Expectation of the time until all 6 balls in an urn become white

Question

An urn contains 6 black balls and 0 white balls. One ball is drawn at random, taken out and replaced by a ball of the other color.

What is the expectation of the time (turns) until all 6 balls become white?

Progress

I'm thinking of calculating the expectation by definition but I can't get my hand on the exact Formula for $P(T=t)$ to put in the sigma. I have tried to find the pattern with $P(T=8)$ and $P(t=10)$ but I'm not sure about what I got.

Answer 1

Let the expected distance from the situation where you have $r$ white balls in the urn be $d(r)$.

If there are no white balls, then after one draw there will be one black ball, and having taken one turn, you have to reach $6$ white balls from the situation where you have one white ball in the urn.

$$d(0)=1+d(1)$$

With one ball in the urn, the next turn will take you back (probability $\frac 16$) or forwards (probability $\frac 56$). Don't forget you've taken another turn.

$$d(1)=1+\frac 16 d(0)+\frac 56 d(2)$$

Continuing on, and noting that $d(6)=0$ $$d(2)=1+\frac 13d(1)+\frac 23d(3)$$$$d(3)=1+\frac 12d(2)+\frac 12d(4)$$$$d(4)=1+\frac 23d(3)+\frac 13d(5)$$$$d(5)=1+\frac 56d(4)$$

You have six equations in six unknowns. These can be systematically solved.

Answer 2

You can simplify the calculation somewhat by using the symmetry. First you need to get from a monochromatic situation to the symmetric situation. Then you need to get from the symmetric situation to a monochromatic situation. Then with probability $\frac12$ you're done and with probability $\frac12$ you need to start over.

Let $E_k$ be the expected number of steps to reach the symmetric situation from a situation with discrepancy $0\le k\le3$. Then

\begin{align} E_0&=0\;,\\ E_1&=1+\frac23E_0+\frac13E_2\;,\\ E_2&=1+\frac56E_1+\frac16E_3\;,\\ E_3&=1+E_2\;. \end{align}

Substituting $E_1$ and $E_3$ into the equation for $E_2$ yields

$$ E_2=1+\frac56\left(1+\frac13E_2\right)+\frac16(1+E_2)=2+\frac49E_2\;, $$

so $E_2=\frac{18}5$ and thus $E_3=\frac{23}5$.

Now let $F_k$ be the expected number of steps to reach a monochromatic situation from a situation with discrepancy $0\le k\le3$. Then

\begin{align} F_3&=0\;,\\ F_2&=1+\frac56F_1+\frac16F_3\;,\\ F_1&=1+\frac23F_0+\frac13F_2\;,\\ F_0&=1+F_1\;. \end{align}

Substituting $F_0$ and $F_2$ into the equation for $F_1$ yields

$$ F_1=1+\frac23(1+F_1)+\frac13\left(1+\frac56F_1\right)=2+\frac{17}{18}F_1\;, $$

so $F_1=36$ and $F_0=37$.

Thus a round trip from monochromatic to symmetric back to monochromatic is expected to take $37+\frac{23}5=\frac{208}5$ steps, and since each such round trip has success probability $\frac12$, the total expected number of steps is $2\cdot\frac{208}5=\frac{416}5=83.2$.