It may help to ask yourself what would have to happen for each of the statements to be false.
(A) If it is raining, then the home team wins.
The only way this can be false is for it to be raining, and the home team loses. (For the same of simplicity I’ll assume that there are no ties.)
(B) If the home team wins, then it is raining.
The only way this can be false is for the home team to win while it is not raining.
Those are two completely different events:
If the two statements were equivalent, exactly the same circumstances would render them both false. As you can see, that’s not the case, so they cannot be equivalent.
In fact, there are four possible states of affairs:
As we’ve seen, statement (A) is inconsistent with state (1); what about the other three states? It’s obviously consistent with state (3). It’s true but less obvious that it’s also consistent with states (2) and (4): this is because statement (A) says nothing at all about who wins if it’s not raining, so no outcome of the game can disprove it in that case. Thus, statement (A) boils down simply to saying that state (1) isn’t the case: either it’s not raining, in which case (A) says nothing about the outcome of the game, or it is raining, and the home team wins.
I’ll leave it to you to apply similar reasoning to see that statement (B) is consistent with each of the states except state (2).
Alternatively, you can show logical equivalence with a truth table. Let $r = \{\text{Rains}\}, w = \{\text{Home team wins}\}$. Then
\begin{array}{c|c|c|c} r&w&r\to w&w\to r\\\hline T&T&T&T\\\hline T&F&F&T\\\hline F&T&T&F\\\hline F&F&T&T \end{array} Notice that the conditional statements $r\to w$ and $w\to r$ have different truth tables, and hence, they are not logically equivalent.
The original statement says that if it's raining, the home team wins. It is also possible, based on this statement, that even if it's not raining, the home team still wins.
The converse statement says that if the home team wins, then it's raining. It is also possible for it to rain without the home team winning.
However: Under this converse, it is not possible for the home team to win without it raining. Conversely (!), under the original statement, it is not possible for it to be raining without the home team winning.
The statement only says that if it is raining then that's enough to know that the home team wins.
So raining guarantees home win, but it does not say that this is the only condition under which the home team wins.
It could be true that when it snows the home team also wins. So you can see that the fact that the home team wins does not guarantee anything about rain.
An intuitive example:
If I have more than 10 coins, I have more than 5 coins.
If I have more than 5 coins, I have more than 10 coins.
You wouldn't regard the above as equivalent, would you?
The first sentence is always true, no matter what. The second can be false, for instance when you have 7 coins.
Consider the following:
If you have studied, then you will pass the test.
Let's say you accept that it is true. Still, you might believe that some people don't have to study but will pass the test anyway! Such people would provide counter-examples to the following:
If you will pass the test, then you have studied.
Geometrically, you can think of implication ($\implies$) as inclusion ($\subseteq$). If we're working in some first-order structure $M$, such that $$M \models \forall x \left( \varphi(x) \rightarrow \psi(x) \right),$$ ("$\models$" is to be read as "satisfies") then for any $a \in M$, if $M \models \varphi(a)$, then $M \models \psi(a)$.
This means: the definable set cut out by $\varphi(a)$ $$\varphi(M) \overset{\operatorname{df}}{=} \{a \in M \operatorname{ | } M \models \varphi(a)\}$$ is contained in the definable set $\psi(M)$ cut out by $\psi(a)$. That is, $$M \models \forall x \left(\varphi(x) \rightarrow \psi(x)\right) \iff \varphi(M) \subseteq \psi(M).$$
(For example, let $\varphi(x)$ be the formula in the language of rings which says that $x$ is a root of a polynomial $p(t)$, and let $\psi(x)$ say that $x$ is a root of $p(t)q(t)$ for some other polynomial $q(t)$.)
To answer the question in your title, if all first-order implications were equivalent to their converses, then there would be no proper inclusions of definable sets in any first-order structure $M$. Which is not the case---in any nonempty structure, the empty set (falsehood) is always properly contained in (hence implies) the entire structure (truth).
consider p:it is raining and q:home team wins so the compound statement is if p then q.that is p->q and converse is q->p.so by identities p->q = ~p V q and q->p = ~q V p = p V ~q
so ~p V q is never equivalent to p V ~q thus the statement is not equivalent to its converse.
In this type of statement, causation works in one direction. The result does not do the causation. Perhaps the raining/wins combination is not truely an example of a "statement" and its "converse."
\implieswith MathJax and the previous slash was driving me nuts) – pjs36 Mar 06 '16 at 07:08