Tuesday, October 11, 2016

The Peril of the Baseball Wild Card Game

I didn't even realize it until this year with the Toronto Blue Jays. Apparently baseball expanded the playoffs to five teams from four. Previously it was the three division winners and a wild card. Then it seems in 2012 the playoffs were expanded to a second wild card team, which plays the higher wild card team in a 1 game playoff.

The addition of a wild card game definitely changes the dynamics of the playoffs. There is more emphasis now on winning the division and avoiding the jeopardy of a 1 game playoff. Anything can happen in one game.

Right now the Blue Jays are in good shape having swept their division series. But they were in a lot of danger in the wild card round. They were only 1 game better than Baltimore during the season, they were 11-16 in September, and Baltimore had won 2 of 3 off Toronto in Toronto just a week before. I'd say going in this game was pretty much a toss up or maybe with home field Toronto was at best a slight favourite to win this game.

It did get me thinking about the 1, 5, and 7 game series. The wisdom is that a longer series favours the stronger team. How much of a cushion does the 5 and 7 game series provide the better team. With a couple of reasonable assumptions it's not too difficult to actually calculate it.

Suppose x is the chance that the team we are hoping for will win any given game over the other team. So x then is a real number between [0, 1].

We know then that the other team's chance to win any game in the series is (1 - x).

For a 1 game series it's easy. The chance of our team winning the series is x.

For a best 3 out of 5 it is a bit more to think about. Our team can win the series in 3, 4, or 5 games. If you notice that our team always wins the last game, then you don't really care about what happened in the earlier games, you just count how many ways to get to that point before the decisive game. We only have to determine the number of ways to win the series. In any other outcome the other team wins and we lose.

So for a 3 out of 5, the chance of our team winning the series is
f(x) = x3(6x2 -15x +10)

In a 4 out of 7, it's the same process, this time we can win in 4, 5, 6, or 7 games. In a 4 of 7, the chance our team wins is
f(x) = x4(35 - 84x + 70x2 - 20x3)

Remember x is the chance that we win any given game.

With the formulas, we can plug them into a spreadsheet and see how much if any the longer series helps the better team.

This table shows the percentage chance that our team wins the series of that length for a given x.

x 1 game 5 game 7 game
0 0.0 0.0 0.0
0.50 50.0 50.0 50.0
0.55 55.0 59.3 60.8
0.60 60.0 68.3 71.0
0.65 65.0 76.5 80.0
0.70 70.0 83.7 87.4
0.75 75.0 89.6 92.9
0.80 80.0 94.2 96.7
1 100.0 100.0 100.0

So the longer series in fact does help the stronger team and there are fewer upsets. This is especially true when the better team is 60% or more to win a given game. There are a couple of things to note about this table.

Going from a 5 to a 7 game series doesn't help the better team a whole lot. Only about 3 times more per hundred series than they would win a 5 game series. So basically when the NHL and NBA made the first round 7 games it was just greed to have more games, not to help ensure that the better team advanced.

Also we can see why baseball changed the world series from 9 games to 7. The extra games don't materially affect the "discovery" of which team is actually better.

We can quick check the equations with the known cases of x= 0, 1, ½. When x is 0 the team has no chance and f(x) is of course 0. When we plug in x=1 the team is a lock and of course wins every time and the chance to win the series is of course 100%, as they win every game.

When x is one half, then each game is a coin flip. Playing extra games doesn't change anything. The result f(x) for the 1, 5, and 7 game series is exactly 50%.


From this we can see that there are cases such as x=0, that the chances in a 5 or 7 game series is the same as a 1 game series. that is f(x) = x.

Now for f(x) = x then f(x) - x = 0. For the 3 of 5 series this is

x3(6x2 -15x +10) - x = 0

This is a quintic equation so there are 5 roots. I was wondering about the other two roots. Since we know 0, 1, and ½, we can factor them out and with not too much work determine the remaining quadratic. It comes to

x(x - 1)(x - ½)2(3x2 -3x -1) = 0

The quadratic formula can be plugged in to yield the other two roots (3 ± √21) / 6. The roots are all symmetrical about x = ½.

These other two roots are in decimal approximately -0.26 and 1.26. So real numbers, but outside the defined probability range [0, 1].

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