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Say I have a 50% probability of winning a game if I play against person 1, and a 50% probability of winning a game against person 2. I will play with both people, one after another. Before the matches, what was the probability that I would win at least 1 match? It has to be something between 90-99% I think. Any suggestions? Thank you!

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How many trials are you doing? Are the trials independent? –  Narut Sereewattanawoot May 13 '13 at 20:50

3 Answers 3

up vote 5 down vote accepted

Note: as Austin Mohr's helpful follow-up comment explains, the answer relies on the fact that the likelihood of a win and a loss are the same: we have $50\%$ chance of a win $= 50\%$ chance of a loss in each game, so every outcome is equally likely.

W: win... L: lose

Four possible outcomes: (Game 1 vs. person 1) followed by (Game 2 vs. person 2)

W W <--

W L <--

L W <--

L L

In $3$ of $4$ outcomes, you win one or both (at least one) of the two games. In only one outcome, will you lose both (and hence not win at least once).

Therefore: The probability of winning at least one game: $P(\text{win at least one game}) = \dfrac 34 = 75\%$

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thank you for your help, really appreciate it –  user1505027 May 13 '13 at 20:58
    
You're welcome! –  amWhy May 13 '13 at 21:03
    
Nice write - up +1 –  Amzoti May 14 '13 at 0:44

person 1 person 2
win win
person 1 person 2
win lose
person 1 person 2
lose win
person 1 person 2
lose lose

I see the answer being 75% for winning at least one match.

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If you are playing just two games, then I think the probability is given by the complement of the probability that you don't win either match, i.e.:

$P$(at least 1) = $1 - (1-.5)(1-.5) = 1 - (.5)^{2} = .75%.$

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thank you for your help, too, really appreciate it –  user1505027 May 13 '13 at 20:58
    
My pleasure! :) –  AWertheim May 13 '13 at 21:04

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