From: jacobs@cs.utah.edu (Steven R Jacobs)
Subject: Re: a few bj questions
In-reply-to: cantor@star.enet.dec.com's message of 8 Apr 92 15:21:02 GMT
Organization: University of Utah CS Dept
	<1992Apr8.153206.19973@engage.pko.dec.com>
Date: 8 Apr 92 11:28:26

In article <1992Apr8.153206.19973@engage.pko.dec.com> cantor@star.enet.dec.com (Dave Cantor) writes:
> In article <1992Apr7.224222.25111@pencom.com>, eric@pencom.com writes...
>>[A good mathematical proof removed for brevity.  Dave C.]
>>Therefore, it never pays to double less than the full amount.
>
> You proved that it never pays to double less than the full amount when
> you have the full amount available to double with.

The proof is flawed, since it does not account for pushes.  However, as
Jacko pointed out, there are many cases in which the fact that you double
does not change the probabilities for win/push/lose, and you have the
advantage (e.g. 10/11 vs. 4,5,6).  These situations are the times when
you would only draw _exactly_ one card if you couldn't double.  In these
situations, your expected gain increases by an amount that is proportional
to the amount that you increase your bet.  It you double your bet, you
double your expected gain.  If you increase your bet by a factor of 1.1,
you increase your expectation by a factor of 1.1.  Thus, in these cases,
it _always_ pays to double for whatever amount that you can.  Doubling for
the full amount is best, but you are better off to double for less rather
than hit.

These are the "best" kinds of doubling situations, because doubling
down also doubles your expected gain.  You are essentially placing two
bets in a favorable situation, while still being allowed to hit in an
optimal fashion.

> I alleged that it is advantageous to double for less when you don't
> have the full amount to double with, rather than just hitting.  I don't
> think your proof applies to the conditions I proposed.  Here they are
> again:
>
>	You have 5 units and bet 3 units.  You receive an 8,3 against
>	the dealer's 5.  You do not have 3 units to double with; you have
>	only 2.  I claim it is more advantageous to double for less with
>	the 2 units then to hit without doubling.  

This particular case is covered by the discussion above, and you are
correct that it is better to double for less rather than hit.

> I suspect (but can't prove) that for every hand in which full doubling
> is advantageous, there is some fraction x, 0 < x < 1, such that doubling
> for less than x*the_flat_bet is disadvantageous compared to just hitting,
> and doubing for less than the_flat_bet, but for more than x*the_flat_bet
> is more advantageous than just hitting, but less so than full doubling.

Now we get to the second type of double-down situations.  In situations
where correct hitting would sometimes require you to draw more than one
card, your expected gain from doubling will be _less_ than twice the
expected gain from hitting.  In some cases, your expectation from doubling
will be only slightly higher than your expectation from hitting.  In
these cases, you are placing two bets but giving up some of your expected
return by not being allowed to hit in an optimal way.  At the extreme
(when doubling is just barely better than hitting), your expectation for
drawing only one card is only half as much as your expectation from
hitting "correctly".

In these cases, there will be some fraction of a bet for which doubling
for less would be less profitable than hitting.

If

   E1 is the expectation from drawing one card,
   EH is the expectation from drawing optimally,
   K is the fraction of a bet that is used for doubling

Doubling for less is more profitable than hitting when

   E1 * (1+K) > EH.

or

   K > (EH/E1) - 1
   K > (EH - E1)/E1

If the fraction of your doubled bet is less than this ratio, it
is better to hit rather than double.
--
Steve Jacobs  ({bellcore,hplabs,uunet}!utah-cs!jacobs, jacobs@cs.utah.edu)
"Limit congressmen to two terms.  One in office.  One in jail."
Newsgroups: rec.gambling