Fingers Crossed

Next Monday, Jools Holland is filming the 200th episode of Later. Radiohead are playing. I believe both those statements are accurate, but the internet wasn't hugely helpful on the subject. I and three of my friends entered the random draw for tickets to the filming of the show, and tomorrow we find out if we were successful. It's one of these things where you can't be sure how much of a chance you have; all four of us applied for four tickets, so I think I'm right in saying the night would have to be eight times over-subscribed before our chances of getting tickets dropped below 50% (I sense some bad logic in there but let's ignore it). Then again, not many people seem to be in the audience when you watch Later, so maybe the demand is a hundred times greater than the capacity? I don't know. Watch this space.

No, hang on, I'm not going to ignore my potentially dodgy statistical analysis of the random draw system - I'm going to use my brain to work it out properly. If every entrant can ask for between one and four tickets, each entry is treated equally (regardless of how many tickets the entrant is requesting), and my group of friends has a total of four entries, each time a random draw is made for a space at the event, we have a 4-in-X chance of getting that space, where X is the total number of entries. Actually, when the draw for the nth seat (starting with zero) is made, our chance is 4-in-(X-n). I sense I have made this too complicated, but sod it. Hang on, I can make this a lot easier by taking out this 'four' stuff. Since there's a bunch of seats, let's assume that basically every person has one entry for one ticket, because groups of two, three or four will all have done the same as us and done an entry each - if they haven't, that's a bonus. Ah, I did cock it up to start with. If we effectively have one entry each, my probability of getting each seat is 1-in-(X-n). If there are N seats, the average probability of me getting each seat is 1-in-(X-(N/2)), and hence the total probability of me getting any of the N seats is N-in-(X-(N/2)), or 1-in-((X/N)-0.5). Really, I've made this far too complicated, but I don't care. Let's stick in some numbers to see if this works. No! I've just done it in my head and it doesn't work. Crap! Clearly a Master's degree, even one that features at least three courses in statistical physics, is irrelevant once you spend a few months not thinking about maths. I'm going back to my two-word title's suggestion and crossing my fingers.