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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 Hi-Lo, which has 40 non-neutral cards per deck) as a full deck.
I suspect that 40 is not the best approximation for the number of non-neutral 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
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