> It's a nice game and a simple concept.  You're not dealing with odds
> right, though.  3:1 against means the same as 1/4 -- in general, when
> the odds against are m:n the probability is n/(m+n).  The point is
> that for every m failures you'd expect n successes.  A failure on the
> first rabbit is 1:24, not 1:25 (or 4:100, which the game gives).
> While you're on the first level, the odds against that failure are
> just x:(25-x), where x is the number of rabbits you've tried to seize.

> The odds against dying on the second level given that you made it at
> all are y:(19-y).  But when you've died after x rabbits on level 1 and
> y on level 2, what were the odds against that particular failure?  In
> terms of probability it would be x/25 + ((25-x)/25)*(y/19).  In terms
> of odds, let r=19*x + (25-x)*y, then it's r:(25*19 - r) against.  Now
> you can plug in r and (25*19 - r) instead of x and 25, and repeat this
> for every level.  It'll be even better if you divide both by the GCD
> each time.

> Or maybe you'll find it simpler to report your results in terms of
> probability: just say "4 in 100" instead of "4:100 against".

> - Joshua