
David Spence: How many aces equal a full deck?
This problem was inspired by the otherwise useless iPhone card counting application. In that application, neutral cards are not counted, but a TC is still determined. People have hypothesized that this is done by treating 40 cards dealt (in the case of HiLo, which has 40 nonneutral cards per deck) as a full deck.
I suspect that 40 is not the best approximation for the number of nonneutral cards dealt to equal a full deck. Consider a similar question: how many aces of clubs must be dealt before it's likely that a full deck has been dealt? The same logic that gives 40 as the answer above would give 1 as the answer here. With a single deck, it's very likely that less than a full deck will have been dealt by the time you see the ace of clubs, so 1 ace of clubs must equal something less than a full deck.
So what fraction of a deck does a single ace of clubs indicate? My initial thought is that this is similar to the median/mean distinction. That is, what is the median number of cards dealt before you see the ace of clubs? A rough approximation for this is given by multiplying the mean by ln2. In this case, 52 x ln2 = 36. So a single ace of clubs indicates that 36/52 of a full deck has been dealt.
Clearly, this is just the start of a solution to the original iPhone problem, and it may very well be wrong. But I was wondering whether others have given this problem any thought.
David

Don Schlesinger: Re: How many aces equal a full deck?
> This problem was inspired by the otherwise useless
> iPhone card counting application. In that application,
> neutral cards are not counted, but a TC is still
> determined.
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 00! :)
> People have hypothesized that this is done
> by treating 40 cards dealt (in the case of HiLo,
> which has 40 nonneutral cards per deck) as a full
> deck.
That would be the mean, but not the median, or over/under.
> I suspect that 40 is not the best approximation for
> the number of nonneutral cards dealt to equal a full
> deck.
Define "best."
> Consider a similar question: how many aces of
> clubs must be dealt before it's likely that a full
> deck has been dealt?
That's an ambiguous question.
> The same logic that gives 40 as
> the answer above would give 1 as the answer here.
Not necessarily. See above.
> With a single deck, it's very likely that less than a full
> deck will have been dealt by the time you see the ace
> of clubs,
Of course. It isn't going to be in the 52nd position all the time!
> so 1 ace of clubs must equal something less
> than a full deck.
Should be about 0.69 times 52, or about 36, for the under/over, or median.
> So what fraction of a deck does a single ace of clubs
> indicate? My initial thought is that this is similar
> to the median/mean distinction.
There you go! One day, I'm going to actually read an entire post before I begin answering, line by line! :)
> That is, what is the
> median number of cards dealt before you see the ace of
> clubs? A rough approximation for this is given by
> multiplying the mean by ln2. In this case, 52 x ln2 =
> 36. So a single ace of clubs indicates that 36/52 of a
> full deck has been dealt.
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?
> Clearly, this is just the start of a solution to the
> original iPhone problem, and it may very well be
> wrong. But I was wondering whether others have given
> this problem any thought.
Not until now, but I wouldn't worry too much about it. If they use 40 nonzerotag 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.
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?
Don

David Spence: Re: How many aces equal a full deck?
> 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 00! :)
Yes, neutral cards are definitely ignored. I saw screenshots of the application, and, for HiLo, 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 underestimate of the number of nonneutral 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 nonneutral 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 nonzerotag 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 nonneutral cards (40) by ln2 either. I believe the solution for the 40cards problem requires either a better approximation for the median (the original ln2 method suggests that seeing 40 nonneutral 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

David Spence: Re: How many aces equal a full deck?
One other thought: as the number of tracked cards increases, the ln2 approximation becomes less accurate, as the median approaches the mean.
Consider the degenerate case: what's the median number of cards dealt from a single deck by the time you see all 52 cards? Certainly, the median=mean=52, and ln2 x 52 is a horrible estimate. I think the estimate works best when you're looking for a single result (e.g. a single ace of clubs, a single 28 in roulette, a single 3 on a die, etc.).
So I'm still leaning towards the median solution for this problem, but with an accurate calculation of it.

Don Schlesinger: Re: How many aces equal a full deck?
> One other thought: as the number of tracked cards
> increases, the ln2 approximation becomes less
> accurate, as the median approaches the mean.
> Consider the degenerate case: what's the median number
> of cards dealt from a single deck by the time you see
> all 52 cards? Certainly, the median=mean=52, and ln2 x
> 52 is a horrible estimate. I think the estimate works
> best when you're looking for a single result (e.g. a
> single ace of clubs, a single 28 in roulette, a single
> 3 on a die, etc.).
Well, that question isn't interesting, but if you had, say, 6 decks, and you asked the same question, the answer would be somewhat different.
> So I'm still leaning towards the median solution for
> this problem, but with an accurate calculation of it.
I know where to go to get the right answer. :)
Don

David Spence: Re: How many aces equal a full deck?
> I know where to go to get the right answer. :)
I'm not surprised :)
David

Don Schlesinger: Re: How many aces equal a full deck?
Mike Shackleford did an Excel study and got 40 cards for both the mean and the median!
So, while not perfect, and surely not the equivalent of counting all the cards, the iPhone methodology isn't so bad after all.
Don

David Spence: Re: How many aces equal a full deck?
> Mike Shackleford did an Excel study and got 40 cards
> for both the mean and the median!
But I think this is for an infinite pack; the median is certainly not 40 for a single deck, and may not be so for small numbers of decks either.
Thanks for finding out this result.
David

Don Schlesinger: Re: How many aces equal a full deck?
> But I think this is for an infinite pack; the median
> is certainly not 40 for a single deck, and may not be
> so for small numbers of decks either.
It's supposed to be for 6deck. We don't care about SD, and infinite deck has no practical application.
Don
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