Score

Nynefingers

Well-Known Member
#41
Kasi said:
Sure, if u mean EV & SD are in $'s.

If so, "doubling betting unit" means orig $ roll has half as many min-units.

Which, in turn means, your risk is now square root of orig risk.
I wasn't trying to get into RoR at that point. I was only trying (and failing, as it were...) to explain what SD is actually trying to quantify and why it is useful compared to just using variance.
 
#42
SCORE is a DERIVED measure....

LC,

There is a short answer and a rather long answer.

The short answer is for any advantage player, at a given time, will be betting with a given spread, say $B_min to $B_max, with intermediate amounts for counts in between. There are also the given rules, and it depends on whether they leave at certain negative counts - these will all affect the result.

But lets break things up: [assume they true count Hi-Lo in a 6-deck game]

1/ Divide everything by $B_min - so that a $20 to $240 spread (assuming play-all), starting to rise at TC=+2 (say $40) up to $240 at TC=+8 [a bad spread] becomes:

Bet 1 unit at +1 or less
Bet 2 units at +2
.
take your pick < +8
Bet 12 units at +8 or greater. [ok so far...right]

This is known as a 1-12 unit spread (or 1-12u for short).

2/ Say the EV per 100 rounds is about $25 (I said is a bad spread), and the SD per 100 rounds is about $750.

Firstly convert this to per round, remembering that the EV is divided by 100, but the SD is divided by 10 (ie VAR is divided by 100).

So EV/r = $0.25 and SD/r = $75.00.

Then doing the division gives: 'ev' = 0.0125u ; 'sd' = 3.75u .

The quantities 'ev' and 'sd' are the unit EV and unit SD per round. If you think the 'ev' looks small you are right, but since the bets range between 1-12u (mostly at the low end) the 'sd' is reasonable.

In general if B_min is just called $B ($20 in this case), and keeping things per round (the 100 round thing is just confusing) then we simply have:

EV = $B * ev
SD = $B * sd
VAR = $B * $B * (sd)^2

3/ Probability theory for large N, says that you have about one 33% chance of being 1 standard deviation below expectation after SQRT(N)*SD. Now this works for either EV(ev) or SD(sd) as the $B cancels out, so an unlucky player(33%) could be zero (or less!) after N hands if

0 = N * EV - SQRT(N)*SD or N*N*EV*EV = N*SD*SD .

Canceling the N and rearranging gives N = (SD/EV)^2 = VAR/(EV)^2 where we get the (famous?) value

N0 = (sd/ev)^2 = (SD/EV)^2

which is called the long run index.

4/ You will have to take my word on this, but Kelly Theory says that for a game using unit 1-M spread, with a given 'ev' and 'sd' has what is called an Equivalent Kelly Bankroll (ekb) given by

'ekb' = sd^2/ev = var/ev.

It is no coincidence that if we put back the $B, we get the dollar EKB

EKB = (SD^2/EV) providing the EV is per round, not 100.

So for the example above we get

N0 = [(3.75)/(0.0125)]^2 = 90000 rounds (lousy)
ekb = (3.75*3.75)/(0.0125) = 1125

and EKB = $20 * 1125 = $22500
(which means if this is your bankroll you have a risk of ruin of 13%. Not a good return on investment. Also notice (it drops out from the math) that

EKB/N0 = $0.25 = EV - not a coincidence!

Therefore your win rate per hand is simply your Equivalent Kelly Bankroll divided by your long run index.

Optimal Betting Theory says that simply by reducing your high bet down to TC=+4, you can halve N0, lower your EKB and increase your win rate at the same time.

But there is another trick - for a given 'ekb', you can choose your unit bet $B so that you can match your Kelly Bankroll. Now what happens if you optimise your betting spread (minimise N0), and adjust your $B such that your EKB is $10000???

You guessed it - your win rate EV/r = $10000/N0 !

Then if you wanted to have a win rate per 100 rounds you get

EV(100) = ($10000/N0)*100 which just happens to be the SCORE!!

I originally came up with N0 with my work on Optimal Spread Theory. Any N0 more than around 20000 is probably unplayable, unless you Wong which really can lower it, and good single deck games should be less than 10000.

Don felt that players relate better to win rate rather than long run chances of being ahead, so he invented SCORE - which is really just (1/N0) multiplied by $1,000,000. The main difficulty I have with it is just that players may just consider it a win rate, which can be calculated for any non-Optimal spread (that is to really 'KNOW' a game, you you need to know the ev and sd (or equivalently NO and ekb, not just N0 which you get from SCORE).

In other words, you should know the $B Don has used to convert the unit 'ekb' to $10000. He may well do this in his book, you need to check.

I believe QFIT has programs like CVCX to compute Optimal Betting Spreads - doing a Google is a good idea.

If all this has been a bit messy, you can always go to:

http://www.blackjackforumonline.com/content/TOClibrary.html#Bettingsystem

and find my two articles under 'Blackjack Betting for Card Counters'. The first 'How-To' article explains the stuff I said above, the second one is really only for the mathematically inclined who may have access to simulator data.

Cheers,
Dr. Brett Harris.
 
#43
Welcome Dr. Harris

Thank you for adding to the discussion. If you have the time I have a related question.

Is minimization of NO always the goal?

If one through winning is approaching table max bets there seems to be the thought that you just place your max bets sooner. However, this raises NO even though it increases EV. My thoughts are this would be acceptable if your overall ror or risk of losing a certain percentage of bank is very low. I am thinking 1/4 kelly optimal betting or perhaps 1/5 or 1/6th fixed kelly ror. I would be very interested in your thoughts. Thank you for your time.

:joker::whip: Think I am the joker this time LOL
 

bjcount

Well-Known Member
#44
Brett_Harris said:
LC,

There is a short answer and a rather long answer.

Cheers,
Dr. Brett Harris.
Dr. Harris,

Not being too knowlegable in advanced mathematics, statistics, or probability I have to thank you for that explanation. It was so simply explained I found it no harder to understand then counting down a deck of cards! .... and no greek letters or 3 dimensional formulas either! :)

BJC
 

assume_R

Well-Known Member
#45
Brett_Harris said:
The first 'How-To' article explains the stuff I said above, the second one is really only for the mathematically inclined who may have access to simulator data.

Cheers,
Dr. Brett Harris.
Dr. Harris,

Thank you for this very informative post; there was so much information packed into this, and it is very much appreciated.

I have 2 questions:
1. In your paper, I noticed that you said "As such, [N0] is a measure of how many rounds must be played to overcome a negative fluctuation of one standard deviation with a fixed spread." My question is that what about trying to overcome 2 standard deviations? How does that affect the EKB and N0 values. The reason I ask is that 1SD by definitely should occur, on average, 67% of the time, but what if somebody wants to overcome negative fluctuations that would occur 95% of the time (i.e. 2 standard deviations?)

2. I assume there is probably a very simple relationship, but could you comment on how N0 relates to the statistical value of standard error of SD/sqrt(N), which I'm sure you know can be defined as the chance that your mean will be close to the expected value, as well as the statistical concept of confidence intervals?

I know this is probably getting off topic from LC's original post about score, but I couldn't help but ask ;)
 
#46
Not Dr. Harris

assume_R said:
Dr. Harris,

Thank you for this very informative post; there was so much information packed into this, and it is very much appreciated.

I have 2 questions:
1. In your paper, I noticed that you said "As such, [N0] is a measure of how many rounds must be played to overcome a negative fluctuation of one standard deviation with a fixed spread." My question is that what about trying to overcome 2 standard deviations? How does that affect the EKB and N0 values. The reason I ask is that 1SD by definitely should occur, on average, 67% of the time, but what if somebody wants to overcome negative fluctuations that would occur 95% of the time (i.e. 2 standard deviations?)

2. I assume there is probably a very simple relationship, but could you comment on how N0 relates to the statistical value of standard error of SD/sqrt(N), which I'm sure you know can be defined as the chance that your mean will be close to the expected value, as well as the statistical concept of confidence intervals?

I know this is probably getting off topic from LC's original post about score, but I couldn't help but ask ;)
for fixed bets
1sd NO
2sd NO is 4NO
3sd NO is 9NO

If you continously resize NO increases.
1 sd, 4NO
2 sd, 16NO
3 sd, 36NO
If you resize less frequently then NO is slightly less.
 

Nynefingers

Well-Known Member
#47
Brett_Harris said:
an unlucky player(33%) could be zero (or less!) after N hands
Thanks for the great explanation. My only issue is, if I'm understanding you correctly to be talking about N0 at this point in your post, this statement should read 16%, not 33%. I may be reading it wrong or you may have made a simple error, as obviously you know that ~33% of the results will fall outside +/- 1SD, so half will fall on the bad side and half on the good side. I think most people here know that as well, but for the benefit of the newer folks that might be reading this, I wanted to make sure it was clear.
 

assume_R

Well-Known Member
#48
blackjack avenger said:
for fixed bets
1sd NO
2sd NO is 4NO
3sd NO is 9NO

If you continously resize NO increases.
1 sd, 4NO
2 sd, 16NO
3 sd, 36NO
If you resize less frequently then NO is slightly less.
Thanks, avenger! I assume most AP's would fall between these 2, as the bets will be variably sized, but using some cover would prevent extreme varying of the bets.
 
#49
Real World Trouble

assume_R said:
Thanks, avenger! I assume most AP's would fall between these 2, as the bets will be variably sized, but using some cover would prevent extreme varying of the bets.

If you resize you will be much closer to the larger NO's
If you don't always bet with your advantage NO increases
Take any money out of the bank NO increases:joker::whip:
 

muppet

Well-Known Member
#50
i guess i'll take a stab at these

assume_R said:
1. In your paper, I noticed that you said "As such, [N0] is a measure of how many rounds must be played to overcome a negative fluctuation of one standard deviation with a fixed spread." My question is that what about trying to overcome 2 standard deviations? How does that affect the EKB and N0 values. The reason I ask is that 1SD by definitely should occur, on average, 67% of the time, but what if somebody wants to overcome negative fluctuations that would occur 95% of the time (i.e. 2 standard deviations?)
i would assume it would simply be (2SD/EV)^2 instead of (SD/EV)^2. using the number in dr. harris' example we arrive at 360000 versus the n0 of 90000. i'm not sure what practical application this has though

assume_R said:
2. I assume there is probably a very simple relationship, but could you comment on how N0 relates to the statistical value of standard error of SD/sqrt(N), which I'm sure you know can be defined as the chance that your mean will be close to the expected value, as well as the statistical concept of confidence intervals?
hmm i'm not sure i understand this question correctly. how is 'standard error of the mean' or confidence intervals applicable? these are based on some sampling distribution whereas n0 is derived through theoretical numbers/calculations
 
Top