Problem 1300

1. Determine the probabilities for the series lasting 3, 4, and 5 games:

- The probability that the series lasts exactly 3 games:
$$P(\text{3 games}) = x^3 + (1-x)^3$$
This is because either the Heat wins all three games, or the Spurs win all three games.

- The probability that the series lasts exactly 4 games:
$$P(\text{4 games}) = 3x^3(1-x) + 3(1-x)^3x$$
This is because one team wins three games, and the other team wins exactly one game. The factor of 3 accounts for the different positions the single win can occur in the sequence of four games.

- The probability that the series lasts exactly 5 games:
$$P(\text{5 games}) = 6x^2(1-x)^2$$
This is because the first four games must be split 2-2, and then the fifth game determines the winner. The factor of 6 accounts for the different ways to arrange two wins for each team in the first four games.

2. Calculate the expected number of games:

The expected number of games, $E$, is given by:
$$E = 3 \cdot P(\text{3 games}) + 4 \cdot P(\text{4 games}) + 5 \cdot P(\text{5 games})$$

Substituting the probabilities:
$$E = 3 \left( x^3 + (1-x)^3 \right) + 4 \left( 3x^3(1-x) + 3(1-x)^3x \right) + 5 \left( 6x^2(1-x)^2 \right)$$

3. Simplify the expression:

$$E = 3 \left( x^3 + (1-x)^3 \right) + 12 \left( x^3(1-x) + (1-x)^3x \right) + 30x^2(1-x)^2$$

$$E = 3x^3 + 3(1-x)^3 + 12x^3(1-x) + 12(1-x)^3x + 30x^2(1-x)^2$$

4. **Evaluate $f(-1)$:**

Substitute $x = -1$ into the polynomial $f(x)$:
$$f(-1) = 3(-1)^3 + 3(1-(-1))^3 + 12(-1)^3(1-(-1)) + 12(1-(-1))^3(-1) + 30(-1)^2(1-(-1))^2$$

Simplify each term:
$$f(-1) = 3(-1) + 3(2)^3 + 12(-1)(2) + 12(2)^3(-1) + 30(1)(2)^2$$

$$f(-1) = -3 + 3 \cdot 8 + 12(-2) + 12 \cdot 8(-1) + 30 \cdot 4$$

$$f(-1) = -3 + 24 - 24 - 96 + 120$$

$$f(-1) = -3 + 24 - 24 - 96 + 120 = 21$$

The final answer is $\boxed{21}$.