
I could only figure out that $\mathbb{P}(N=0) = (\frac{5}{6})^3$, which is quite obvious, but can't figure out the rest.
Please any help or direction.
On
The possible valuse for $Y$ are $1, 2, 3, \ldots$.
Now $$ P(Y=1) = \left( \frac{1}{6} \right)^3, $$ which is the probability that each of the three dice shows a 1.
Note that on the first round, we can have no 1's, one 1, or two 1's; and, corresponding to these outcomes, we must have, respectively, all three 1's on the three dice available, two 1's one the two dice available, or one 1 on the one die available to us on the second round. All of these outcomes are mutually exclusive. Moreover, the two rounds are independent. So $$ \begin{align} P(Y = 2) &= \left( \frac{5}{6} \right)^3 \left( \frac{1}{6} \right)^3 + { 3 \choose 1 } \frac{ 1 }{ 6 } \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^2 + { 3 \choose 2} \left( \frac{1}{6} \right)^2 \frac{5}{6} \left( \frac{1}{6} \right)^1 \\ &= \left( \frac{1}{6} \right)^3 \left[ \left( \frac{5}{6} \right)^3 + { 3 \choose 1 } \left( \frac{5}{6} \right)^2 + { 3 \choose 2} \frac{5}{6} \right] = \left( \frac{1}{6} \right)^3 \left[ \left( 1 + \frac{5}{6} \right)^3 - 1 \right]. \end{align} $$
Can you now continue from here?
Let $\Bbb P(N=0)=(\frac 56)^3=p$. The point is that you have probability greater than $p$ that the game lasts at least two rounds because if you don't throw any $1$s you are back where you started. The game will also last at least two rounds if you throw one or two $1$. You have probability at least $p^2$ that the game lasts at least three rounds because that represents the chance you don't throw any $1$s in the first two rounds. The expected number of rounds is the greater than $$1+p+p^2+\ldots = \frac 1{1-p}$$