# how to calculate the real probability

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

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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