February 21st 2015
It's the Oscars tomorrow, and if Oscar season isn't a time for self-congratulation then I don't know what is, so forgive me for taking the time to tell you all how right I am. Or, I guess, how right I was. On August 5th 2013 I wrote an entry on this page about the actress Felicity Jones, in fact quoting another entry I'd written in March 2009 (this is very much the Inception of self-congratulation), and the 2013 instalment was summed up in the last line: "The point is: she's gonna be huge. And you heard it here third". Now, I may have been wide of the mark when I suggested that her upcoming role in The Amazing Spider-Man 2 was going to be the game-changer, because in the end she only appeared in a couple of scenes (she still did better than Shailene Woodley, who was cut altogether) and, given that the Spider-Man franchise is in turmoil again, that might be all we get of her there. But! She has still hit the big time, a little later than scheduled, as she has been nominated for a Best Actress Oscar for her performance as Stephen Hawking's first wife in the biopic The Theory of Everything. I think it is her best performance to date and, while she's unlikely to win this time round, it's good to see that the world in general has caught up to the fact that she is one of the most exciting young actresses around. Well, she's older than me. But I think we can all agree that she's looking better for it.
In other self-referential news, I set y'all a probability problem not too long ago, asking if you could tell me a formula for R(n,k) where R(n,k) := the number of ways of putting n balls in order 1 to n such that k of them are in the right position. Well, Katy had a crack, and I know that my good friend Rich has been working on it, so look away if you don't want spoilers... I'm actually going to cheat a bit in my answer, because the value for R(n,k) is not actually that interesting: it follows directly from knowing the formula for R(n,0) [i.e. R(n,k) = nCk*R(n-k,0)]. And, in fact, even the formula for R(n,0) isn't as interesting as the probability of getting zero balls in the right position [i.e. R(n-k)/(n!)] which is equal to the power series formula for 1/e. That is, the probability of putting n balls in order 1 to n such that none of them are in the right position tends to 1/e as n tends to infinity (and, in fact, by the time n=5 you've got a very decent approximation). Fascinating, eh?
I have written the full proof out neatly in a Word doc, so if anyone wants to see it, please do get in touch...
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what was I listening to?
The Gilded Palace of Sin - The Flying Burrito Bros |
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what was I reading?
Travelling to Work - Michael Palin |
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what was I watching?
Family Business |
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