LogFAQs > #931747839

Not exactly. It's just that when your probability is 50% on a single go, it just so happens that it halves every time.

If you go twice, here are your possible outcomes:

WW
WL
LW
LL

Each has a 25% chance of success, since there's no bias toward one or the other. Now, as you can see, first go winning has a 50% chance, just like we expected, and so does second go winning.

The trick is that one of those four cases is an overlap. The WW case is a part of both the first go wins and the second go wins sets, so if you just add the 50%'s, which would be the intuitive thing, you've counted that case twice. To remedy this, just subtract one out. In this case, you add the 50%'s and subtract out a 25%.

This is an idea taken from set theory, on calculating the number of elements in the union of two sets. Which makes sense because what probability really boils down to is a number of desirable possibilities compared to the total number of possibilities.

Count(A union B) = Count(A) + Count(B) - Count(A intersect B)

Again, you subtract out the intersect once because you already added it twice.

A = (A intersect not B) union (A intersect B)
B = (B intersect not A) union (B intersect A)

Since A and not A are complementary, and B and not B as well, you don't need to worry about their intersections. The term is still there, just always an empty set.

But if we then union these together we, we can see why we need to subtract out one instance of the intersect.

A union B = ((A intersect not B) union (A intersect B)) union ((B intersect not A) union (B intersect A))

A union B = (A intersect not B) union (A intersect B) union (B intersect not A) union (B intersect A)

A union B = (A intersect not B) union (B intersect not A) union (B intersect A) union (A intersect B)

Since B intersect A is equal to A intersect B, and any set unioned with itself is itself, this simplifies to:

A union B = (A intersect not B) union (B intersect not A) union (A intersect B)

Thing is you're usually given A and B so it's easier to work with those than to figure out what the sets look like without each other and with each other and stuff, so the easy thing to do is just manually subtract out the count of the intersection.

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