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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??

ryang
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Supreme Poudel
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  • I am confused about two things in your question: [1] The impression that I got is that you are assuming that neither $A$ nor $B$ can score $2$ or more goals. This implies that the only way that their scores can sum to $2$ is if they both score $1$ goal. I did not see this constraint in the posted question. Did you leave this constraint out? [2] I fail to see why it is relevant which team, $A$ or $B$ plays their game first. Just focusing on your intuition for a moment, why would it seem that this is relevant? – user2661923 Sep 23 '21 at 02:30
  • [1] Yes sir, the question assumes that Both teams can score at maximum 1 goals each , so the only way to have to 2 goals combined is for them to score 1 goal each. Its my fault I forgot to mention this, I am sorry. [2] My thought process was in a way that the order in which it done matters. For example, with die rolling. lets say we want sum of 3 while rolling a die 2 times consecutively, so P(sum=3)=p(1 in the first roll)p(2 in the second roll) + p(2 in the first roll)p(1 in the second roll) – Supreme Poudel Sep 23 '21 at 02:36
  • Why do you think that it is relevant who plays first $A$ or $B$? Ignoring the math, the intuitive definition of independent events, is that neither $A$ nor $B$ is affected by how the other team does. – user2661923 Sep 23 '21 at 02:39
  • As far as your dice roll example, for having the dice sum to $3$, suppose that one die is red and the other green. There are two combinations that sum to $3$: red/green $= 1,2$ or red/green $= 2,1$. Again, why would it matter if the red die is rolled before or after the green die? – user2661923 Sep 23 '21 at 02:42
  • So i dont have to consider the possibilities in which the events might happen since they are independent ? [I am sorry if I am sounding weird with these questions] – Supreme Poudel Sep 23 '21 at 02:44
  • Exactly. That is one of the consequences of the events being independent of each other. – user2661923 Sep 23 '21 at 02:46
  • Ohh , thank you so much sir, I think I understood what you are trying to say, The dice roll example explains it , thank you sir. – Supreme Poudel Sep 23 '21 at 02:46
  • You may further condition on who plays first, but doing so, [assuming independence] you'd need to multiply by the probabilities of who plays first. That is, scenario 2 would be written $$\begin{align} P(A+B=2) = ,, &P(A=1)\cdot P(B=1)\cdot P[\text{Team A plays its game before team B}]\ &+ P(B=1)\cdot P(A=1) \cdot P[\text{Team B plays its game before Team A}]\end{align}$$ Adding them together and simplifying gives the same result as the first scenario. – Brian Moehring Sep 23 '21 at 03:04
  • AHH yess, if we are to take the way events take place in consideration too, we need to include the probability that the event occurs first ( probability of A playing first or B playing first in this case which sum up to 1 ) and so adding them gives scenario 1 . Thank you so much Brian. I think I have understood the concept completely thanks to you and user2661923. – Supreme Poudel Sep 23 '21 at 03:30

1 Answers1

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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.

ryang
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