Playoff Probabilities
At certain times of the year attention turns to a playoff series where two teams engage in a bestofthree or bestoffive or even a bestofseven series.
The normal approach of using a probability tree can be used, but with a slight modification to the usual tree which has all the terminal branches at the same level. In the case of a playoff scenario some branches are shorter than others. As an example, a bestofthree series can end after two games or after three games.
It turns out this raises no special difficulties—we need to make certain the sum of the probabilities of all the terminal branches regardless of their level is still one.
First, we need to compute the probability that a given team will win. Generally, we known the winning percentage for both teams—given as w_{A} and w_{B} for the two teams, A and B respectively.
The simple appropriate probability for team A to win a single game is then:
This probabilty does not take into account other factors such as "homefield" (or "homecourt") advantage, or the relative strengths of the two teams's opponents during the regular season, or special strengths and weaknesses of the two teams in a headtohead situation.
Note: if both teams have the same winning percentage, then this formula gives p = 1/2 as should be expected.
In the unlikely situation that the winning percentage for team B is 0.000, then p = 1.

The probability of team A losing (which is the same as team B winning) a single game is q = 1 − p.
BestofThree
A bestofone is a rather simple case. The probability that team A wins is just p.
Therefore, the first "nontrivial" case is the bestofthree playoff series. A tree for this looks like:
game 1  game 2  game 3  outcome  probability 

A wins p  A wins p  —  team A wins series  p^{2} 
B wins q  A wins p  team A wins series  p^{2}q 
B wins q  team B wins series  pq^{2} 
B wins q  A wins p  A wins p  team A wins series  p^{2}q 
B wins q  team B wins series  pq^{2} 
B wins q  —  team B wins series  q^{2} 
There are 6 different playoff series branches or scenarios. Three of these have team A winning and three have team A losing the series.
Notice that the probability for each branch contains a factor of p^{k} where team A wins exactly k times and a factor of q^{m} where team B wins exactly m times.
Further, the probabilities for each scenario that results in the same overall outcome (e.g. team A wins series 21 in 3 games) are the same.
Thus, the bestofthree table becomes:
team A wins series 20 in 2 games  1 scenario  P = p^{2} 
team A wins series 21 in 3 games  2 scenarios  P = 2p^{2}q 
team A loses series 12 in 3 games  2 scenarios  P = 2pq^{2} 
team A loses series 02 in 2 games  1 scenario  P = q^{2} 
The probability that team A wins the series is the sum of the terminal branches that have team A winning:
P_{A} = p^{2} + p^{2}q + p^{2}q = p^{2} + 2p^{2}q = p^{2}(1 + 2q)
And using q = 1 − p:
P_{A} = = p^{2}(3 − 2p)
If you add the probabilities of team A winning and team A losing (which is the probability of team B winning replacing p with q and vice versa), you obtain:
P_{A} + P_{B} = p^{2}(1 + 2q) + q^{2}(1 + 2p)
With q = 1 − p and a little algebra, this can be shown to be exactly 1 regardless of p!
BestofFive
The bestoffive table is similar except there are now additional winning scenarios and a total of 20 possible different series of which 10 are a series win for team A. Listing only the winning branches for team A:
team A wins series 30 in 3 games  1 scenario  P = p^{3} 
team A wins series 31 in 4 games  3 scenarios  P = 3p^{3}q 
team A wins series 32 in 5 games  6 scenarios  P = 6p^{3}q^{2} 
Thus the probability of team A winning the series is given by:
P_{A} = p^{3}(1 + 3q + 6q^{2})
BestofSeven
The bestofseven table is similar except there are now additional winning scenarios and a total of 70 possible different series. The 35 winning scenarios for team A are:
team A wins series 40 in 4 games  1 scenario  P = p^{4} 
team A wins series 41 in 5 games  4 scenarios  P = 4p^{4}q 
team A wins series 42 in 6 games  10 scenarios  P = 10p^{4}q^{2} 
team A wins series 43 in 7 games  20 scenarios  P = 20p^{4}q^{3} 
Thus the probability of team A winning the series is given by:
P_{A} = p^{4}(1 + 4q + 10q^{2} + 20q^{3})
One factor is common in all the probabilities. The p^{4} appears in all the probabilities—this is the probability of team A winning 4 games overall (and hence winning the series).
Further, the power of q corresponds to the number of games team A loses (or team B wins).
The coefficients are all related to the combination or different ways of winning r games out a total of n games.
If you look at the values in Pascal's Triangle, you will find the coefficients above quite prominent.
Keep in mind that for team A to win the series, team A will win the last game played; therefore, we need to examine the combinations involved for all but the last game.
As a specific example, for a 41 series outcome there are 4 ways team A could lose a game in the first 4 games which is _{4}C_{1}.
In a 42 series outcome there are 10 ways team A could lose 2 games in the first 5 games—and the combination _{5}C_{2} = 10.
Finally, for the 43 series outcome we note _{6}C_{3} = 20.
BestofNine
Bestofnine is not one of the more common series; however, we can still work out the probabilities.
The table is similar to the simplier ones except there are now additional winning scenarios and a total of 252 possible different series. The 126 winning scenarios for team A are:
team A wins series 50 in 5 games  1 scenario  P = p^{5} 
team A wins series 51 in 6 games  5 scenarios  P = 5p^{5}q 
team A wins series 52 in 7 games  15 scenarios  P = 15p^{5}q^{2} 
team A wins series 53 in 8 games  35 scenarios  P = 35p^{5}q^{3} 
team A wins series 54 in 9 games  70 scenarios  P = 70p^{5}q^{4} 
Thus the probability of team A winning the series is given by:
P_{A} = p^{5}(1 + 5q + 15q^{2} + 35q^{3} + 70q^{4})
These coefficients come from the number of different ways team B could win in n−1 games.
As an example, for team B to win 3 games in 7 (on the way to a team A's 53 in 8 games series win), the number of ways of acheiving this is _{7}C_{3} = 35.
Example
The 2009 New York Yankees ended the season with a record of 103 wins and 59 losses for a winning percentage of 0.636,
and their opponent in the American League Division Series was the Minnesoto Twins who compiled a record of 87 wins and 76 losses (the Twins needed one extra game beyond the normal 162 to determine the division winner) for a winning percentage of 0.534.
The probability for the Yankees winning a single game is p = 0.636/(0.636+0.534) = 0.5435, and the Twins winning is therefore q = 0.4564.
They play a bestoffive playoff series.
Yankees win series 30 in 3 games  P = p^{3} = 0.1606 
Yankees win series 31 in 4 games  P = 3p^{3}q = 0.2199 
Yankees win series 32 in 5 games  P = 6p^{3}q^{2} = 0.2008 
Therefore, the total probability that the Yankees win the series is the sum: P_{Y} = 0.581 (and a probability of losing 0.419).
More Fun
Things really get more interesting after the first game is played. This trims the complete playoff tree in half and makes the number of branches of the remaining tree one less as well as changing the number of wins (of the remaining games) required for a given team to win the series. It actually simplifies the analysis!
And, of course, you can construct a series tree after the second game is played, and after the third game, and .....
Continuing the example, the Yankees actually won the first game. The remaining tree contains 10 branches of which 6 constitute a Yankees series win.
(Recall that the original tree had 20 branches with 10 having the Yankees win the series.)
Yankees win 2 of 2 and win series 30 in 3 games  1 scenario  P = p^{2} = 0.2955 
Yankees win 2 of 3 and win series 31 in 4 games  2 scenarios  P = 2p^{2}q = 0.2697 
Yankees win 2 of 4 and win series 32 in 5 games  3 scenarios  P = 3p^{2}q^{2} = 0.1847 
Given the fact that the Yankees have won the first game, the probability they win the series is thus P_{Y} = 0.750.
As an aside, if the Twins had won the first game, the remaining tree has 10 branches of which only 4 represent a series win for the Yankees.
The probability for a Yankees series win is reduced to P_{Y} = p^{3}(1+3q) = 0.381.

The Yankees did indeed win the second game! The remaining tree has only 4 branches with 3 ending in a series win for the Yankees.
Yankees win 1 of 1 and win series 30 in 3 games  1 scenario  P = p = 0.5436 
Yankees win 1 of 2 and win series 31 in 4 games  1 scenario  P = pq = 0.2481 
Yankees win 1 of 3 and win series 32 in 5 games  1 scenario  P = pq^{2} = 0.1132 
Therefore, since the Yankees won the first two games, the probability they win the series becomes P_{Y} = 0.905.
If the Twins had won the second game and thereby tied the series (or if the Twins had won the first game and the Yankees the second game), then the series would be reduced to a bestofthree situation, and we know how to compute that.
There are 6 branches to the tree, and the Yankees win the series for 3 of these branches.
The probability for a Yankees series win becomes P_{Y} = p^{2}(1+2q) = 0.565.

By the way, if the Twins had won the first two games, then the Yankees would need to win the next three games without any additional loss to win the series.
The probability for a Yankees series win becomes P_{Y} = p^{3} = 0.161.

The Yankees won the third game to sweep the series!
On the other hand, if the Twins had won the third game (or any scenario that led to the Yankees leading the series 2 games to 1), then the remaining tree has only 3 branches with 2 branches for the Yankees to win the series.
The probability for a Yankees series win becomes P_{Y} = p(1+q) = 0.792.

Note that if you ever get to the last game of the series (or any series) with the series tied; that is, the tree has only two branches left representing the last game, then the probability of the Yankees winning the game and the series is just P_{Y} = p = 0.5436!
