Preferrrrrred food

There’s a show on TV called 8 Out of 10 Cats which is one of those ubiquitous BBC panel shows designed to keep the UK’s comedians from starving. There are loads of shows like it and some of them have made their way to the colonies, sometimes even striking an Australian clone, complete with the same theme tune.

It took me a while for the title of the Cats show to make sense.

The format is based on public surveys with panellists having to guess which answers were most popular. Pointless and Family Feud are similar.

But it turns out the Cats show takes its name from an old TV advert which used to claim that eight out of 10 cats preferred Whiskas cat food.

I’m pretty sure I remember seeing the ad and blindly accepting that Whiskas was pretty good because most cats preferred it. I didn’t ever consider that the company’s survey technique might have been flawed. I naively assumed they had conducted thorough research, without questioning how you’d even go about surveying cats? That sounds much harder than herding them, which is notoriously difficult.

Fortunately there were viewers more discerning than I who took Whiskas to task, and, with a bit of gentle persuasion by the local advertising standards body, the company changed the wording in ads to, “eight out of 10 owners who expressed a preference said their cat prefers it”.

Not very snappy, but that probably delighted the marketing people at rival cat food behemoth Snappy Tom.

Puss-cat probability:
Fred walks into the dark room where his cat has given birth to a litter of six kittens.
They are a mixture of black kittens and white kittens.
If Fred lifts two kittens from the litter, the probability that he is holding a pair of white kittens is 2/3. But what is the probability that they are a pair of black kittens?

– ABK

Congratulations to Steve Rose for his correct answer, below

2 Comments on "Preferrrrrred food"

  1. The answer is 0 probability.
    If the probability of Fred lifting 2 white kittens is 2/3 then there must be 5 white kittens (a 5/6 probability of picking up a white one first then a 4/5 probability of picking up a white one second giving an overall probability of picking up 2 white kittens being 20/30 or 2/3).
    If there are 5 white kittens then there can only be one black one hence Fred can’t pick up 2 black ones.

  2. Another way of looking at the problem is to name our six kittens A, B, C, D, E, and F.
    That means there are 15 possible pairs of kittens:
    AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF and EF.
    If you eliminate any 10 of these pairs (which is 2/3 of the 15 pairs) as two white kittens, you will see that each of the remaining five pairs contains at least one of the white kittens.
    The conclusion is that only one of the six kittens is black and that there is no way of picking out a pair of black kittens.

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