Archive for Mathematics
Facebook competitions. Lovely aren’t they? The sort where you’re invited to supply some of your contact details in exchange for a token in a pile, one of which will be randomly picked and its owner be declared the winner. The prizes are often rather nice, rather desirable, so you’re tempted. “Facebook already has my contact details“, you think, “… this company isn’t getting any more than I’ve already chucked in Facebook’s general direction anyway“. What’s the harm?
But then they invite you to be a virus. They give you an opportunity to increase your chances of winning. “Invite five friends“, they cry. For every friend that joins in, we’ll give you an extra vote. So the deal here is that, in exchange for you doing their work for them (increasing the likelihood of their getting the large amount of contactage they’re after), you get more tokens in the pile.
Of course you instinctively ‘know’ immediately that this is bollocks. Obviously the more people in the game, the worse your chances are, but you’re thinking “yes but I get more votes, so maybe …?”
Let’s do some sums.
Clearly the best way to win is to be the only player. One participant, one token in the pile, it’s the only one that can be drawn, chance of winning – 100%. Easy. Your strategy – kill all those you know who’ve entered. Which means, pretty much – since you don’t know who’s entered, kill everybody but the guys with the prize.
But that sort of behaviour is deemed unacceptable. We know, it’s health and safety gone mad, but what are you to do?
So, like an idiot, you invite your five friends. Let’s assume the worst – they all take up the offer. You’ve now won an extra token for each friend so you have six tokens in the pile. And there are now six people in the game, most – five – with only one token. So you’re better off than they are, for sure. There are eleven tokens in the pile – your original one plus the extra five you just earned, and the five new ones from your friends. Your chance of winning just dropped from 1 in 1 (certainty) to 6 in 11 (about a half) and theirs went up from zero to 1 in 11.
Suppose that each of these friends is as equally idiotic as you. They aren’t going to invite you back because you’re already in, so they each invite five new friends. That’s twenty five brand new people. There are now thirty one people in the game – you, plus five, plus 25 – and there are your six (the number didn’t increase, sadly) tokens sitting in a pile with twenty five new ones from twenty five new people, plus thirty tokens – six each now – from your five successful friends. Your chance of winning is now down to 6 in 61, about 10%. And you’ve lost control over what happens next.
Except it might not be twenty five new people of course. Because your friends’ friends may each be shared by more than one of your friends. That’s quite likely. In the extreme case, your little coterie may well comprise only you, your five friends, and one other person you don’t know but who happens to be the common friend of each of your friends. The competition runner is only ever going to get seven sets of contact details out of this sorry bunch – so you have that satisfaction – but what are your chances of winning now? They can’t be quite as bad because your universe is so small.
Well, your five friends each invite the maximum five friends, but four of those friends were (unbeknownst to them – they knew only not to bother inviting you) already in the game (you brought them in) and it’s only the fifth person (the one you don’t know) who is a new joiner, and only one of your five friends is going to succeed in trapping that individual. So there are seven individuals and a pile of thirteen tokens. That’s the eleven before the new round of invitations, plus one for the new individual, plus one introducer’s token for your successful friend inviter. Your chance of winning dropped a tad, from 6 in 11 to 6 in 13. One of your friends has a 2 in 13 chance of winning and your other four friends, and your ‘friend-in-law’, have a 1 in 13 chance. The universe can have a little snigger at this though since it knows that your friends believe – erroneously – that their chances of winning may still increase at any time when their untaken invitations succeed (which of course they never will). So there’s that.
You invited your own competition. How silly of you. If you’d invited only one friend – you had a gun pointed at you, the devil made you do it, whatever – you’d’ve been better off with a 2 in 3 chance.
But you know you’re not the one in control here. Others are going to come in regardless of your meanness in attempting to keep knowledge of this competition to yourself. So having more of your tokens in the pile is always going to be better. In fact, if you’re the only one keeping your friends out, and everyone else is being all lady bountiful with their invitations, then you’ve no chance. So you have to bring in more competition. It’s really annoying.
Had an idea to find my dad’s Bacon Number. Though not in the film business, he had featured (as half of the double act ‘Johnnie and Ronnie’)- playing the accordion alongside his drummer chum – in Stewart Mackinnon’s 1993 film Border Crossing. This is a difficult film to track down. Amber Styles was in that, so she’s the obvious link, and indeed it turns out that her Bacon Number is three and so his is four.
However, Gerhard Garbers – also in Border Crossing – was in a 1990 film called Werner – Beinhart!, which also featured Ludger Pistor, who was in this year’s X-Men First Class with the man himself, old Kev. So Gerhard (also in the highly enjoyable Run Lola Run) beats Amber (sorry Amber) with a Bacon Number of two, so my dad’s is three.
Except – there’s a route with even more cachet. Border Crossing’s Les Wilde was in Stormy Monday with Tommy Lee Jones, who was of course with ‘the Bake’ in JFK. So, though my dad’s Bacon Number is still three (which is pretty good for a non-player), that’s a fairly decent path.
But whoa hang on there and hold your horses just a cotton’ pickin’ minute – Stormy Monday? My sister was in that! OK, it was a bar scene and she didn’t say anything and she’s uncredited, but still – that means she’s got a Bacon Number of two.
So this woman is standing with a clipboard inside the shop, which is Maplin’s, and I’m on my way out and I haven’t bought anything (but she wouldn’t necessarily know that) and she asks (yes she does), like, “Would you mind taking part in a survey?” and I say “It’s ninety pounds an hour” and she laughs and says “No thanks” and lets me go without further ado.
Congratulations to Mrs Ivy duPorc of East Gleaming, Hants for proving that the complex function z(s) = 1+2-s+3-s+4-s+… has all of its non-trivial zeros symmetrically distributed about the line Re(s) = ½. A book token for £4.50 is on its way!