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ZFR: Something quick then, it's fairly easy but the result is pretty interesting.

A̶ ̶c̶a̶m̶e̶l̶ ̶m̶e̶r̶c̶h̶a̶n̶t̶... A football (that's soccer for you Americans) team plays better at home than it does away. One year, in order to qualify for the finals they have to beat a team from across the country twice (at least) in three matches. They are given a choice of playing either
Home then Away then Home
or
Away then Home then Away
Naturally they select the first option in order to play twice at home.

Next year however the rules change. They have to win two games (at least) in a row in order to qualify. They are given the same choices. Which one should they choose?

Since there is a 50% chance of getting the right answer by guessing (and since the way the problem is posed seems to suggest what the correct answer is), I'll need some sort of explanation or calculation besides the answer.
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?
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ZFR: A̶ ̶c̶a̶m̶e̶l̶ ̶m̶e̶r̶c̶h̶a̶n̶t̶...
Lol

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ZFR: A football (that's soccer for you Americans) team plays better at home than it does away. One year, in order to qualify for the finals they have to beat a team from across the country twice (at least) in three matches. They are given a choice of playing either
Home then Away then Home
or
Away then Home then Away
Naturally they select the first option in order to play twice at home.

Next year however the rules change. They have to win two games (at least) in a row in order to qualify. They are given the same choices. Which one should they choose?

Since there is a 50% chance of getting the right answer by guessing (and since the way the problem is posed seems to suggest what the correct answer is), I'll need some sort of explanation or calculation besides the answer.
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.)
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zeogold: I'm not sure how it matters, unless they have some greater chance of winning depending on where they're at.
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dtgreene: I actually think the problem is poorly stated, as it doesn't explain what the difference between playing at home versus away is.
Hmmm... I really thought the first sentence explains the difference:

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ZFR: A football (that's soccer for you Americans) team plays better at home than it does away.
So yes, it means that probability of winning at home is bigger than winning away.
Edit: Nevermind. Adalia beat me to it.
Post edited January 31, 2016 by zeogold
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adaliabooks: Well I would go with the second one.
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
Post edited January 31, 2016 by ZFR
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dtgreene: 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.)
I'm going to take a stab at the answer: You should switch doors.

Am I right? :D
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dtgreene: 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.)
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Bookwyrm627: 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)
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adaliabooks:
Right. Now about your question...
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adaliabooks:
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ZFR: 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.
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bler144: 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. ;)
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 ;).
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zeogold: I could jump in to provide a question, if anybody needs me to. I can't guarantee it'll be all that hard, though.
Out with it already.
Post edited February 02, 2016 by ZFR
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zeogold: I could jump in to provide a question, if anybody needs me to. I can't guarantee it'll be all that hard, though.
Any time you're ready...
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dtgreene: How about this one:

A 9 Volt battery is connected to a superconducting wire (0 resistance) which is turn connected to a LED. The circuit is complete (that is, it is not broken).

How much current flows through the LED? (Hint: Use Ohm's Law.)
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.