A chess player, X, plays a series of games against an opponent, Y A chess player $X$ plays a series of games against an opponent $Y$. For each game, the probability that $X$ wins is $p$, independently of the results of other games. If $X$ plays 4 games against $Y$, show that the probability that this series of games contains at least 2 consecutive wins by $X$ is $p^2(3-2p)$.
I know this can be done by listing all the possible ways and then summing the probabilities, but is there another way of doing this?
 A: To generalize, let $p_n$ be the probability that player $1$ wins at least $2$ games when $n$ games are played. Then,
$$
p_n=(1-p)p_{n-1}+(1-p)^2p_{n-2}+p^2, \quad p_2 =p^2, \quad p_3=2p^2-p^3
$$
From here you can find any $p_n$ by recursively substituting in.... alternatively, an exact solution can be found by standard methods that solve the cubic equation,
$$
0=x^3+(p-2)x^2+p(p-1)x+(1-p)^2
$$
A: Ok, this is simpler ...
You want all cases with a "WW" in them.  e.g. WWLL, WWWL, etc.  Let "A" mean either win or lose Prob(A)=1 of course.  The cases are:
WWAA:   $p^2$  
LWWA:   $(1-p) \times p^2$ 
ALWW:   $(1-p) \times p^2$ 
$= 3 p^2 -2 p^3$ 
$= p^2 (3 -2 p)$ 
Note - to avoid counting cases twice, I listed the cases by doing the WW moving from left to right, allowing anything to the right of the WW but no '2 consecutive wins' to the left of the WW.

For a larger number of games, this method is much better than the basic method, e.g. for 6 games, the answer would be: 
WWAAAA:  $p^2$  
LWWAAA:  $(1-p) \times p^2$  
ALWWAA:  $(1-p) \times p^2$  
XXLWWA:  $(1-P_2) \times (1-p) \times p^2$ 
XXXLWW:  $(1-P_3) \times (1-p) \times p^2$ 
where $P_2 = p^2$ is the probability of having at least 2 consecutive wins in 2 games, and $P_3 = p^2 (2-p)$ is the probability of having at least 2 consecutive wins in 3 games.
$P(6) = 5 p^2 -4 p^3 -3 p^3 +4 p^5 -p^6$

The general answers are:
$P_2 = p^2$ 
$P_3 = 2 p^2 -p^3$ 
$P_4 = 3 p^2 -2 p^3$ 
$P_5 = 4 p^2 -3 p^3 -p^4 +p^5$ 
$P_6 = 5 p^2 -4 p^3 -3 p^4 +4 p^5 -p^6$ 
$P_7 = 6 p^2 -5 p^3 -6 p^4 +9 p^5 -3 p^6$ 
$P_8 = 7 p^2 -6 p^3 -10 p^4 +16 p^5 -5 p^6 -2 p^7 +p^8$ 
$P_9 = 8 p^2 -7 p^3 -15 p^4 +25 p^5 -6 p^6 -9 p^7 +6 p^8 -p^9$ 
$P_{10} = 9 p^2 -8 p^3 -21 p^4 +36 p^5 -5 p^6 -24 p^7 +18 p^8 -4 p^9$ 
There are some patterns to the coefficients of $P_n$ e.g. the $p^2$, $p^3$, $p^4$ coefficients, but I have not found a pattern for all of them.
A: Is it really that tedious?
The cases are: 
4 wins:  $1 \times p^4$ 
1 loss, can happen in 4 ways:  $4 \times p^3 (1-p)$ 
2 wins if WWLL or LWWL or LLWW, so can happen in 3 ways:  $3 \times p^2 (1-p)^2$ 
$p^4  +(4 p^3 -4 p^4) +(3 p^2 -6 p^3 +3 p^4)$  
$= -2 p^3 +3 p^2$ 
$= p^2 (3 -2 p)$
