I did a quick back-of-the-envelope calculation to determine the approximate odds of winning such a contest and came up with 1 in 45,193,226,156,719,200 which is about "45 followed by 15 zeros". Here's how I got this number:

First, let's choose the bye week. Last year no teams had a bye in weeks 1-3 or 13-17. This leaves

**choices for the bye week.**

__9__Next, we can place the 16 opponents among the remaining 16 weeks. This can be done

**ways (reminder: 16! = 16 * 15 * 14 * ... * 2 * 1). That's a lot of choices. However, the Rams play the 3 teams in their division twice each, and we can assume that they will not play the same team in consecutive weeks. Accounting for the bye week, this**

__16!__**possibilities (although there is some double counting that I'm ignoring).**

__removes about 15 * 14 * 3 = 630__Finally, we need to select the day of the game. On average, each team plays one Sunday night game and one Thursday night game. For simplicity, we will assume that the Rams will play exactly 1 Thurs. night and one Mon. night game. Ignoring the bye week, this leaves 16 * 15 =

**possibilities. (In reality, games that are thought to be "better" are on Monday night and games that are thought to be less interesting are played on Thursday. Therefore, the odds are probably slightly different).**

__240__To get the final number of possibilities, we need to multiply the different possibilities together: 9 * (16! - 630) * 240 = 45,193,226,156,719,200. Although this is slim chances, I still took 5 minutes to fill out the schedule. So here's to hoping I'm $100,000 richer when I write my next post!

P.S. - I'm sure I made some silly mistake, so please comment if you notice one and I'll get it fixed.