Team $A$ facing Team $B$. Team $A$ Hosts the first and Second and if needed fifth and seventh games in the best of seven contest. Team $A$ has a 65% chance of beating Team $B$ at home in the first game. After that they have a 60% chance of a win at home if they won the previous game but a 70% chance if they are bouncing back from a loss. Similarly team $A$ chances of victory at Team $B$ are 40% after a win and 45% after a loss. What is the probability that Team $A$ will win in exactly 6 games?
problem solving with Conditional Probability
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probability
problem-solving
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0@gr6989b I am sorry. I misread your question. I was sure it said "win 6 games" when it said "win in 6 games". So you are asking for the probability of the cases where the 4th win of A is in game six. – 2012-09-21
1 Answers
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The sequences that lead to a 4th win by A in game 6 are
WWWLLW 1
WWLWLW 2
WWLLWW 3
WLWLWW 4
WLWWLW 5
WLLWWW 6
LWWWLW 7
LWWLWW 8
LWLWWW 0
LLWWWW 10
These are 10 sequences because the 6th slot is fixed as a win and the arrangements of the other 5 slots is $^5$C$^3$ =5! /(3! 2!)=5 (4)/2 = 10.
The game H = home for team A and R = road for team A sequence for 6 games HHRRHRH Then assuming the Markov property these 10 probabilities are
p(1)= (.65)(.60)(.40)(.60)(.30)(.45)=0.012636 based on the stated conditional probabilties for wins and loss on home and road games after a win and after a loss and the probability of a win in the first game.
p(2)= (.65)(.60)(.60)(.45)(.40)(.45)=0.018954
and you can do the rest.
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0Mine is a brute force approach that is not the best and is tedious but can be demonstrated in this case. – 2012-09-21