Warp <war### [at] tagpovrayorg> wrote:
>   3) Assume two people, person A and person B, who want to decide who gets
> a price by tossing a coin.

>   Person A is a bad loser and a bully, so if he loses he says "I said it's
> two out of three". So they play it like that. If A loses again, he says
> "I said it's three out of five", and so on, until he wins.

>   How many tosses is this game expected to last, in average?

  Actually this problem isn't so "relatively easy" is it first seemed.
It's actually quite complicated.

  AFAIK the answer lies on whether the sum of probabilities for a game
lasting a certain number of rounds forms a convergent or a divergent series.
If it forms a divergent series, then the average is infinite. However,
proving that the series is divergent is, AFAIK, not trivial at all.

  Anyways, maybe you could want to consider two variants with finite
solutions:

  3a) Like above, but after n losses the bully gets tired and declares himself
a winner anyways. In other words, rather than saying "I said it's <n+1> out
of <(n+1)*2-1>" he just ends the game.

  Question: How many coin tosses does a game have in average, with an upper
limit of n round losses for the bully?

  3b) Like 3a, but instead of tossing a coin, they play rock-paper-scissors.
How many hand throws in average does a game last with an upper limit of n
round losses for the bully?

-- 
                                                          - Warp

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