John and Ken have each have an urn with $N$ colored balls. They are playing a game in which Player-1 *(see figure below)* will name a color, and Player-2 has to pick a ball from his urn. If he picks the named color ball, he will earn a point (success) and an opportunity to challenge the other.

The winner will be decided after the following three steps. The game will always reset other to step 1.

Let's play the game now **(Note that this game is played with a replacement of balls)**.

**Step 1.** John calls GREEN, and Ken picks the GREEN with probability $P_1$, so now Ken gets to call the color. (Ken earns 1 point) (else reset)

**Step 2.** Ken calls BLUE, and John picks the color BLUE probability $P_2$, now John gets to call the color: (John earns 1 point) (else reset)

**Step 3.** John again calls a color (e.g. RED), and Ken picks up the RED ball, with probability $P_3$ (Ken earns 10 points and the game ends).

**Total points earned:**

John = 1

Ken = 11

What is the probability of Ken winning?

My initial work:

$P(Ken\ earn\ 11\ points) = P(Ken\ wins) = P_{kw}$

$P_{kw}= P(John\ call\ and \ loose). P(Ken\ call\ and \ loose). P(John\ call\ and \ loose)$

$P(John\ call\ and \ loose) = P_{JL} = \frac{1}{N}$

$P(Ken\ call\ and \ loose) = P_{KL} = \frac{1}{N}$

Now the probability that first step is taken by John is $P_x = 1/2$. So the actual probability of Ken winning would be $P_{k} = P_x.P_{kw} \, . $

$P_k = P_x(P_{JL}.P_{KL}.P_{JL})$

$P_k = \frac{1}{2N^3}$

Is it correct?

Second, what would be the expected number of times we have to play this game to decide a winner?

Third, how many failures can we expect? By failure I mean if we reset during step 1 or step 2?