A chess competition in which <A> is the present champion
<B> and <C> are the challengers
Furthermore,following:
• 0.6. = probability that B will Beat C in any match
• 0.5. =probability that A will Beat B in any match
• 0.7. = probability that A will Beat C in any match
<B> AND <C> are made to play a two games against each others max
If
one of them wins both games,he gets to play a two-game
second round
with A ,the current champion
.A retains his championship unless a second round is required and the challenger
beats A in both games.
If A wins the initial game of the second round,no more games are needed
work out the probability that
A will retain his championship this year.
i want a direct approach
I mean { 1- P(A asCHAMP) - P(B asCHAMP) }
is also fine and gives the answers
but i need a direct method
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-spring-2005/recitations/tut01.pdf
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http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-spring-2005/recitations/tut01_sol.pdf
Rahul K Rai
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