WWDC 2013

WWDC tickets (https://developer.apple.com/wwdc/) are on sale at 3am tomorrow morning Sydney time! Last year they sold out in under 2 hours. The year before it took 12 hours and the year before that 8 days. Seems to be an exponential progression, I’m sure @Andrew_Stacey could work out a formula that would predict how quickly they will sell out this year.

I’ve been trying to get tickets for 3 years now. Fingers crossed.

I signed up to all sorts of automated alerts to tell me when the tickets went on sale so I wouldn’t miss out again. Normally the exact date of sale is a surprise but this year they announced it ahead of time. In fact, they announced it last night at 11pm - I almost had a heart attack as all the alerts on my phone went off.

Anybody else planning to attend or has attended in the past?

Up at 3am - couldn’t even log into the server due to the traffic. Sold out in 2 minutes.

Not happy!!!

Bummer

Assuming this is an exponential function, f(x)=a*b^x we need to solve for a and b, to predict next years ticket sales availability. The earliest sell out data I have is 2009, so let’s call that year 1 in our progression.

Year Time
2009 2 weeks = 1,209,600 secs
2010 8 days = 691,200 secs
2011 12 hours = 43,200 secs
2012 2 hours = 7,200 secs
2013 2 minutes = 120 secs

Taking the last 2 data points (1, 7200), (2, 120) as being representative of our function, we call 2012 year 1 and 2013, as year 2 (this simplifies the calculation), we get:

ab^1=7200 → ab=7200
a
b^2=120 → ab^2=120

which implies (ab^2)/(ab)=120/7200

and thus b=0.02
and a=7200/b=7200/0.02 = 432,000

Which provides a solution for our WWDC sell out formula:

f(x)=432,000*0.02^x

Consequently, next year (2014) will be year 3 and will sell out in 3.5 seconds!

Last change https://twitter.com/danielpunkass/statuses/327538348883324928 or eBay if have enought money

Unfortunately I didn’t make it as far as the cart…