Western States 100 Odds!

UPDATE: GOOD NEWS AND BAD NEWS

• The good news is I heard from my old man that all this math is indeed correct.
• Bad news is I heard from the email the WS 100 folks sent out this afternoon that my assumption of 280 available spots was way overestimated.  Sounds more like 230 instead.  The numbers below are updated with this latest info.

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So I just took a little break to do some math. Sounds really exciting I know…hang with me though, there’s a chance a small number of you might be very interested in this!

Before I get to what I know you all are waiting for, I am dedicating this post to my Dad – he’s been a math professor at U of M – Crookston for over 40 years…taught me everything I know.

Hold on to your boot straps…here we go: 
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 Ultrasignup.com site provides quite a bit of these important details for us:

• One-time lottery entrants: 1,115 
• Two-time lottery entrants: 498 (Note: two-time entrants get two names in the hat) 
• Total lottery entrants: 1,613
• Total names in the lottery hat: 2,111 (1,115 + 498*2 = 2,111) 
• Number of people selected in lottery: 230 spots (approximately) 

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 230/2,111, which is 10.8953%. 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 2,111 names in the hat and 230 pulls, that equals 1,881/2,111 – or 89.1047%. Now the odds of not being selected with your second name are a very tiny bit more favorable at 1,880/2,111 – or 89.0995%. Now to find the odds that one of your two names is selected, we multiply both and subtract from 1. So, since .891047 * .890995 = .793919, after you subtract that from 1, that means all of us two-time entrants have a whopping 20.6081% chance of getting selected! Better than you thought, huh?

Now this guy's good at math!

Let’s take it to the next level… 
I know you all are wondering to yourself by now, “Since there are 21 Minnesotans in the lottery (including honorary Minnesotan Joe Z), what are the odds that at least one can get in?” Now that one’s a piece of cake. So of these 21 Minnesotans, 14 are two-time lottery entrants, and 7 are one-time entrants. This gives us a total of 35 Minnesotan chances to get drawn. Now we can use the same concept as for the two-time entrants, but since the odds of not get drawn are almost the same each time, let’s just take a little shortcut and do 1,881/2,111 to the 35th power and subtract from 1. That gives us a 98.2359% of having at least one Minnesotan selected! 

How about Joe’s Crew? 
There are seven of us in a little group of TCRC runners who train together in the lottery…so for the handful of you guys out there, here are the odds that at least one will get selected. Same concept as for the Minnesotans, but here we have 5 two-time entrants and 2 one-time entrants, which gives us 12 chances total. So we just take 1,881/2,111 to the 12th power and subtract from 1, and we’ve got a 74.9501% chance of at least one getting in! 

And for the odds that all seven of us get in? That would be .0004412%...I think. So I’m telling you there’s a chance.  Good luck to all in the lottery on Saturday!

Oh, and thanks Dad! Sorry if it’s all completely wrong.