Warp wrote:
>   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?

I won't spoil it since I found this to be a very fun little problem, but 
once you get the answer in the right form it's not too bad to show that 
the series diverges.  In fact, I derived my previous answer by noting 
that (unless I made algebra errors) for epsilon > 0 the value:

	1/2 + 1/(2*sqrt(pi)+epsilon) * sum_{i=1:n} (2*i+1) / i**(3/2)

lower bounds the answer to 3a for sufficiently large values of n (I 
don't think this is too much of a hint).

The main problem is that it requires either significant derivation or 
some background knowledge to approximate the sum in that form.  Of 
course you can still compute the answers efficiently for any given n 
without this, and I found deriving even this part of the answer pretty 
satisfying.

Basically: fun problem! (although I agree that it wasn't "relatively 
easy", nor is #4, as I don't think it's possible).


>   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?

It should be 1.5 times as many rounds as the answer to part 3a I would 
think.

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