Monday, November 28, 2011

2012 Western States 100 Odds!


The original post on the initial assumptions is below, but there seems to be a bit more information out there now.  First of all, the number of people in the lottery is down to 1,946 and total tickets is down to 2,941.  Also, there's some chatter out there that there could be closer to 340 names pulled in the lottery this year...I think that's a bit optimistic though, so I'm going to assume 320.  I still haven't figured out how to account for the fact that once someone's name gets pulled, any additional names they have in the hat would be removed.  I don't think it's significant, but this would slightly increase these odds below...maybe by a percent or so I think. This all slightly improves these odds for each:

  • Odds with one name in the hat: 10.54%.
  • Odds with two names in the hat: 20.61%.
  • Odds with three names in the hat: 29.3%.
There's also a bunch of VERY CONFLICTING math being posted on the forum.  Some are coming up with some "logic" that having extra names in the hat doesn't help hardly at and decide for yourself!  I guess we can analyze to the end of the day, but we'll all find out soon enough anyway...good luck to all!

Below is my original post:


First, let me preface this by once again dedicating this post to my old man – he’s been been a math professor for something like 41 or 42 years now. This is all based on the same math from last year's WS odds post…from what I remember, I think he said there might be a few minor factors being left out, but this is a good estimate without getting too complicated. Thanks Dad!

On with the show:
The good news here is that they should be pulling more spots this year than last year since the two-time losers are done.  The bad news is there are quite a few more people in the lottery this year.

Some key assumptions:
We first have to make a really big assumption about how many names they are going to pull in the lottery. They say that they need a five-year average of 369 starters and they usually select about 400 total (planning for some attrition of course) Within that, there are a few more assumptions:
  • Montrail Cup Series Winners - 36 spots reserved according to the website. 
  • Raffle spots - 5 were drawn last June, and I’m assuming another 5 were drawn last December for the 2012 race, but not sure 
  • Top 10 runners - That's 20 spots, but not sure how many will do it (13 maybe?) 
  • International preference - They say they preference foreign entrants not drawn in the lottery – let’s assume 6 (complete guess) 
  • Gordy and Cowman - 2 more starters 
  • Groups that sponsor and staff aid stations and some sponsors can designate one runner – maybe 15 -20 spots? Let’s assume 18 (another complete guess) 
  • 9-time finishers going for their 10th - maybe 5 spots 
  • Finally, not sure where they are they at on their 5-year average starters? They have to average 369. I believe they had more starters two years ago than last year, which might help us this year. 
Now with all these things considered, assuming they select 400 total, and subtracting all those preferences, that would leave about 310 names pulled in the lottery. Last year they pulled approximately 230 (and I think there were about 80 two-time losers), so this would make sense.

If anyone has an inside track on the actual number of names being pulled in the lottery, let me know!

Laying out the facts:
Before we get to the real exciting part, we first need to lay out some of the facts. Most of these important details are on the site provides quite a bit of these important details for us:
  • One-time lottery entrants: 1,251 
  • Two-time lottery entrants: 497 (two-time entrants get two names in the hat) 
  • Three-time lottery entrants: 269 (three-time entrants get three names in the hat) 
  • Total lottery entrants: 2,017 
  • Total names in the lottery hat: 3,052 (1,115 + 497*2 + 269*3 = 3,052) 
  • Number of people selected in lottery: 310 spots (very estimated, remember?) 
Now we get to do some math!
Let’s start with the easy one – the chances of a one-time lottery entrant getting selected. Pretty basic, it’s just 310/3,052, which is 10.1573%. Good luck to you guys…you’ll need it.

Now for all you people with two names in the hat, pay attention! We start by figuring the odds that you are not selected with your first chance. With 3,052 names in the hat and 310 pulls, that equals 2,742/3,052 – or 89.8427%. Now the odds of not being selected with your second name are a very tiny bit more favorable at 2741/3,052 – or 89.8100%. Now to find the odds that one of your two names is selected, we multiply both and subtract from 1. So, since .898427 * .898100 = .806877, after you subtract that from 1, that means all of you two-time entrants have a 19.3123% chance of getting selected!

Now as you should have figure out by now, it’s really easy to determine your odds if you have three names in the hat. Just multiply the odds of not getting drawn after two pulls (.806877) by the odds of not getting drawn on the third pull (2,740/3,052 = .897772) and you get .724392. Subtract that from 1 and your odds of getting drawn with three names in the hat are 27.5608%. Approximately. I think.

Good luck to all in the lottery!

Let it be noted:
Last year, a couple people (Kurt N and Steve Quick), pointed out some potential minor flaws. Since I never really understood their explanations of a fully correct method, I’m just going to assume they didn’t actually know what they were talking about and this is pretty close to accurate (just kidding guys, I know you are both real smart).