Expectation value of a coin flipping game (the tools of a mathematician)

was recently asked how I might go about working out the expectation value of the game where an unbiased coin is thrown n times where you win nothing each time a head is thrown and you win $n if you throw a tail on the nth throw. 

I hadn’t thought of this problem for a long time and didn’t have a pen and paper on hand and couldn’t give an answer (!) – but when I got home, quickly playing with the problem with pen and paper showed that, of course, the expectation value is
$n(n + 1)/4.

Pen and paper – the tools of a mathematician! It would be nice to have a mental whiteboard on which to do calculations, but paper is much more reliable!

2 Responses to Expectation value of a coin flipping game (the tools of a mathematician)

  1. Steve says:

    Any problem of this type which involves the sum of a series, I always remember the story about gauss as a child being asked to sum the numbers 1 to 100, allegedly to slow him down and him coming back immediately with with 5050, on the basis that you can sum the pairs from either end of the series to equal 101, and then there is 50 of them. – this is the n(n+1)/2 bit with the extra half for the expectation.

  2. Sacha says:

    Yes, it’s a nice story about Gauss and a nice early example of people using abstract symmetry to solve problems for kids (although one probably wouldn’t describe it in those terms). The best part of the story is the way young Carl got around the teacher’s silly time-consuming exercise with his own immediately obvious solution.

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