I’m not a big watcher of television but, as I mentioned yesterday, I do enjoy Pointless for its reverse Family Fortunes format that rewards obscure knowledge, while also allowing people to also give obvious answers.
It would be pointless to explain Pointless when you can swot up on it in detail on Wikipedia but what has bothered me on and off for ages is this: what is the optimum number of contestants for a day’s recording?
Here I do have to go into some detail. The show starts with four pairs of contestants and one couple is eliminated each round until the last remaining pair reach to the Pointless Final.
The thing is, the eliminated couples get to have another go in the next show but if they fail again then off they go home. So say there are four couples A, B, C and D and A get to the final then B, C and D come back next time and A is replaced by couple E.
Now B, C or D might reach the final, but then so might E so the next lot of contestants would be E, F, G and H or F, G, H and I. By now the potential permutations get pretty hazy in my head.
So going back to my original question, there must be an algorithm to work the optimum number of couples on standby for the three or four recorded shows in a day without some having to go home and come back again the next day. The trouble is, my maths isn’t up it.
And it’s this sort of pointless puzzling that can spoil my Pointless viewing pleasure.