3. A coin-flipping game, starting from the number $n$, if the coin lands heads up, subtract 1; if it lands tails up, subtract 2. Let $E_{n}$ be the expected number of coin flips before the number becomes zero or negative. If $\lim _{n \rightarrow \infty}\left(E_{n}-a n-b\right)=0$, then the pair $(a, b)=$ $\qquad$
3. $\left(\frac{2}{3}, \frac{2}{9}\right)$.
From the problem, we have the recurrence relation
$$
\begin{array}{l}
E_{n}=\frac{1}{2}\left(E_{n-1}+1\right)+\frac{1}{2}\left(E_{n-2}+1\right) \\
\Rightarrow E_{n}=1+\frac{1}{2}\left(E_{n-1}+E_{n-2}\right)(n \geqslant 2) .
\end{array}
$$

Let $F_{n}=E_{n}-\frac{2}{3} n$. Then
$$
F_{n}=\frac{1}{2}\left(F_{n-1}+F_{n-2}\right),
$$

The roots of the characteristic equation are $1$ and $-\frac{1}{2}$.
Thus, $F_{n}=A+B\left(-\frac{1}{2}\right)^{n}$, where $A$ and $B$ depend on the initial conditions.
$$
\begin{array}{l}
\text { Clearly, } E_{0}=0, E_{1}=1 . \\
\text { Hence } F_{0}=0, F_{1}=\frac{1}{3} \\
\Rightarrow A=\frac{2}{9}, B=-\frac{2}{9} \\
\Rightarrow E_{n}=\frac{2}{3} n+\frac{2}{9}-\frac{2}{9}\left(-\frac{1}{2}\right)^{n} \\
\Rightarrow(a, b)=\left(\frac{2}{3}, \frac{2}{9}\right) .
\end{array}
$$