(Conditional?) probability If a football team won the last match then the probability that the team will win the next match is 0.6. If the team lost the last match, then the probability of winning the next game is 0.5. If they won the next-to-last match, then the probability that they will win the next match is 0.55. If they lost, the probability of winning the next match is 0.45. 
We know that the team has won the last two matches. Find the probabbility that they will win the next match.
I have tried to use the conditional probability for three events but I think it's more complicated.
Give me some hint, please. 
 A: One approach might be to assume that you have a Markov Chain, with steady state probabilities for the last two matches having been WW, WL, LW, LL.  Each of these states could have a probability of winning the next match, so giving you a transition matrix.  You could then find expressions for the information in the question.  Then you need to solve for the eight unknowns (the four steady state probabilities and the four probabilities of winning the next match) subject to constraints.
I suspect this may not be possible with the particular data you give: I think you may have eight unknowns and nine constraints (the four numbers in the question, preserving the four steady state probabilities over time, and the steady state probabilities needing to add up to $1$).   
As an example of the problem, "if a football team won the last match then the probability that the team will win the next match is 0.6, and if the team lost the last match, then the probability of winning the next game is 0.5"  implies for a steady state that the probability the team won the last match was $\frac59$ and that it lost was $\frac49$, so the probability it wins the next match is $\frac59 \times 0.6 + \frac49 \times 0.5 = \frac59$, so ensuring a steady state.  But the probability it wins the match after next is $\frac59 \times 0.55 + \frac49 \times 0.45 = \frac{91}{180}$ which is not steady.  
I suspect there would be a solution if $\frac{1-0.6}{0.5}$ were equal to $\frac{1-0.55}{0.45}$ (or more sensibly, if the four numbers in the question were adjusted to make these equal), but they are not. 
A: This is exploratory. I would check if our system has any kind or degree of "markovity". 
I use the notation for the possible states in each period win/lose = $w/l$, and by $\pi_1(i, j) $ I denote the 1st-order (one game ahead) conditional (transition) probability that if in current period we are at state $i$ next period (game) we will be in state $j$. $\pi_2(i,j) $ is the corresponding 2nd-order (two games ahead) conditional probability. I also note $\Pi_1$ the 1st order transition matrix,
$$ \Pi_1=\begin{bmatrix} \pi_1(w,w) & \pi_1(w,l) \\ \pi_1(l,w) & \pi_1(l,l)  \end{bmatrix} = \begin{bmatrix} 0.6 & 0.4 \\ 0.5 & 0.5  \end{bmatrix}$$
(we have been given the first column, the second comes form the fact that in the transition matrix rows sum up to unity).  
Analogously, for the 2nd-order transition probabilities we have
$$ \Pi_2=\begin{bmatrix} \pi_2(w,w) & \pi_2(w,l) \\ \pi_2(l,w) & \pi_2(l,l)  \end{bmatrix} = \begin{bmatrix} 0.55 & 0.45 \\ 0.45 & 0.55  \end{bmatrix}$$
If we had a 1st-order stationary Markov chain, the following relation should hold:
$$\Pi_{m} = \Big (\Pi_{1}\Big)^m $$
In our case 
$$ \Big (\Pi_{1}\Big)^2= \begin{bmatrix} 0.6 & 0.4 \\ 0.5 & 0.5  \end{bmatrix}\begin{bmatrix} 0.6 & 0.4 \\ 0.5 & 0.5  \end{bmatrix} = \begin{bmatrix} 0.6\cdot0.6+0.4\cdot0.5 & 0.6\cdot0.4+0.4\cdot0.5 \\ 0.5\cdot0.6+0.5\cdot0.5 & 0.5\cdot0.4+0.5\cdot0.5  \end{bmatrix} = \begin{bmatrix} 0.56 & 0.44 \\ 0.55 & 0.45  \end{bmatrix} \neq \Pi_{2}$$
...but, the first row of  $\Big (\Pi_{1}\Big)^2$  is "almost" equal to the first row of $\Pi_2$. This first row contains the probabilities conditional on winning. This tells us that conditional on continuously winning, the sequence appears to have "no memory" beyond the previous game.  A winning streak does not add to the probabilities of winning the next game (while a loss two periods previously would matter). Since we are only asked to give the probability conditional on a winning streak, I could say that they are "pretty much" the same as the probability of winning given only that the last match was won, i.e. $\approx 0.6$.  
