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Originally Posted by battists
So, do we think the game should attempt to mimic the existing MLB version?
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To a point, yes. The use of head-to-head records to decide which club gets designated what in a three-way tie need not be used though, just do it randomly (which is what MLB does in the case of two-way and four-way ties; I don't know why they seed three-way ties using head-to-head record).
I was going to post this little treatise on the Beta forums, but I'll do it here. Consider this my beginner's primer on tiebreaking playoff game procedures which mostly follow MLB's rules but have been somewhat simplified where necessary.
Single elimination scenarios
(one loss and a team is eliminated)
Two team tie
The teams are randomly designated A and B. The playoff game is then:
Day 1: B at A
Three team tie
The teams are randomly designated A, B, and C. The playoff games are then:
Day 1: B at A
Day 2: C at A/B
Four team tie
The teams are randomly designated A, B, C, and D. The playoff games are then:
Day 1: B at A, D at C
Day 2: C/D at A/B
Five team tie
The teams are randomly designated A, B, C, D, and E. The playoff games are then:
Day 1: B at A, D at C
Day 2: E at A/B
Day 3: C/D at A/B/E
Double elimination scenarios
(two losses and a team is eliminated)
Double elimination playoffs are the way MLB used to resolve ties before divisional play was introduced in 1969. Note that from about 1929 up to 1956 the AL used the single elimination method; in other years it used the double elimination method. The NL used the double elimination method right up until 1969.
Two team tie
The teams are randomly designated A and B. The playoff games are then:
Day 1: A at B
Day 2: B at A
Day 3: B at A*
*if necessary
Three team tie
The teams are randomly designated A, B, and C. The first three playoff games are:
Day 1: B at A
Day 2: C at B
Day 3: A at C
After these three games, there are two main possibilities. The first is that one of the teams may have lost twice, in which case it is eliminated and the remaining two teams continue to play the tiebreaker; the other is that all of the teams may have won once and lost once in which case all three clubs continue on in the tiebreaker.
If one of the teams has been eliminated after the first three games, the possible matchups are:
If A was eliminated:
Day 4: B at C
Day 5: C at B*
If B was eliminated:
Day 4: C at A
Day 5: A at C*
If C was eliminated:
Day 4: B at A
Day 5: A at B*
*if necessary
If none of the teams have been eliminated after the first three games, then the remaining games are:
Day 4: B at A
Day 5: C at A/B
Four team tie
The teams are randomly designated A, B, C, and D. The playoff games are then:
Day 1: A at B, C at D
Day 2: B at A, D at C
Day 3: B at A*, D at C*
Day 4: A/B at C/D
Day 5: C/D at A/B
Day 6: C/D at A/B*
*if necessary
I'll discuss the permutations in regards to wildcard and division ties in a three-division league in a subsequent post.