I'm not sure how it matters, unless they have some greater chance of winning depending on where they're at. Otherwise, wouldn't they just go for the first one again so they can play at home as much as possible/get the ticket money?

Lol

Hmmm.

Well I would go with the second one. If they choose the first one then they have a good chance of winning either the first or last game but a poor chance of winning the second game; meaning they have a poor chance of winning two games in a row.

In the second option they have a good chance of winning the second game so they only need to win one of the more difficult matches before or after to win two in a row. So in the second option they have two chances to win the more difficult match as opposed to only one in the first option.

I actually think the problem is poorly stated, as it doesn't explain what the difference between playing at home versus away is. If there is no difference, then the correct answer is "it doesn't matter". If there is a difference, then the problem is unfair as currently stated.

As a side note, I have a really evil problem, one that is famous for people (and sometimes mathematicians) insisting that their incorrect answer is the correct one. (The problem is fully specified, so no outside information is needed to solve it.)

Hmmm... I really thought the first sentence explains the difference:

So yes, it means that probability of winning at home is bigger than winning away.

Edit: Nevermind. Adalia beat me to it.

adaliabooks got it right. Calculations:

p = probability of winning at home
q = probability of winning away

p > q

In order to qualify you have to win two games in a row, so you have to:
(win the second one) AND (win the first OR win the third)

let probabilites be p1, p2 and p3 respectively

probability of winning the second one = p2
probability of winning the first OR the third = p1 + p3 - p1*p3

so probability of qualifying = p1*p2 + p3*p2 - p1*p3*p2

First choice gives you:
pq + pq - p^2*q = (2-p) * pq

Second choice gives you:
pq + pq - q^2*p = (2-q) * pq

and since p>q then (2-q) * pq > (2-p) * pq

Hence second choice gives a larger probaility

I'm going to take a stab at the answer: You should switch doors.

Am I right? :D

Either that or the aeroplane will take off anyway... ;)

(also, the answer you've given is the wrong one :P)

Right. Now about your question...

I still don't have one... I can solve them, but I'm not great at coming up with them :/

I can go find something on Google though...

While the math holds, as a practical matter no one would ever design a tournament that way, as it becomes quite possible to have neither team from that side of the bracket advance to the finals. ;)

I could jump in to provide a question, if anybody needs me to. I can't guarantee it'll be all that hard, though.

While no one would design a tournament that way (you have to win 2 matches in a row and get to choose where to play), it wasn't said that both teams need to win 2 matches in a row. So it could have been: you need to win 2 matches in a row, otherwise the other team qualifies ;).
Out with it already.

I am an electronic technician, so I have a more than few thoughts on this.

Okay, an LED is a type of diode, so it has polarity. Then there are two cases to consider:

Case 1: The LED is forward biased (installed forwards). With a limiting resistor in series an LED will typically drop its rated voltage, usually 2V for a LED. The remaining voltage is carried through the resistor. Without that resistor, all the voltage will try to pass though the LED. Assuming the LED does not immediately burn out, then it will try to carry a very large current, potentially as much as the battery can produce (the current a diode carries grows exponentially as the voltage goes over its rating).

Case 2: The LED is reverse biased (installed backwards). There will be a few microamps of current due to reverse bias leakage (treated as 0 amps in practice). This is true unless the reverse breakdown voltage is exceeded; then the voltage overwhelms the p-n junction and potentially large amounts of current will flow, similar to the overvoltage scenario in the forward biased case.

The practical result: either the diode is carrying effectively no current due to either being reverse biased or burnt out.