Math Demonstrates Just How Unlikely it is that the Padres Will Pick #1

In my previous post, I concluded: "theoretically it is possible the Padres will end up with the worst record, but the odds are not that great." Well, my curiosity got the best of me, and I decided to calculate just how small those odds actually are...

Using Bill James log5 method of estimating one-game winning percentages, I calculated the odds of both Washington and San Diego beating their opponents in any given game. I included an adjustment for home-field advantage, but did not account for pitching match ups. Although there is an extremely good chance Seattle actually winds up with the worst record in MLB, for the sake of simplicity, I am excluding them in this analysis.

The Padres are expected to beat the Dodgers 37% of the time, and Pittsburgh 51% of the time. Washington is expected to beat Florida 39% of the time, and Philly in 31% of their games. In short, over their next 6 games, ignoring pitching matchups, the Padres are expected to win 2.664 games, and Washington is expected to win 2.121 games.

In order to end up with a worse record than Washington, the Padres must lose 4 games more than Washington. Three situations exist which satisfy this requirement. The Nats can lose 0, and the Padres lose 4 or more games, the Nats can lose 1 game, and the Pads lose 5 or more games, or the Nats can lose 2 games, and the Padres lose 6 games.

The first situation has a probability of 0.00112214. The probability of the Padres losing 4 or more games is nearly 60%, but the probability of the Nationals losing 0 games is under 1%. Situation 2 has a probability of 0.0054376. Situation 3 has a likelihood of 0.002737. Taken as a whole, the probability of San Diego finishing with a worse record than Washington is under 1%. Combined with the fact that the Padres must also overtake Seattle for the worst record, it is extremely unlikely the Padres will select first in the 2009 draft.