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I was wondering if there is a simpler or more straight forward way to go about solving probablity questions that envolve many events, without having to make gigantic tree diagrams or tables.

For example:

In a sports tourney between team A and B, 3 games will be played.
Victory earn 2 point, draw 1, and defeat 0.
Both teams being equal, what is the probability of:

a) Team A scoring 3 points in all games.
b) Both team finishing with the same total points.

Is there to do it or at least finding the number of favourable events?

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You don't have enough information to solve this problem. Since both teams are "equal", maybe they draw every time? Or maybe this is a game where a draw is very unlikely? The answer depends on this information, but the statement doesn't include it... – Jonathan Christensen Jan 23 '13 at 19:56
I interpreted as uniform probability of win, draw, or loss. – pre-kidney Jan 23 '13 at 20:03
IMHO, the tree size for this problemdoes not yet qualify as "gigantic". – Hagen von Eitzen Jan 23 '13 at 20:07
pre-kidney is right, it just means they all have the same chance of w/d/l. As for the size of the tree, it's still rather big. and it the number of games were, say 5, even more so. – T23 Jan 24 '13 at 10:38
up vote 1 down vote accepted

Taken your interpretation that probability of a team winning, losing and drawing is 1/3.

a) There are 3 games, each game has 3 endings, therefore 3^3 = 27 as the base.

A scores 3, could be only with 1 or with {0,1,2}
only draw is one case, 1+1+1
{0,1,2} has permutations 3! = 6   
So in total 7 cases A scores 3 points, probability is 7/27

b) let (x,y) denote points of A then B, following is probability.

(0,0) : 0
(1,1) : 0
(2,2) : 0
(3,3) : 7/27 by above
(4,4) : 0
(5,5) : 0
(6,6) : 0
sum is 7/27 

Maybe write down the probability space ($\Omega$,$F$,$P$), where $\Omega$ is the sample space, a family $F$ of subsets of $\Omega$, and a probability measure $P$. For a single event A, if the probability of events is uniform (each event happens with same probability), then use cardinality of A over cardinality of $\Omega$ to calculate the probability.

In problem a), $\Omega$ = {(a,b,c)}|a,b,c $\in${0,1,2}}, $F = 2^\Omega$,A ={(1,1,1),(0,1,2),(0,2,1),(1,0,2),(1,2,0),(2,0,1),(2,1,0)}

For a discrete random vector of more than 1 component, use joint distribution table in general, problem b) is a special case where some thoughts is enough.

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You can use the binomial distribution in such a case.

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Since there are three possible outcomes in the example he gives, the multinomial distribution would probably be more helpful. – Jonathan Christensen Jan 23 '13 at 19:57
Thanks for pointing this out. You're correct, we should be using a trinomial distribution. I just wanted to point the user in the right direction (that probability distributions should be used instead of tree diagrams) and I neglected to read the problem more carefully on my first pass. – pre-kidney Jan 23 '13 at 20:06

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