Expert: Josh - 5/17/2009

Question
Team A is competing against Team B in a best-of-7 series (whoever wins 4 games). Games #1,2,5 and 7 are at Team A location (home), games #3,4 and 6 are at Team B location (away). Assume Team A has a 67% probability of winning each of their home games, and a 50% probability of winning their away games. How do I determine the probabilities of Team A winning the series, winning in exactly 5 games, and winning in exactly 6 games? I can only calculate Team A winning in 4 games- 11.2225% (.67x.67x.5x.5).

Answer
Hi Jonathon,

This is going to be more like a tutorial than regular Q&A. Before we discuss the problem in detail, let's introduce some notations.

1) Use "H" and "A" to denote home and away games.
2) Concept of combination: The number of ways for selecting (winning) k of N (games) is given by C(N,k)=N!/[k!(N-k)!], where N!=N*(N-1)*(N-2)*...*1 is called N factorial. By definition, 0!=1.
3) Let P(E) represents the probability that an event E occurs. [Note: we can think of E as a particular win/loss sequence, e.g., WWLLWLW]

To simplify our analysis, we reorder the games without loss of generality. Even though games 1,2 5 and 7 are actual home games, we reorder {1,2,3,4,5,6,7} as {1,2,5,7,3,4,6} so that the season is partitioned nicely into a block of four H games and a block of three A games. Specifically, the set S(Home) consists of {1,2,5,7} and the set S(Away) consists of {3,4,6}.

Team A can win the series by winning 4, 5, 6 or 7 games. We consider each scenario one by one.

============================
Scenario 1: 4 wins, 3 losses

There are four distinct possibilities. Using (xH,yA) to represent "x" victories at home and "y" victories away from home, the four possibilities may be summarized neatly as:
(4H,0A), (3H,1A), (2H,2A) and (1H,3A).

Our general strategy may be divided into two parts. First, count how many ways that each combination (xH,yA) can occur. Second, calculate the probability associated with each event.

For instance, with (3H,1A), three wins from four home games {1,2,5,7} can happen in C(N,k)=C(4,3)=4!/[3!1!]=4 ways. [If you are not convinced, you can enumerate all possibilities -- these include, in this instance, winning game "1,2,5", "1,2,7", "1,5,7" or "2,5,7". But the formula always work as long as you know what you are doing]. In addition, the team must get 1 win from three away games {3,4,6}, this is equivalent to a "choose k=1 from N=3 problem". For this, there are C(3,1)=3 different combinations. Overall, the event (3H,1A) can happen in C(4,3)*C(3,1)=4*3=12 ways.

Alright, enough of the explanation. Now, we must consider the remaining three events. I'll leave this as an exercise for you. Once this is done you should be able to fill the following table, which is incomplete.

Table 1: 4 wins, 3 losses
event E  |  number of combinations N(E) | associated probability P(E)
(1H,3A)  |  C(4,1)*C(3,3)= ??           |
(2H,2A)  |  C(4,2)*C(3,2)= ??           |
(3H,1A)  |  C(4,3)*C(3,1)= 12           |
(4H,0A)  |  C(4,4)*C(3,0)= ??           |

Now, we deal with the second aspect. Consider the case of 3 home victories and 1 victory away from home. We figured that (3H,1A) can happen in 12 ways. So, what is the chance that this will occur in a single realization. For instance, what is the probability of winning game "1,2,5,6" in a given tournament. Note, this carries the same probability as any of the eleven combinations, such as (winning game) "1,2,7,6", or "2,5,7,4" and so forth. This probability is given by [P(H)^3*P(A)] * [1-P(H)]*[1-P(A)]^2 = (0.67)^3(0.5) * (0.33)*(0.5)^2. This may be understood in two parts: [P(H)^3*P(A)] has to do with WINNING 3 home and 1 away games whereas [1-P(H)]*[1-P(A)]^2 has to do with LOSING 1 home and 2 away games. This formula is valid provided the outcome for each game is independent of previous games.

IMPORTANT: We will call L(3H,1A)=(0.67)^3(0.5)*(0.33)*(0.5)^2 the likelihood for a single realization of "3 home and 1 away victories". Referring to the last column in Table 1, the associated probability P(E) = N(E)*L(E). viz., the probability associated with a particular event -- such as E=(1H,3A) -- is given by the number of ways that E can occur, multiplied by the likelihood that E can occur in a single observation.

e.g., P(3H,1A)=N(3H,1A)*L(3H,1A)=12*(0.67)^3*(0.5)*(0.33)*(0.5)^2=0.14887768...

e.g., P(2H,2A)=N(2H,2A)*L(2H,2A), where
N(2H,2A)=C(4,2)*C(3,2)=6*3=18 and
L(2H,2A)=(0.67)^2*(0.5)^2*(0.33)^2*(0.5)^2=0.003055326
Thus, P(2H,2A)=0.05499586...

TO DO: Your task now is to calculate the associated probability for each event, P(E), in Table 1 for the case of "4 wins, 3 losses".

Then, repeat for the case of "5 wins, 2 losses" in Table 2.
Repeat for the case of "6 wins, 1 loss" in Table 3.
Repeat for the case of "7 wins, 0 loss" in Table 4.

Table 2: 5 wins, 2 losses
event E  |  number of combinations N(E) | associated probability P(E)
(2H,3A)  |  C(4,2)*C(3,3)= ??           |             ??
(3H,2A)  |  C(4,3)*C(3,2)= ??           |             ??
(4H,1A)  |  C(4,4)*C(3,1)= ??           |             ??

Table 3: 6 wins, 1 losses
event E  |  number of combinations N(E) | associated probability P(E)
(3H,3A)  |  C(4,3)*C(3,3)= ??           |             ??
(4H,2A)  |  C(4,4)*C(3,2)= ??           |             ??

Table 3: 7 wins, 0 losses
event E  |  number of combinations N(E) | associated probability P(E)
(4H,3A)  |  C(4,4)*C(3,3)= 1            |      (0.67)^4*(0.5)^3

The probability that team A wins the series is given by the sum of all the P(E) for all the events.

I am going to give you some of the answers and encourage you to work out the rest for yourself.

ANSWERS: (some parts are omitted for you to do)
E | N(E) | L(E) | P(E)
Table 1:
(3H,1A)   12 0.012406474  0.148877685
(1H,3A)   4  0.003009724  0.012038895
Table 2:
(4H,1A)   3  0.025188901  0.075566704
(2H,3A)   6  0.006110651  0.036663908
Table 3:
(3H,3A)   4  0.012406474  0.049625895

Final probability: I think the sum of all the P(E)'s is 0.652591