[bushindo]

Here's a question inspired by the NBA finals. While waiting for the Lakers game to start, I made up this problem for fun.

Assume that there are four teams in semi-finals, Lakers, Nuggets, Cavaliers, and Magic. Lakers would play with the Nuggets in a best of 7 series. Cavaliers would also play with Magic in a best of 7. The winner of those semi-final rounds would advance to the final and compete in a best of 7 for the championship.

Lets say that each game played is a binomial event, with the following probability distribution per game for each team match-up

P( Laker win over Nugget ) = .6

P( Laker win over Magic ) = .65

P( Laker win over Cavalier) = .55

P( Nugget win over Magic ) = .45

P( Nugget win over Cavalier ) = .4

P( Cavalier win over Magic ) = .4

1) Lets say that the semi-finals just started. What is the chance that Lakers will win the championship? ( Bonus: give the championship chance for all other teams as well)

2) Let's start from the current team score, where Lakers are up 3-2 versus Nuggets, and Magic up 3-2 versus Cavalier. What is the chance that Lakers will win the championship? Chance that Cavaliers will win the championship? (Bonus: give the championship chance for the other two teams too)

Edited May 30, 2009 by bushindo

[plasmid]

First I'll describe an approach that does not use a shortcut:

You can describe the odds of winning a series between teams A and B as a function of the probability of team a winning a game (P(A)). You would also have to account for the number of different ways that teams A & B could win a specified number of games (e.g. team A sweeps it in 4, team B wins it in 7). This would NOT simply be _nC₄ = n! / 4! (n-4)! as you might expect. For example, ₆C₄ would include the combination: A wins games 1-4 and B wins games 5-6, but clearly there would never be a game 5 or 6 if A had won games 1-4. You would have to eliminate possibilities where the losing team wins the last game, which (if n>4) would be equal to the number of ways that team B could win (n-5) games (the total number of games it would win excluding the last game) in the preceeding (n-1) games in the series. For n>4, there would be _(n-1)C_(n-5) = (n-1)! / 4! (n-5)!

Putting that all together: the probability that A wins it in 4 would be

₄C₄ x P(A)⁴ = 1 x P(A)⁴

The probability that A wins it in 5 would be

(₅C₄ - ₄C₀) x P(A)⁴ x P(B) = (5-1) x P(A)⁴ x (1-P(A)) = 4 x (P(A)⁴ - P(A)⁵)

The probability that A wins in 6 would be

(₆C₄ - ₅C₁) x P(A)⁴ x P(B)² = (15-5) x P(A)⁴ x (1-P(A))² = 10 x (P(A)⁴ - 2 P(A)⁵ + P(A)⁶)

And the probability that A wins in 7 would be

(₇C₄ - ₆C₂) x P(A)⁴ x P(B)³ = (35-15) x P(A)⁴ x (1-P(A))³ = 20 x (P(A)⁴ - 3 P(A)⁵ + 3 P(A)⁶ - P(A)⁷)

So adding all of those probabilities to get the sum probability of team A taking the series would be

35 P(A)⁴ - 84 P(A)⁵ + 70 P(A)⁶ - 20 P(A)⁷

Now about that shortcut I mentioned. Suppose we lived in a hypothetical world where the NBA held all 7 games of the series regardless of whether or not someone had already clinched it with 4 wins. Clearly that should not affect the overall outcome of the series, but it might make the math slightly easier. The probability that A takes the series would be the sum of the probabilities of A winning all 7, 6 of 7, 5 of 7, or 4 of 7.

Probability of winning all 7

₇C₇ x P(A)⁷ = 1 x P(A)⁷

Probability of winning 6 of 7

₇C₆ x P(A)⁶ x P(B) = 7 x P(A)⁶ x (1-P(A)) = 7 x (P(A)⁶ -P(A)⁷)

Of winning 5 of 7

₇C₅ x P(A)⁵ x P(B)² = 21 x P(A)⁵ x (1-P(A))² = 21 x (P(A)⁵ - 2 P(A)⁶ + P(A)⁷)

Of winning 4 of 7

₇C₄ x P(A)⁴ x P(B)³ = 35 x P(A)⁴ x (1-P(A))³ = 35 x (P(A)⁴ - 3 P(A)⁵ + 3 P(A)⁶ - P(A)⁷)

So the sum is

35 P(A)⁴ - 84 P(A)⁵ + 70 P(A)⁶ - 20 P(A)⁷

Reassuringly, the two approaches give the same result.

So the probability that the Lakers win the championship is

P(Lakers beat Nuggets) x P(Cavs beat Magic) x P(Lakers beat Cavs) + P(Lakers beat Nuggets) x P(Magic beat Cavs) x P(Lakers beat Magic) = 0.71 x 0.29 x 0.61 + 0.71 x 0.71 x 0.8 = 0.125599 + 0.40328 = 0.528879

The probabilities for the other 3 could be calculated similarly.

For part 2, the probability that the team that's behind will win the series is P(B)², so the probability that the team that's ahead will win is 1-P(B)². Then

P(Lakers beat Nuggets) x P(Cavs beat Magic) x P(Lakers beat Cavs) + P(Lakers beat Nuggets) x P(Magic beat Cavs) x P(Lakers beat Magic) = 1-0.4² x 0.4² x 0.61 + 1-0.4² x 1-0.4² x 0.8 = 0.84 x 0.16 x 0.61 + 0.84 x 0.84 x 0.8 = 0.081984 + 0.56448 = 0.646464

[bushindo]

First I'll describe an approach that does not use a shortcut:

You can describe the odds of winning a series between teams A and B as a function of the probability of team a winning a game (P(A)). You would also have to account for the number of different ways that teams A & B could win a specified number of games (e.g. team A sweeps it in 4, team B wins it in 7). This would NOT simply be _nC₄ = n! / 4! (n-4)! as you might expect. For example, ₆C₄ would include the combination: A wins games 1-4 and B wins games 5-6, but clearly there would never be a game 5 or 6 if A had won games 1-4. You would have to eliminate possibilities where the losing team wins the last game, which (if n>4) would be equal to the number of ways that team B could win (n-5) games (the total number of games it would win excluding the last game) in the preceeding (n-1) games in the series. For n>4, there would be _(n-1)C_(n-5) = (n-1)! / 4! (n-5)!

Putting that all together: the probability that A wins it in 4 would be

₄C₄ x P(A)⁴ = 1 x P(A)⁴

The probability that A wins it in 5 would be

(₅C₄ - ₄C₀) x P(A)⁴ x P(B) = (5-1) x P(A)⁴ x (1-P(A)) = 4 x (P(A)⁴ - P(A)⁵)

The probability that A wins in 6 would be

(₆C₄ - ₅C₁) x P(A)⁴ x P(B)² = (15-5) x P(A)⁴ x (1-P(A))² = 10 x (P(A)⁴ - 2 P(A)⁵ + P(A)⁶)

And the probability that A wins in 7 would be

(₇C₄ - ₆C₂) x P(A)⁴ x P(B)³ = (35-15) x P(A)⁴ x (1-P(A))³ = 20 x (P(A)⁴ - 3 P(A)⁵ + 3 P(A)⁶ - P(A)⁷)

So adding all of those probabilities to get the sum probability of team A taking the series would be

35 P(A)⁴ - 84 P(A)⁵ + 70 P(A)⁶ - 20 P(A)⁷

Now about that shortcut I mentioned. Suppose we lived in a hypothetical world where the NBA held all 7 games of the series regardless of whether or not someone had already clinched it with 4 wins. Clearly that should not affect the overall outcome of the series, but it might make the math slightly easier. The probability that A takes the series would be the sum of the probabilities of A winning all 7, 6 of 7, 5 of 7, or 4 of 7.

Probability of winning all 7

₇C₇ x P(A)⁷ = 1 x P(A)⁷

Probability of winning 6 of 7

₇C₆ x P(A)⁶ x P(B) = 7 x P(A)⁶ x (1-P(A)) = 7 x (P(A)⁶ -P(A)⁷)

Of winning 5 of 7

₇C₅ x P(A)⁵ x P(B)² = 21 x P(A)⁵ x (1-P(A))² = 21 x (P(A)⁵ - 2 P(A)⁶ + P(A)⁷)

Of winning 4 of 7

₇C₄ x P(A)⁴ x P(B)³ = 35 x P(A)⁴ x (1-P(A))³ = 35 x (P(A)⁴ - 3 P(A)⁵ + 3 P(A)⁶ - P(A)⁷)

So the sum is

35 P(A)⁴ - 84 P(A)⁵ + 70 P(A)⁶ - 20 P(A)⁷

Reassuringly, the two approaches give the same result.

So the probability that the Lakers win the championship is

P(Lakers beat Nuggets) x P(Cavs beat Magic) x P(Lakers beat Cavs) + P(Lakers beat Nuggets) x P(Magic beat Cavs) x P(Lakers beat Magic) = 0.71 x 0.29 x 0.61 + 0.71 x 0.71 x 0.8 = 0.125599 + 0.40328 = 0.528879

The probabilities for the other 3 could be calculated similarly.

For part 2, the probability that the team that's behind will win the series is P(B)², so the probability that the team that's ahead will win is 1-P(B)². Then

P(Lakers beat Nuggets) x P(Cavs beat Magic) x P(Lakers beat Cavs) + P(Lakers beat Nuggets) x P(Magic beat Cavs) x P(Lakers beat Magic) = 1-0.4² x 0.4² x 0.61 + 1-0.4² x 1-0.4² x 0.8 = 0.84 x 0.16 x 0.61 + 0.84 x 0.84 x 0.8 = 0.081984 + 0.56448 = 0.646464

Well done, plasmid. Your logic is spot on. I like your shortcut, it is a novel way to think about probability, at least for me. Posts like yours are the reason why I come here.