Ordering of independent events 
Probability of Team A scoring $1$ goal is $P(A=1)=0.7,$
Probability of Team $B$ scoring $1$ goal is $P(B=1)=0.5.$
Events A and B are independent, and the teams do not score more than 1 goal each.
On match day we need to find the probability $P(A+B=2).$

On some days Team $A$ plays before Team $B,$ on other days Team $A$ plays after Team $B.$ So,
Scenario 1: $P(A+B=2)=P(A=1)P(B=1)$   [simple as that]
Scenario 2 : We need to take into consideration which team plays first. So,
$$P(A+B=2)\\=   P(A=1)P(B=1)  \quad\text{[Team A plays before team B]}                       \\+     P(B=1)P(A=1) \quad\text{[Team B plays before Team A]} \\=2P(A=1)P(B=1)$$
Which scenario is right, and why is the other one wrong??
 A: 
Scenario 2 : We need to take into consideration which team plays
first. So, $$P(A+B=2)\\=  P(A=1)P(B=1)  \quad\text{[Team A plays before Team B]} \\ + P(B=1)P(A=1) \quad\text{[Team B plays before Team A]}\\=2P(A=1)P(B=1).$$

Formulations and rules in classical logic (so, mathematical reasoning and probability) are agnostic to time/tense.
In particular, events in a probability experiment are specific collections of outcomes, and don't have an inherent sequence. For example, when considering pairwise independence, (informally: For $P(A)\neq0,$ events $A$ and $B$ are independent iff the probability of $B$ is unchanged by the knowledge that $A$ occurs), it doesn't matter if $B$ occurs before $A,$ or if they occur concurrently, or if their sequence is undefinable (e.g., $A$=getting at most three Heads, $B$=getting at least three Heads).
And two dice successively thrown are equivalently analysed as two dice concurrently thrown, because the trials of a probability experiment can even be sequenced reverse-chronologically.
So, it's unnecessary to consider whether Team A or B plays first. But if we insist on conditioning on this, then the correct presentation for Scenario 2 is
$$P(A+B=2)\\=  P(A=1)P(B=1)  P(\text{Team A plays its game before team B})+P(B=1)P(A=1) P(\text{Team B plays its game before Team A})\\=P(A=1)P(B=1)\bigg(P(\text{Team A plays its game before team B})+P(\text{Team A plays its game before team A})\bigg)\\=P(A=1)P(B=1),$$ which unsurprisingly ends up being the same as in Scenario 1.
