Cacarulo: Composition-dependent indices for Insurance

[Igor]

Can you point me in the right direction? If I wanted to calculate the ten insurance break-even true counts for the above count, how would I?

I've never been happy about being an idiot and my resignation to this fact has led to some lazy habits. I only know, which isn't much at all, that insurance has a zero expectation at a composition where Q(T) = Q(NT)/2, or the NT/T ratio is exactly 2. Uncounted ranks should be assumed to consist of 1/13 per rank of the total cards in the undealt subset. The problem comes in the two tags that share the low half. A high count can presume an undealt subset that holds fewer Fours and Fives among the low cards, raising the NT/T ratio. I trust departure determination algorithms based upon simulation of the undealt subset (meaning SBA) more than algorithms assuming composition of the undealt subset (meaning BCA and PBA) for this reason.

Is there a quick and dirty approach to this? Accuracy two places to the right of the decimal is unnecessary. Even one place would be overkill; I would be ecstatic with accuracy to a fifth, a third or even a half of a point.

Thanks in advance and, although I have always been impressed with your work, I am literally astounded by this development.

Congratulations on your insight.

[Cacarulo]

> Can you point me in the right direction? If
> I wanted to calculate the ten insurance
> break-even true counts for the above count,
> how would I?

It's not that simple. You can find some info on bjmath where Pete Moss developed an algebraic formula for determining indices based on EORs.

> I've never been happy about being an idiot
> and my resignation to this fact has led to
> some lazy habits. I only know, which isn't
> much at all, that insurance has a zero
> expectation at a composition where Q(T) =
> Q(NT)/2, or the NT/T ratio is exactly 2.
> Uncounted ranks should be assumed to consist
> of 1/13 per rank of the total cards in the
> undealt subset. The problem comes in the two
> tags that share the low half. A high count
> can presume an undealt subset that holds
> fewer Fours and Fives among the low cards,
> raising the NT/T ratio.

> I trust departure
> determination algorithms based upon
> simulation of the undealt subset (meaning
> SBA) more than algorithms assuming
> composition of the undealt subset (meaning
> BCA and PBA) for this reason.

You're right here although for insurance, which is a linear function, this is not a problem.

> Is there a quick and dirty approach to this?
> Accuracy two places to the right of the
> decimal is unnecessary. Even one place would
> be overkill; I would be ecstatic with
> accuracy to a fifth, a third or even a half
> of a point.

See Bjmath.

> Thanks in advance and, although I have
> always been impressed with your work, I am
> literally astounded by this development.

You're welcome.

> Congratulations on your insight.

Thank you!

Sincerely,
Cacarulo