> So, are you sure that the cards are IGNORED, as
> opposed to being counted as zero? Sometimes people
> misunderstand what the words are saying. I always joke
> when someone says that there is "no score"
> in a baseball game, when what he means is that there
> IS a score, and it is 0-0! :-)

Yes, neutral cards are definitely ignored. I saw screenshots of the application, and, for Hi-Lo, it just has one button for high cards and one for low.

> Define "best."

> That's an ambiguous question.

True enough. But, with a single deck at least, 40 is definitely an under-estimate of the number of non-neutral cards needed before you can expect the full deck to have been dealt. There are many cases where less than a full deck will have been dealt by the time you see 40 non-neutral cards, but no cases where more than a full deck will have been dealt. So, if we can't find a "best" answer, at least find one that's not clearly wrong :-)

> More precisely, the probability of seeing the ace of
> clubs after viewing 36 cards is, roughly, 50%. Do you
> understand that this is not quite the same thing as
> what you're saying?

I do...it was just a first thought at a solution, since it seems reasonable at least with the ace of clubs/single deck problem. The "mean" answer is clearly wrong in this case, and the median answer may or my not be :-)

> Not until now, but I wouldn't worry too much about it.

I realize this is a primarily a theoretical problem, since no one in his right mind will use the iPhone app, but it got me thinking (if not worried) about it nonetheless.

> If they use 40 non-zero-tag cards to estimate when one
> deck has been dealt, they will be correct as to the
> mean number of cards required to see the full deck.
> But, the median would surely be fewer -- in this case,
> around 28, I suppose.

Before, we multiplied 52 by ln2 to get 36 for the median number of cards before you see the ace of clubs. Here, you're multiplying 40 by ln2. This isn't the same pattern. We didn't multiply the number of aces of clubs (1) by ln2, so we shouldn't multiply the number of non-neutral cards (40) by ln2 either. I believe the solution for the 40-cards problem requires either a better approximation for the median (the original ln2 method suggests that seeing 40 non-neutral cards is equivalent to seeing ln2 x 52 = 36 cards of a full deck, which is clearly illogical), or suggests that this is the wrong path entirely. I'm still leaning towards the median method, but with the appropriate calculation of it.

> I'm not sure which would be a better way to estimate
> the true count, but I'm leaning towards the mean.
> After all, there will times when it will take a lot
> MORE than 40 cards, and you want to be correct, in the
> average, no?

Absolutely. As the number of decks approaches infinity, I think the best estimate approaches the mean. With one deck, and one tracked card, however, the median (or at least something OTHER than the mean) seems best.

I think this problem must have been solved before. Unbalanced counts implicitly approximate a TC, based on an incomplete track of the cards dealt. Whatever method is used for such counts, I suspect, could be used to solve this problem, unimportant though it may be :-)

David