Question for the math gurus

A while back, in the Green Chip section, someone was asking about true count conversions in DD.

This got me to thinking, and I have come up with a very interesting and possibly moderately useful observation. This is most easily described by example:

Taking the number of cards in the discard pile as a fraction of a deck, let "N" be the numerator of that fraction and "D" be the denominator.

Let "S" be the result of D minus N, and "F" be the true count conversion factor.

For example, the are 2/3rds of a deck in the discard pile. This means there are 4/3rds decks unplayed; so the TC conversion factor is 3/4 (IOW, multiply the RC by 3/4 to get the TC).

It seems that for all cases in the two deck game: F = ( N + S ) / ( D + S ).

In the above example, N = 2; D = 3; S = 1; so F = (2+1)/(3+1) = 3/4.

Another example: 3/5ths of deck in the discard pile. This would leave 7/5ths unplayed, thus a conversion factor of 5/7. Using the formula, N = 3; D = 5; S = 2, F = (3+2)/(5+2) = 5/7.

Finally, an example of >1 decks in the discard pile: 6/5ths leaves 4/5ths unplayed and a factor of 5/4.

Using the formula: N = 6; D = 5; S = -1; so F = (6-1)/(5-1) = 5/4.

Here's my question: is this observation/postulation mathematically provable? Would any of you math gurus out there care to post the proof?

N/D is the number of decks in the discards, so 2-N/D is the number of decks remaining in the shoe. If you want to multiply, you have to take the reciprocal, so F = 1/(2-N/D). Multiply top and bottom by D, and you have F = D/(2D-N) = (N+(D-N)/(D+(D-N)) = (N+S)/(D+S).

But I don't think this is the simplest way to do it. It's simpler to just look at the remainder of the shoe, make a fraction, and invert to get the reciprocal.

ETF

Taking the number of cards in the discard pile as a fraction of a deck, let "N" be the numerator of that fraction and "D" be the denominator.

Let "S" be the result of D minus N, and "F" be the true count conversion factor.

For example, the are 2/3rds of a deck in the discard pile. This means there are 4/3rds decks unplayed; so the TC conversion factor is 3/4 (IOW, multiply the RC by 3/4 to get the TC).

It seems that for all cases in the two deck game: F = ( N + S ) / ( D + S ).

In the above example, N = 2; D = 3; S = 1; so F = (2+1)/(3+1) = 3/4.

Are you serious you want us to add that to everything else you need to pay attention to.What am I RAIN MAN ?

> It's simpler to just look at the remainder of the shoe, make a fraction, and invert to get the reciprocal. <

That is what I normally do; and have done these calculations so often that I just *know* the TC in 99% of situations.

But, using simple fractions, where the difference between numerator and denominator is equal to 1, it would be easy for a newbie to memorize things like:

1/2 : 2/3;
3/4 : 4/5;
7/8 : 8/9; etc.

What does the newbie do when there are 5/3 decks left? Better to just keep 10 in the numerator all the time. 10/5 decks left, multiply by .5, 10/6 left, multiply by .6 ... etc. Plenty accurate enough. Or do it DD's way. I think my way is simpler, but most like his better. I'm sure it's archived below somewhere.

ETF

But, using simple fractions, where the difference between numerator and denominator is equal to 1, it would be easy for a newbie to memorize things like:

1/2 : 2/3;
3/4 : 4/5;
7/8 : 8/9; etc.

> What does the newbie do when there are 5/3 decks left? <

The formula relates to the # of cards in the discard pile, not how many are left (well, it does, but only indirectly). 5/3 left means 1/3 in discard pile, which has a conversion factor of 3/5ths [ (1+2)/(3+2) ].

Fortunately, that early in the shoe (anything less than half a deck in the discard pile), it's pretty safe to use a factor of 0.6 and floor the result.

E.g.:
1/4 : 4/7 : 0.57;
1/5 : 5/9 : 0.55;
3/7 : 7/11 : 0.64; etc.

The formula relates to the # of cards in the discard pile, not how many are left (well, it does, but only indirectly). 5/3 left means 1/3 in discard pile, which has a conversion factor of 3/5ths [ (1+2)/(3+2) ].

Yeah, that's what I meant.

Fortunately, that early in the shoe (anything less than half a deck in the discard pile), it's pretty safe to use a factor of 0.6 and floor the result.

Sure, with two decks going in. My system works with ANY number of decks at the start.

You mean to say you're actually using this goofy formula? ;) ;)

ETF

As I said earlier, this might provide a newbie an easy way to memorize conversion factors.

Also, as I mentioned earlier, I have done the "invert and multiply" calculations so often that it is rare situation indeed when I don't just "know" the TC and actually have to do the mental calculation.