REPOST: Bad Players Hurt You and Other True Count Myths
Abdul Jalib M'hall
===============================================
Abdul Jalib M'hall
Often, I hear counters and others complaining that bad
players at third base hurt the expectation of the others at
the table, because these bad players hit when they should
not, tending to "steal" the dealer's bust card.
Another thing many people say is that in European style
no hole card blackjack, you can attempt to give or not give
the dealer a hole card, by varying your hitting or standing
at third base.
And in the past some have published articles suggesting
that because a blackjack round is more likely to end in a
ten than not, that basic strategy should be different even in
a constant number of rounds situation because of this.
All these things are false for the same reason, unless you
have specific information about the order of the cards
remaining to be dealt. For the rest of the article, I will
assume that the cards are randomly shuffled as far as you
can tell and that the dealer does not run out of cards.
The reason in a nutshell is that you can't drive the true
count up or down on average, no matter how hard you try.
Stealing the Dealer's Bust or Hole Card
---------------------------------------
Bad players, or good players for that matter, cannot
change the expectation of the hands hands already com-
pleted that round. Even if they maliciously try to steal the
dealer's bust card or hole card, or hit in high true counts in
an attempt to drive the count down, they can't affect any-
one else at the table who has acted before them. Think of it
this way. Every card in the rest of the deck is random. It
doesn't matter whether the dealer gets the first card from
the top or the fifth card from the top, those cards are equiv-
alent as far as you are concerned beforehand.
You could argue that bad players at third base will some-
times hurt you. Sometimes after the fact, you can see that a
bad player has stolen the dealer's bust card and let the
dealer make a good total with a small card. But, some-
times after the fact, if you were observant you'd see that a
bad player had helped you. They will sometimes steal the
dealer's small hit card and leave him with a bust card.
When you average it out, bad players have absolutely no
net effect on you! The same goes for good players.
Of course, if you are counting cards and bad players are
standing "too often", then you see fewer cards before it is
your turn to play and bet, and so the bust-scared players
are costing you in that sense. The converse of this is that
the ones that split and hit "too often" are helping you by
letting you see more cards before it is your turn to play or
bet next, so long as they don't make the dealer shuffle a
round earlier. When the dealer is dealing a constant num-
ber of rounds dealt, "excessive" hitting and splitting by
other players will have no affect on other players and can
only help card counters on average.
The underlieing mathematical reason is that removing and
revealing games does not affect the true count. An inter-
esting thought experiment called "Say Red" will demon-
strate this.
Say Red
-------
Say Red is a game played as follows. First we each bet a
dollar, winner takes all. We start with a standard deck, half
red, half black, randomly shuffled. Now, I let you turn over
cards one at a time, until you are ready to bet that the next
card will be red. If you do not decide until the last card,
then you are forced to settle the bet on that last card.
One strategy you might choose is to draw cards until you
draw a black card, stopping at that point. With this strat-
egy, you often create a mildly red-rich deck, but you also
increase the probability of a heavily black-rich deck. In
one extreme, you draw a black card right away and create
a mildly red-rich deck, and in the other extreme, you draw
every red card in the first half of the deck and create a
remaining deck that is 100% black. "Amazingly," these
situations balance out exactly, and you can't make any
money with this strategy.
Even more "amazingly," you can be as devious as you
want, but no strategy can change your expectation of get-
ting exactly your $1 back on average, that is, you have a
50-50 shot at winning the $2. This is because you can
never change the expected fraction of red cards from the
original 50%.
It is a mathematical theorem that no such removal strategy
affects the expected (average) proportion of red cards to
black cards. Rather than prove this theorem, I'll show and
prove a more general one, which applies to blackjack.
The True Count Theorem
----------------------
The following is a mathematical theorem:
Theorem: the expected value of the true count after a
card is revealed and removed from any deck composi-
tion is exactly the same as before the card was
removed, for any balanced count, provided you do not
run out of cards.
See the proof at the end of the article. Expected value is a
precise mathematical term defined as the mean average, which
is computed by summing the probability of an event times
the value of that event, over all possible events. So the
expected value of the true count after drawing a card is the
summation of the probability of drawing each card times
the value of the true count after drawing that card.
One consequence of this theorem is that the expected
value of the count after any number of cards have been
revealed and removed will be the same as before, and so
the expected value of the count after a round has been
dealt will be the same as before. And when you get a con-
stant number of rounds, the expected true count after each
is zero. Since the true count starts at zero, the overall
expected value of the count is zero, when you get a fixed
number of rounds.
The Running Count
-----------------
The theorem applies only to the true count, not to the
running count. The running count does not obey the same laws
of as the true count.
With regards to the effects of other players at the table and
the tendency of the round to stop with a big card, much
confusion stems from a mistaken assumption that the
behavior of the true count is the same as the behavior of
the running count.
The running count must be zero at the end of the deck.
Therefore, drawing cards in high counts tends to cause the
running count to fall, and drawing cards in low counts
tends to cause the running count to fall. But the expected
true count is unchanged.
The Proof
---------
Theorem: the expected value of the true count after a
card is revealed and removed from any deck composi-
tion is exactly the same as before the card was
removed, for any balanced count.
Proof:
Let Wi be the Weight of the card of rank i, i.e., the
count value.
Let Ni be the Number of cards of rank i already
revealed, counted, and removed from the deck.
Let Si be the Starting number of cards of rank i in a full
deck.
Let Li be the number of cards of rank i currently Left in
the deck, i.e., Li=Si-Ni
Let C be the number of Cards remaining in the deck,
i.e., C=sum_over_i{L[i]}
Let R be the Running count, i.e., sum_over_i{W[i]*S[i]}=0.
Let T be the True count, i.e., T=R/C
Assume that the count is balanced, i.e., sum_over_i{W[i]*S[i]}=0.
Need to show that: T = sum_over_j{((R+W[j])/(C-1))*(L[j]/C)},
that is to say that the true count after a card j is
revealed, removed, and counted, averages out to the
same as the true count before. The average is computed
by adding up the true counts after each card is drawn
weighted by the probability of each card being drawn.
sum_over_j{((R+W[j])/(C-1))*(L[j]/C)} =
(1/(C(C-1)))sum_over_j{(R+W[j])L[j]} =
(1/(C(C-1)))sum_over_j{(RL[j]+W[j]L[j])} =
(1/(C(C-1)))sum_over_j{(RL[j]+W[j](S[j]-N[j])} =
(1/(C(C-1)))sum_over_j{(RL[j]+W[j]S[j]-W[j]N[j])} =
(1/(C(C-1)))(sum_over_j{RL[j]} + sum_over_j{W[j]S[j]} - sum_over_j{W[j]N[j]}) =
(1/(C(C-1)))(R*sum_over_j{L[j]} + 0 - R) =
(1/(C(C-1)))(R*C-R) =
(1/(C(C-1)))(R(C-1)) =
R/C =
T
.
. . QED
Again...
Theorem: the expected true count after a card is revealed and removed from
any deck composition is the same as before the card was removed,
for any balanced count, provided you do not run out of cards.
Corollary: the expected true count after any number of cards are revealed
and removed from any deck composition is the same as before the
cards were removed, for any balanced count, provided you do
not run out of cards.
Corollary: the expected true count after a round is the same as before
the round, for any balanced count, provided you do not run out
of cards.
--
Abdul |"Probability theory is... common sense reduced to calculation."-Laplace
Jalib |"randomness morphs into predictability."-Sarnelli
M'hall |"Insane lotto theory is calculation reduced to nonsense."-M'hall
Paul R. Pudaite
M'hall) wrote:
> Bad Players Hurt You and Other True Count Myths
> ===============================================
> Abdul Jalib M'hall
>
> Often, I hear counters and others complaining that bad
> players at third base hurt the expectation of the others at
> the table, because these bad players hit when they should
> not, tending to "steal" the dealer's bust card.
Gretchen and I played our first semi-serious session of
blackjack on Monday -- the $2 minimum single deck game
at Binion's Horseshoe -- trying to get down basic
strategy.
We had a guy at our table (5th seat of 6 -- short stop?) who
would sometimes stand under 10 if the dealer had a bad card
up and 3rd base was also standing. Normally, I suppose,
the table would have been chewing this guy out, but
it worked repeatedly, and even the elderly woman sitting
next to us was giving him high 5's after yet another
dealer bust.
Paul R. Pudaite
pud...@msu.edu
cmo...@sedona.intel.com
Abdul Jalib M'hall <mh...@netcom.com> wrote:
>
>Theorem: the expected value of the true count after a
>card is revealed and removed from any deck composi-
>tion is exactly the same as before the card was
>removed, for any balanced count, provided you do not
>run out of cards.
Hi again Abdul, I'm back from sabbatical and have my simulator
running again. I'm still seeing the phenomenon that I reported
months ago. The *mean* of the *true count* is more positive *during
one round* of blackjack than it is at the end of the round. I
lost access to internet when I took off on vacation and if you
gave an explaination, I missed it.
This is not to say that anything you have posted is wrong. Just
wondering if the theorum allows for what actually happens in reality.
The mean of *all* the true counts after each card is dealt in one
round of blackjack is positive, like a DC bias. Is it OK for the true
count to have a positive mean? Is the expected value positive during
the time when a positive mean exists and then go back to zero at the
end of the round?
Best Regards, cmoore (not speaking for my employer)
Abdul Jalib M'hall
>In article <mhallD9...@netcom.com>,
>Abdul Jalib M'hall <mh...@netcom.com> wrote:
>>
>>card is revealed and removed from any deck composi-
>>tion is exactly the same as before the card was
>>removed, for any balanced count, provided you do not
>>run out of cards.
>
>running again.
Oh no. :)
>I'm still seeing the phenomenon that I reported
>months ago. The *mean* of the *true count* is more positive *during
>one round* of blackjack than it is at the end of the round. I
>lost access to internet when I took off on vacation and if you
>gave an explaination, I missed it.
"During one round of blackjack" is not a well-defined stopping point.
It is well known that the last card of the round is more likely to
be a ten than not, and so I don't doubt that the mean of the true
count is more positive during one round of blackjack (or one card
before the end) than it is at the end of the round. So what?
This is not what the theorem is talking about, and this little
factoid has no practical relevance that I can think of. There
are an infinite number of such tautological useless factoids, such
as that the number of remaining cards in a single deck is always
factorable using the numbers 1..52.
>This is not to say that anything you have posted is wrong. Just
>wondering if the theorum allows for what actually happens in reality.
Of course the theorem allows for what actually happens in reality.
Read the theorem again. Turn over cards from a fresh randomly
shuffled deck, stopping any time you wish and computing the true
count there; now turn over additional cards one at a time
without peeking, stopping any time you wish; when you stop, the mean
true count is exactly the same as the true count before you turned
over additional cards. The stopping times do have to be well-defined.
>The mean of *all* the true counts after each card is dealt in one
>round of blackjack is positive, like a DC bias. Is it OK for the true
>count to have a positive mean?
You can make your tautological factoids infinitely complex and at some
point you will succeed in confusing me, and if that is your point, then
I concede now.
A round of blackjack tends to either chew up a lot of small cards,
creating many very positive true counts, or a few big cards,
creating a few mildly negative true counts. At the end of the
round, or at any other well-defined stopping point, the true count
will average exactly the same as before.
However, one cannot average different numbers of true counts across
different decks and expect it to come out to zero. Think of it
this way; suppose you are not playing blackjack, but just drawing
cards at will, and you decide to keep drawing cards when you have
a positive true count, stopping any time the true count goes negative.
With this approach, you can often go much of the way through a deck
with a positive true count, before having to stop. On the other hand,
when you draw a big card right away and the true count goes mildly
negative, you can stop right there. Average all these true counts
together across different decks, and you will obviously have a positive
mean. All this means to me is that it's meaningless to average true
counts across different decks, at least when you are getting different
numbers of true counts from different decks.
>Is the expected value positive during
>the time when a positive mean exists and then go back to zero at the
>end of the round?
See my answer at the start.
The mean true count is zero at the end of the round or any other
well-defined stopping point. If you pick an equal number of
well-defined points in each deck, like the end of each round in
a fixed rounds situation, then it seems to me that you could average
meaningfully across decks and come out to zero.
--
Abdul |
Jalib | Lotto, the choice of an insane generation.
M'hall |
cmo...@sedona.intel.com
>>card is revealed and removed from any deck composi-
>>tion is exactly the same as before the card was
>>removed, for any balanced count, provided you do not
>>run out of cards.
I'm no mathematician and I'm trying to the understand what the
theorum means to us in the real world and compare its expected
values to my measured simulated results.
Assume one million rounds of one round each playing basic
strategy. Add the TRUE_COUNT to the TRUE_COUNT_SUM after every
card is dealt (as you suggested).
I assume the expected value of the True Count will be zero after
the first, second, third, and fourth cards, since they are always
dealt. Also, I assume that the expected value of the True Count
will be zero after the last card of the round. Of course, the
forth card will be the last card about 17% of the time. So the
expected value of the True Count will be zero for 4,827,994 of the
total 5,445,672 cards dealt, right? Question is, what is the
expected value after the other 617,678 cards and what is the
actual value that occurs?
The TRUE_COUNT_SUM is +733,749 after one million hands. Dividing
by 617,678 means that if the True Count is indeed zero after the
first, second, third, fourth, and last cards, then the True Count
must average more than +1 for the rest of the time. Seems to me
the theorum doesn't predict what actually happens. And if it
doesn't, what good is it? Am I missing something? Why do we go
to so much trouble to force casino-unlike conditions, like rounds
instead of number of cards dealt from a deck, so that a
mathematical theorum will be true? ... an engineer wonders about
such things.
cheers, cmoore (not speaking for my employer)
Ken Fuchs
>
> >In article <mhallD9...@netcom.com>,
> >Abdul Jalib M'hall <mh...@netcom.com> wrote:
> >>
> >>card is revealed and removed from any deck composi-
> >>tion is exactly the same as before the card was
> >>removed, for any balanced count, provided you do not
> >>run out of cards.
>
Cecil,
You are mis-applying the theorem. The theorem applies to the expected
value of true count after any "unknown" cards are shown. That is,
the expected value of the true count after one more card or two or
ten is the same as the true count right now. In your example you
look at the true count at a point where the round (or a hand) ends,
not an arbitrary number of cards N. The fact that the round has
ended gives extra information about the cards that WERE PREVIOUSLY
DEALT. Knowing that a hand or round has just ended gives you some
information about the last card (or just looking at the card will give
you information about that card); now, that expected value stuff is
no longer needed because you know exactly what happened in the past.
Since you only know when a round ends after the fact, you
cannot make ANY use of phenomena which you have correctly observed.
I can make other observations and theorems that are very similar in
nature to yours: The running count generally increases after a small card
shows up. The true count goes down everytime I see a ten.
The theorem does accurately model what is happening in a casino, you
are just mis-applying it.
-ken
_______________________________________________________________________
|Ken Fuchs | Secure Design Center |
|e-mail: ke...@comm.mot.com | Motorola, Schaumburg, IL |
| | |
|"The guy that's the best is the guy with the most." -Fast Eddie Felson|
|----------------------------------------------------------------------|
cmo...@sedona.intel.com
Abdul Jalib M'hall <mh...@netcom.com> wrote:
>You can make your tautological factoids infinitely complex and at some
>point you will succeed in confusing me, and if that is your point, then
>I concede now.
Thanks Abdul, if you are confused, I hate to think of the adjective that
describes my mental state. :-) I'm just trying to figure out if I can
gain additional advantage at blackjack by understanding the theorum.