On 10/22/05, Bob Hutchison <hutch / recursive.ca> wrote:
> Well, two aces counting as 11 total to 22 which is bigger than 21 and
> so you can never have more than one ace count as 11. I'd just keep an
> extra bit of information: do we have an ace? If we do then we may add
> 10 (just once) to the total when the total <= 11 (and where the total
> counts each ace as 1). This won't give you *all* totals just the best
> legal one.

This would work too, thanks. I did have similar ideas but for some
reason I thought it might be incorrect. Sometimes hearing it from
someone else just settles it.

--
David Vincelli