The prevailing argument on AN recently has been, unsurprisingly, "are the A's good enough to contend?"
Those labeled as the "optimists" are the ones who say "yes."
Those labeled as the "pessimists" are the ones who say "no."
The truth is that everybody on AN knows that there is a certain probability that the A's will make the playoffs and a certain probability that they won't. In fact, it seems to me that a fair amount of this argument has been entirely semantic, in that if two people were to see that the A's had a 10% chance of making the playoffs, one person might say "the A's are unlikely to make the playoffs" and the other might say "the A's might make the playoffs if things break right!" Of course these are both true, and the difference between these two outlooks actually does depend on whether or not the poster wants to be optimistic or realistic (I'm not going to suggest a false equivalence between optimism at 10% and pessimism at 10% — obviously the "pessimist" here is more likely to be right. A pessimist would be someone who sees as 50% chance of winning and says "the A's probably won't make the playoffs.")
The point here is people should indeed allowed to be as pessimistic or optimistic as they choose, but to make the choice, they should have some grasp of the probability of their desired outcome — i.e. the A's making the playoffs.
The truth is, a really solid formula for playoff odds (like Baseball Prospectus) requires a lot of different variables, or some sort of game simulator, neither of which I (or the regular person around) has access to.
Instead what I will give you is a fairly simple method of calculating playoff odds from a basic set of numbers. The idea here is that, though far from perfect, this methodology will provide a simple total probability of the A's chances of winning the division (I'm leaving the Wild Card out, because it would require a much longer formula and I think the A's chances of winning the WC are pretty slim, but if anybody else would like to do it I encourage them, and I recognize it as a flaw in this formula. Those of you who think the final probability I come up with is too low, for instance, feel free to suggest that it should be higher because the A's might win the wild card.). This formula will also have a fairly basic set of input numbers which are up for debate at all times, and by changing, you can alter the final probability of the A's making the playoffs.
Without any further ado, follow me to after the jump.
For the sake of simplicity in a baseline, I'm going to use an equal distribution of variance, meaning that the "true talent" of the A's becomes the mean outcome, and there is the same likelihood (10% here, because 10 is an easy number to do calculations with) of them being at exactly that many wins as being 1/10 of a standard deviation away, etc. If that didn't make sense, see what I mean by looking at the numbers below this paragraph. But first we need to calculate a standard deviation. In accordance with the binomial theorem, the standard deviation should be calculated as follows: sqrt((win%)*(1-win%)*162)). The number I'm going to be using for the A's true talent is 84 wins, which means an 84-78 record, or a .516 winning percentage. (Let me be absolutely clear—these numbers are variables. The purpose of this fanpost is to provide not only a basic projection based on my own numbers, but to give everyone the chance to argue about talent levels in the AL West and have an easy method of plugging those numbers into a formula to have a nice easy percentage possibility of winning the west, so you can see 20% and think, hey that's not that unrealistic! or see 25% and say, oh, damn, we're probably gonna lose, etc.) Alright, now using the projected .516 winning percentage, the standard deviation comes out to: 6.36. Therefore an 84 win true talent team has a single standard deviation range of 78-90 wins.
But, according to this comment from Colin Wyers at Baseball Prospectus:
"Even if we are correctly predicting everything – everybody’s raw hitting and pitching stats, everyone’s playing time, etc. – correctly, there is still some random variation. How much? I mean, a lot. The standard deviation of win percentage over 162 games simply due to random variation (or “luck,” if you prefer) is a little over six games. Events within one standard deviation occur 68% of the time.
Moreover, events only fall within two standard deviations 95% of the time.
So, I promised to break these numbers into even percentiles, which I will do:
If the A's are an 84 win true talent team, and there is a 68% chance that they fall within one standard deviation at 78-90 wins, staying true to an even variance within each SD, that means the following:
- 78 wins: 5.2%
- 79 wins: 5.2%
- 80 wins: 5.2%
- 81 wins: 5.2%
- 82 wins: 5.2%
- 83 wins: 5.2%
- 84 wins: 5.2%
- 85 wins: 5.2%
- 86 wins: 5.2%
- 87 wins: 5.2%
- 88 wins: 5.2%
- 89 wins: 5.2%
- 90 wins: 5.2%
The 5.2% comes from 68/13 (78 through 90 is 13 different outcomes). Once again, an equal distribution of variance is not accurate—for instance, for any player's projection there is a weight for their chances of being injured, but being injured is a binary outcome most of the time (playing or not playing), thus there is an inherently unequal distribution of variance. However, for simplicity's sake, it should be an acceptable starting place.
Now let us move to the second standard deviation. Once again, 95% of outcomes should fall within two SD, and since 68% fell within the first SD, thus the A's have a 27% chance of having a win total that is two SD away from their true talent level of 84. Now, the 2nd standard deviation is 12.7 wins, meaning a range from 71-77 and 91-97 wins. Using an equal distribution of variance again, we get:
- 71 wins: 1.9%
- 72 wins: 1.9%
- 73 wins: 1.9%
- 74 wins: 1.9%
- 75 wins: 1.9%
- 76 wins: 1.9%
- 77 wins: 1.9%
- 91 wins: 1.9%
- 92 wins: 1.9%
- 93 wins: 1.9%
- 94 wins: 1.9%
- 95 wins: 1.9%
- 96 wins: 1.9%
- 97 wins: 1.9%
Now we reach the very unlikely outcome that the A's win total in 2011 is at 3 standard deviations from their projected true talent level. There is a 5% possibility of this being true (if this seems high to you, think about the chances of the entire A's rotation being injured, for instance. Things like that represent this 3rd SD). This range is from 98-103 wins and from 65 to 70 wins. There is a 0.4% chance of each of these outcomes.
- 65 wins: 0.4%
- 66 wins: 0.4%
- 67 wins: 0.4%
- 68 wins: 0.4%
- 69 wins: 0.4%
- 70 wins: 0.4%
- 98 wins: 0.4%
- 99 wins: 0.4%
- 100 wins: 0.4%
- 101 wins: 0.4%
- 102 wins: 0.4%
- 103 wins: 0.4%
If these numbers seem impossible to you, think about this: Baseball Prospectus projected the Mariners to win 86 games before the 2010 season. They won 61 games. That's 25 games less than their projection, and in the FOURTH standard deviation. Wow.
Simply, here are the numbers I'm using for true talent level:
Rangers: 90 wins
Angels: 81 wins
Mariners: 75 wins
These are all open for discussion, they come from my back of the envelope WAR calculations, but I'd like to see everybody else's.
Using these numbers, here is the variance I get for each team:
- 71-76 wins: 0.4% each
- 77-83 wins: 1.9% each
- 84-96 wins: 5.2% each
- 97-103 wins: 1.9% each
- 104-109 wins: 0.4% each
- 62-67 wins: 0.4%
- 68-74 wins: 1.9%
- 75-87 wins: 5.2%
- 88-94 wins: 1.9%
- 95-100 wins: 0.4%
- 56-61 wins: 0.4 %
- 62-68 wins: 1.9%
- 69-81 wins: 5.2%
- 82-88 wins: 1.9%
- 89-94 wins: 0.4%
A's win 77 or fewer: 15.7% chance. Rangers win 78 or more: 95% chance. Angels win 78 or more: 69% chance. Mariners win 78 or more: 38% chance.
Chances of A's winning the division with 77 or fewer wins: 0.2% chance, which due to my lack of decimal places earlier is effectively 0. Therefore I'm going to start with 78 wins and up, ignoring the .000000x percent chances of the A's winning the division with each win total below.
Chances of A's winning the division with 78 wins: 0.1%
Chances of A's winning the division with 79 wins: 0.1%
Chances of A's winning the division with 80 wins: 0.1%
Chances of A's winning the division with 81 wins: 0.2%
Chances of A's winning the division with 82 wins: 0.3%
Chances of A's winning the division with 83 wins: 0.4%
Chances of A's winning the division with 84 wins: 0.6%
Chances of A's winning the division with 85 wins: 0.9%
Chances of A's winning the division with 86 wins: 1.0%
Chances of A's winning the division with 87 wins: 1.3%
Chances of A's winning the division with 88 wins: 1.6%
Chances of A's winning the division with 89 wins: 1.9%
Chances of A's winning the division with 90 wins: 2.3%
Chances of A's winning the division with 91 wins: 1.0%
Chances of A's winning the division with 92 wins: 1.1%
Chances of A's winning the division with 93 wins: 1.2%
Chances of A's winning the division with 94 wins: 1.3%
Chances of A's winning the division with 95 wins: 1.4%
Chances of A's winning the division with 96 wins: 1.5%
Chances of A's winning the division with 97 wins: 1.7%
Chances of A's winning the division with 98 wins: 0.4%
Chances of A's winning the division with 99 wins: 0.4%
Chances of A's winning the division with 100 wins: 0.4%
Chances of A's winning the division with 101 wins: 0.4%
Chances of A's winning the division with 102 wins: 0.4%
Chances of A's winning the division with 103 wins: 0.4%
TOTAL Percent Chance of the A's winning the division: 22.4%
A couple notes: First, those lines above represent the individual chance of the A's winning exactly that many games and winning the division at the same time, not winning that many games or fewer. Thus the total probability is the sum total of these numbers. Second, that number is higher than I expected.
Ok, let's open this up to discussion!