Simple probability question

umair

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Guessball is a team sport requiring x players on each side. Alex and I are part of 2x players drawn randomly. What's the probability of being on same side?

a) (x-1)/2x
b) 1/2
c) More info needed
d) (x-1)/(2x-1)
e) x/(2x-1)
 
Guessball is a team sport requiring x players on each side. Alex and I are part of 2x players drawn randomly. What's the probability of being on same side?

a) (x-1)/2x
b) 1/2
c) More info needed
d) (x-1)/(2x-1)
e) x/(2x-1)
We need to see your work to know how to help..

The choices for each of you are "random" - perhaps you could think of flipping a coin? What is the chance that two flips are the same, either HH or TT? Is that easier than you thought?

EDIT: this was an oversimplification, leading to the wrong answer! See post by JeffM below.
 
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We need to see your work to know how to help..

The choices for each of you are "random" - perhaps you could think of flipping a coin? What is the chance that two flips are the same, either HH or TT? Is that easier than you thought?

Very elegant solution!!

Here I was going (2x-2)C(x-1) and all such complications ..... and staring at the result and saying "that can't be right ...."
 
The way I figured it we could choose Team A x B, B x A, A x A, B x B = 1/2

Is that right?
 
The way I figured it we could choose Team A x B, B x A, A x A, B x B = 1/2

Is that right?
Dr. Phil and Subhotosh. Please check: I am not coming up with 1/2 as the proper answer.

P(Alex will be chosen by team 1) = \(\displaystyle \dfrac{x}{2x} = \dfrac{1}{2}.\)

P(I will be chosen by team 1 given Alex was chosen by team 1) = \(\displaystyle \dfrac{x - 1}{2x - 1}.\)

P(Alex and I will both be on team 1) = \(\displaystyle \dfrac{x - 1}{2x - 1} * \dfrac{1}{2} = \dfrac{x - 1}{2(2x - 1)}.\)

P(Alex will be chosen by team 2) = \(\displaystyle \dfrac{x}{2x} = \dfrac{1}{2}.\)

P(I will be chosen by team 2 given Alex was chosen by team 2) = \(\displaystyle \dfrac{x - 1}{2x - 1}.\)

P(Alex and I will both be on team 2) = \(\displaystyle \dfrac{x - 1}{2x - 1} * \dfrac{1}{2} = \dfrac{x - 1}{2(2x - 1)}.\)

P(Alex and I will be on same team) = P(both on team 1) + P(both on team 2) =

\(\displaystyle \dfrac{x - 1}{2(2x - 1)} +\dfrac{x - 1}{2(2x - 1)} = \dfrac{2(x - 1)}{2(2x - 1)} = \dfrac{x - 1}{2x - 1}.\)

If I am analyzing correctly, we are not dealing with independent events.
 
Dr. Phil and Subhotosh. Please check: I am not coming up with 1/2 as the proper answer.

P(Alex will be chosen by team 1) = \(\displaystyle \dfrac{x}{2x} = \dfrac{1}{2}.\)

P(I will be chosen by team 1 given Alex was chosen by team 1) = \(\displaystyle \dfrac{x - 1}{2x - 1}.\)

P(Alex and I will both be on team 1) = \(\displaystyle \dfrac{x - 1}{2x - 1} * \dfrac{1}{2} = \dfrac{x - 1}{2(2x - 1)}.\)

P(Alex will be chosen by team 2) = \(\displaystyle \dfrac{x}{2x} = \dfrac{1}{2}.\)

P(I will be chosen by team 2 given Alex was chosen by team 2) = \(\displaystyle \dfrac{x - 1}{2x - 1}.\)

P(Alex and I will both be on team 2) = \(\displaystyle \dfrac{x - 1}{2x - 1} * \dfrac{1}{2} = \dfrac{x - 1}{2(2x - 1)}.\)

P(Alex and I will be on same team) = P(both on team 1) + P(both on team 2) =

\(\displaystyle \dfrac{x - 1}{2(2x - 1)} +\dfrac{x - 1}{2(2x - 1)} = \dfrac{2(x - 1)}{2(2x - 1)} = \dfrac{x - 1}{2x - 1}.\)

If I am analyzing correctly, we are not dealing with independent events.
OOPS - I fear JeffM is right! Whereas the probability that Alex will be on team A is exactly 1/2, we must use CONDITIONAL probability for which team you will be on. Since one of the positions on team A is already taken, your probability of being one of the (x-1) players on the same team, out of (2x-1) yet to be chosen, is (x-1)/(2x-1).

Sorry about my oversimplification. Thanks to JeffM for keeping us precise!
 
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It's not necessary to talk divide this into "team A" and "team B". I am sure to be on a team. That leaves 2n-1 people to be chosen from and n-1 places on my team. The probability that any one of those 2n- 1 persons will be selected for one of the n-1 places on my team is \(\displaystyle \dfrac{n-1}{2n-1}\).
 
It's not necessary to talk divide this into "team A" and "team B". I am sure to be on a team. That leaves 2n-1 people to be chosen from and n-1 places on my team. The probability that any one of those 2n- 1 persons will be selected for one of the n-1 places on my team is \(\displaystyle \dfrac{n-1}{2n-1}\).
Elegant.
 
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