# problem solving with Conditional Probability

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?

• Trick question. The answer is 0 because the series ends when a team wins 4. So A will never get a chance to win six. Maybe you can restate this problem with consistent assumptions? – Michael R. Chernick Sep 20 '12 at 23:35
• Say that you asked what is the probability that A wins the series Look at each possible scenarios such as WWLWLW calculate the probability of that scenario and add up all the cases that lead to A getting 4 wins. That would be the probability that A wins because the scenarios are mutually exclusive. – Michael R. Chernick Sep 20 '12 at 23:41
• Assuming the results form a Markov chain (necessary for there to be a solution given just the conditional probability of the outcome given the previous games result and nothing about earlier games). For the WWLWLW scenario you just multiply the conditional probabilities which would be P(A wins game 1| A is playing at home) P(A wins game 2| A wins game 1 and A is at home)P(A loses game 3|A won game 2 and A is on the road)... – Michael R. Chernick Sep 20 '12 at 23:48
• @MichaelChernick To win in 6 games means that exactly 6 games are played. So the scenarios are specifically 6 long and must have 3W and 2L in the first 5 characters, followed by a W. This nails it down. – gt6989b Sep 20 '12 at 23:53
• @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. – Michael R. Chernick Sep 21 '12 at 3:55

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.

• For my taste, this is (at least) enough of a hint. From here, it is just reading the probabilities and getting home/away and win/lose previous right. – Ross Millikan Sep 21 '12 at 4:17
• Mine is a brute force approach that is not the best and is tedious but can be demonstrated in this case. – Michael R. Chernick Sep 21 '12 at 10:51