This follows a binomial distribution \(B(6, \frac{2}{3})\).

Probability of exactly 4 wins: \( P(X = 4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^{6-4} \).

Calculate: \(\binom{6}{4} = 15\), \(\left(\frac{2}{3}\right)^4 = \frac{16}{81}\), \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\).

\( P(X = 4) = 15 \times \frac{16}{81} \times \frac{1}{9} = 15 \times \frac{16}{729} = \frac{240}{729} \).

Simplify: \(\frac{240 \div 3}{729 \div 3} = \frac{80}{243}\).

Thus, \(\boxed{\frac{80}{243}}\).

Each game has 2 possible outcomes (Alfred wins or loses), and there are 6 games.

Total outcomes = \(2^6 = 64\).

\(\boxed{64}\)

Expand \((1 + x)^6\) using the binomial theorem:

\((1 + x)^6 = \binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \binom{6}{4}x^4 + \binom{6}{5}x^5 + \binom{6}{6}x^6\).

Substitute \(x = 1\):

\((1 + 1)^6 = 2^6 = 64\).

Right side: \(\binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}\).

Calculate: \(\binom{6}{0} = 1\), \(\binom{6}{1} = 6\), \(\binom{6}{2} = 15\), \(\binom{6}{3} = 20\), \(\binom{6}{4} = 15\), \(\binom{6}{5} = 6\), \(\binom{6}{6} = 1\).

Total = \(1 + 6 + 15 + 20 + 15 + 6 + 1 = 64\).

Thus, \(\boxed{64 = \binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}}\).

This equality represents the total number of possible outcomes for the 6 games, which is the sum of the number of ways Alfred can win 0 games, 1 game, 2 games, 3 games, 4 games, 5 games, or 6 games.

\(\boxed{\text{Total number of outcomes = sum of ways Alfred wins 0 to 6 games}}\)

Probability Alfred wins 4 on first day and 2 on second day:

\(P(4, 2) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 \times \binom{6}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^4\).

Combine: \(\binom{6}{4} \times \binom{6}{2} \left(\frac{2}{3}\right)^{4+2} \left(\frac{1}{3}\right)^{2+4} = \binom{6}{4}^2 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^6\).

So, \(r = 4\), \(s = 6\), \(t = 6\).

\(\boxed{\binom{6}{4}^2 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^6}\)

Probability Alfred wins total of 6 games:

\(P(\text{Total} = 6) = P(0, 6) + P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) + P(6, 0)\).

Each term: \(\binom{6}{0}^2 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^6 + \binom{6}{1}^2 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^6 + \cdots + \binom{6}{6}^2 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^6\).

\(\boxed{\frac{2^6}{3^{12}} \left(\binom{6}{0}^2 + \binom{6}{1}^2 + \binom{6}{2}^2 + \binom{6}{3}^2 + \binom{6}{4}^2 + \binom{6}{5}^2 + \binom{6}{6}^2\right)}\)

Probability of 6 wins out of 12: \(\binom{12}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^6 = \frac{2^6}{3^{12}} \binom{12}{6}\).

From (ii), this equals \(\frac{2^6}{3^{12}} \left(\binom{6}{0}^2 + \binom{6}{1}^2 + \binom{6}{2}^2 + \binom{6}{3}^2 + \binom{6}{4}^2 + \binom{6}{5}^2 + \binom{6}{6}^2\right)\).

Thus, \(\binom{12}{6} = \binom{6}{0}^2 + \binom{6}{1}^2 + \binom{6}{2}^2 + \binom{6}{3}^2 + \binom{6}{4}^2 + \binom{6}{5}^2 + \binom{6}{6}^2\).

\(\boxed{\binom{12}{6} = \binom{6}{0}^2 + \binom{6}{1}^2 + \binom{6}{2}^2 + \binom{6}{3}^2 + \binom{6}{4}^2 + \binom{6}{5}^2 + \binom{6}{6}^2}\)

Expected value: \(E(A) = \sum_{r=0}^n r \binom{n}{r} \left(\frac{2}{3}\right)^r \left(\frac{1}{3}\right)^{n-r} = \sum_{r=0}^n r \binom{n}{r} \frac{2^r}{3^n}\).

So, \(a = 2\), \(b = 3\).

\(\boxed{a = 2, b = 3}\)

Differentiate \((1 + x)^n\): \(\frac{d}{dx} (1 + x)^n = n (1 + x)^{n-1}\).

Expansion: \(\sum_{r=0}^n \binom{n}{r} x^r\), derivative: \(\sum_{r=1}^n r \binom{n}{r} x^{r-1} = n (1 + x)^{n-1}\).

Set \(x = \frac{2}{3}\): \(n \left(1 + \frac{2}{3}\right)^{n-1} = n \left(\frac{5}{3}\right)^{n-1}\).

Right side: \(\sum_{r=1}^n r \binom{n}{r} \left(\frac{2}{3}\right)^{r-1}\).

Multiply by \(\frac{2}{3}\): \(\frac{2n}{3} \left(\frac{5}{3}\right)^{n-2} = \sum_{r=1}^n r \binom{n}{r} \left(\frac{2}{3}\right)^r \frac{1}{3^{n-1}}\).

Adjust: \(\frac{2n}{3} = \sum_{r=1}^n r \binom{n}{r} \frac{2^r}{3^n}\).

Including \(r=0\) term (which is 0), \(E(A) = \frac{2n}{3}\).

\(\boxed{\frac{2n}{3}}\)