# Help on events

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If I have say 24 red possibilities and 24 black possibilities and 8 (white) possibilities that fall under breaking a red or black streak what should be the random occurences of 1 in a rows 2 in a rows 3 in a rows for red and black. What about white?

Any help.:33:

at the Crescent Dealer's School

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• Baccarat Hall of Fame Member

Hi Keith,

if by "white" you mean 0/00 they're 2/38 = 1/19

so the prob to find 2 white in a row is 19x19=361

so is 1 on 361.

If this can help...

andrea

bacclover

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Hey Keith,

I don't know of a game that offers 24 red's + 24 black's + 8 white's ( = 56 total )

But lets do it anyway big grin

When calculating possibilities or "chances" you only need two words: AND...OR

With "AND" you multiply chances ( x )

With "OR" you add up chances ( + )

red = 24/56

black = 24/56

white = 8/56

1 red's in a row ( the same goes for black )

( 1st=red AND 2nd=white ) OR ( 1st=red AND 2nd=black )

<=> ( 24/56 x 8/56 ) + ( 24/56 x 24/56 )

<=> ( 192 / 3136 ) + ( 576 / 3136 )

<=> 768 / 3136

<=> 24.50%

2 red's in a row ( the same goes for black )

( 1st=red AND 2nd=red AND 3rd=white ) OR ( 1st=red AND 2nd=red AND 3rd=black )

<=> ( 24/56 x 24/56 x 8/56 ) + ( 24/56 x 24/56 x 24/56 )

<=> ( 4608 / 175616 ) + ( 13824 / 175616 )

<=> 18432 / 175616

<=> 10.50%

3 red's in a row ( the same goes for black )

( 1st=red AND 2nd=red AND 3rd=red AND 4th=white ) OR ( 1st=red AND 2nd=red AND 3rd=red AND 4th=black )

<=> ( 24/56 x 24/56 x 24/56 x 8/56 ) + ( 24/56 x 24/56 x 24/56 x 24/56 )

<=> ( 110592 / 9834496 ) + ( 331776 / 9834496 )

<=> 442368 / 9834496

<=> 4.50%

And so on for 4, 5, 6, ... in a rows.

Now for white:

1 white's in a row:

( 1st=white AND 2nd=red ) OR ( 1st=white AND 2nd=black )

<=> ( 8/56 x 24/56 ) + ( 8/56 x 24/56 )

<=> ( 192 / 3136 ) + ( 192 / 3136 )

<=> 384 / 3136

<=> 12.25%

2 white's in a row:

( 1st=white AND 2nd=white AND 3rd=red ) OR ( 1st=white AND 2nd=white AND 3rd=black )

<=> ( 8/56 x 8/56 x 24/56 ) + ( 8/56 x 8/56 x 24/56 )

<=> ( 1536 / 175616 ) + ( 1536 / 175616 )

<=> 3072 / 175616

<=> 1.75%

3 white's in a row:

( 1st=white AND 2nd=white AND 3rd=white AND 4th=red ) OR ( 1st=white AND 2nd=white AND 3rd=white AND 4th=black )

<=> ( 8/56 x 8/56 x 8/56 x 24/56 ) + ( 8/56 x 8/56 x 8/56 x 24/56 )

<=> ( 12288 / 9834496 ) + ( 12288 / 9834496 )

<=> 24576 / 9834496

<=> 0.25%

That's that

Have fun!

Bye,

Heike

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Heike,

Thanks. Can we figure in a time element such that, in the course of an hour we should see ( or rather an amount within the statistical norm would be )x amount of 1 in a rows x amount of two in a rows etc within an hour with 75 rounds chances an hour.

KS

Edited by Keith Smith

at the Crescent Dealer's School

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Hi Keith,

if by "white" you mean 0/00 they're 2/38 = 1/19

so the prob to find 2 white in a row is 19x19=361

so is 1 on 361.

If this can help...

andrea

Yes it does thank you

at the Crescent Dealer's School

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Hey Keith,

The % chances are for every 100 "rounds"

To know the numbers for any other amount of rounds, you just use "the rule of 3"

For example:

1 red's in a row ( the same goes for black ) --> 24.50 per 100 rounds

FOR 75 ROUNDS

if 100 rounds = 24.50

then 1 round = 24.50 / 100 = 0.245

and 75 rounds = 0.245 x 75 = 18.375

FOR 250 ROUNDS

if 100 rounds = 24.50

then 1 round = 24.50 / 100 = 0.245

and 250 rounds = 0.245 x 250 = 61.25

In your case, with 75 rounds, it is exactly 75% of the normal chances for every 100 rounds...

1 red's in a row ( the same goes for black ) --> 24.50 x 75% = 18.375

2 red's in a row ( the same goes for black ) --> 10.50 x 75% = 7.875

3 red's in a row ( the same goes for black ) --> 4.50 x 75% = 3.375

1 white's in a row --> 12.25 x 75% = 9.1875

2 white's in a row --> 1.75 x 75% = 1.3125

3 white's in a row --> 0.25 x 75% = 0.1875

But... what exactly is it you are inventing?

I mean, 24 red's and 24 black's and 8 white's... is it some kind of win every round casino buster system

Bye,

Heike

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Cool thanks sometime my brain just stops in my old age.

Thanks

KS

at the Crescent Dealer's School

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If we figure that we have an average 7 player game with the dealer we have 16 cards dealt in BJ.

If we add the hit cards we average about 22 cards per round. In 75 rounds per hour we have a total of

1650 events so to speak. From your calculations in a random game we should see about 24% one in a rows. Of course a one in a row is any combination of these subsets, and an ace ace could be a high card or a low card.

( Ace, King, Queen, Jack, Ten, Nine) or ( Ace, Two, Three, Four , Five, Six)

Red Black

Neutral card are 7 and 8

White

So I should see a 3 in a row of anything Red ( High Cards) or Black (Low Cards) etc.

Edited by Keith Smith

at the Crescent Dealer's School

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So in an hour we see 1650 cards aprx. How may times should we observe a 5 in a row, a 6 in a row, 7 in a row etc. And if we see them all night long, too many times we can say that the cards are clumped up. Is this what you are getting at?:261:

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• Users

Well, I have never accepted the counter definition of Hi, Lo, Neutral. It simply is not consistent with the way the game is actually played and it tends to muddy the waters. Aces are actually played low as often or more often than they are played High. So Aces are a poor indicator of high vs. low clumps.

I think youare better off to look at it this way:

2 thru 7 = low

8 thru K = High

Aces swing: Count them Hi when they follow a Hi and Lo when they follow a lo.

This gives you a balanced count with 6 lo cards and 6 Hi cards with Aces swinging. And no neutral cards to muddy up your count.

I know this is brand new to non NBJ players and card counters but I suspect that even card counters would be better off with this count.

What causes clumping in the first place? It is the fact that break cards are picked up as a group and placed into the discard shoe together. Then Non break cards are picked up thus sorting the discard rack into break card clumps and non break card clumps. The challenge for the casino is to maintain this clumping through the shuffle. Machines do that best.

Think about non break cards. What are they MOSTLY? 2 card 20's, 19's, 18's, and 17's in that same order of occurrence. Sure, a player may hit to a good hand but what are they MOSTLY?

Now, what are non break cards? Well, we know his last card must be a 6 or more. Otherwise he wouldn't have broke. But what are the rest of his cards. MOSTLY low right?

Think about 8s and 9s. Are they mostly in the non break group or the break group. Yep. they are high aren't they? When you double on 11 and get an 8 or 9 are you happy? It's a lot better than a 7 or less isn't it.

Think about 7s: You can pair a 7 with only 5 cards 10-A to make a pat hand. But 8 cards, 2-9 make your 7 a hit hand. 7s belong with the break card group don't they?

As I've always maintained. One of the greatest of the many many flaws with card counting math is that it totally ignores the way the game is actually played. If you want to win at BJ, the LAST thing you want to do is ignore the way the game is actually played.

So Keith, I think you need to rearrange your marbles.

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I wasn't really concerned with what card counters think I was arranging them my way, as we play. This way I can try to get a more objective "count" on how clumped the game is. Sorry Kit my shooter just knocked your marble out of the ring big grin But yes I agree the eights tend to clump with the high cards and all what you said I'm trying to get a discussion going here.

Who wants to figure out what the statistical norm for a random game in an hour for 4 in a rows fives and runs greater. If we can do the norm then I was trying to place an objective count on clumping frequency in a game, Not only to survey the game but the casino.

I don't think it matters if you only have a high, low sets and no neutral but i would like to do the math to get an idea how many times in random we should see a ten card run of highs or lows.

If we see it 5 times in an hour then how does that compare to norm and then what if I look in a second game and see it 7 times in an hour.

Edited by Keith Smith

at the Crescent Dealer's School

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Oh, and you naive guys who still think machines produce random cards. Think about this: If the cards were random Basic Strategy, by itself, would win with a 6% advantage. Is that what you think is happening? There would be no BJ, would there. Get real! Why did the casinos throw out the 1st generation machines? the second generation machines? the third generation machines? the 4th generation machines?

Consider this! True random machines would make the casinos lose at the rate of 6%. Casinos are not stupid. Casinos are not going to stick with machines that make them lose. They are not going to stick with random machines are they. But a machine that can maintain clumping even better than the dealers could. That's a whole different story isn't it?

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That's all fine Keith. I'm not arguing with you. I think you are working on a good idea. I'm just suggesting that you will get a much clearer picture of actual clumping if you rearrange your marbles to the count I recommend. That IS how we play.

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Well I am trying to get some counters reading this too anyway ok reaarange them in the order we play. Play Ace high and low. Highs are Ace - Eight and Lows are Ace - seven. How often should we see a 10 in a row in a random game and then once we know what it should be we can jusdge it against what we see. :banghead:

at the Crescent Dealer's School

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Ellis - Keith

Are we going to be using this for NBJ/WCB table selection? :33:

"If you don't think too good, don't think too much!!"

-----------------------

John

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No I just want to figure out a simple concept, in a random game how often should we see a "run" of straight high cards or low cards. After we know how often it should occur in a random game what are we actually seeing.

at the Crescent Dealer's School

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No I just want to figure out a simple concept, in a random game how often should we see a "run" of straight high cards or low cards. After we know how often it should occur in a random game what are we actually seeing.
Still sounds like a table selection method! LOL like counting the dealer ten ups. How many is that supposed to be in a random game? 4 out of 13? How many do we see in real games? Basic non counting questions..... :33:

"If you don't think too good, don't think too much!!"

-----------------------

John

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No i'm not trying to do a table selection I am trying to place numbers of Blackjack events.

Ok so if we group the cards as Ace, 2, 3, 4, 5, 6, 7 = low

and, 8, 9, 10. 10, 10, 10, Ace = High So we have a 50/50 change of a high or low appearing. Now just to give it a simplier perspective, I said lets call Low cards red if any of them follow each other and high cards black.

So this run of cards Ace, Jack, Queen Ace, 10, 9, 10 is a seven in a row of high card so for the sake of events a black seven in a row. Now for a random game how often should I see that . In an Bac shoe I should see it about 1 time every two shoes or about .5 times a shoe on average. I see this about 3 times and even more time a shoe in most multiple deck blackjack games.

By knowinghow often the events should occur in a random game i can gauge by observing how clumped or closer to random games are . Does that make sense?:1zhelp:

at the Crescent Dealer's School

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• Users

Yes it makes perfect sense and yes, you ARE using it for table selection. What I'm trying to get across to you is that there is a much simpler far more accurate way of doing this that not only tells you if the game is clumped or random but also tells you Exactly HOW clumped the game is.

Look, forget Aces altogether because they pretty much fall equally with highs or lows since they are played high or low about equally.

That leaves you:

234567 as your 6 low cards and

8 9 10 10 10 10 as your 6 high cards

Now, looking at your highs vs lows that way, and ignoring aces altogether count +1 every time a hi follows a high or a low follows a low. Count -1 every time a lo follows a hi or a hi follows a lo.

Now a plus count tells you the game is clumped and the higher the count the more the game is clumped.

A 0 or minus count tells you the game is random.

But, in addition, a minus count tells you you're looking at a super random, dealer breaking game. Such a condition signals an UNLOSABLE NBJ 3rd base, jugular opportunity.

This is extremely simple once you get used to the new count values. But it tells you so much more than the way you are trying to do it AND you don't have to remember a bunch of event frequencies.

Mad Dog, you NBJ players, are you watching this stuff?

Carlos, you've been looking for a way to do this for years. Voila!

Neil, How do you like them apples?

Edited by ECD
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Ok you and I understand this and the way to play. My point is that when you try to explain to a card counter the concept, it is easier for that party line to understand that a seven high cards in a row shouldn't occur as often as we see it, rather than trying to show them a count that tracks as you say.

If I say to them, look we just saw 7 high card s in a row in this game 15 times in two shoes they should get that becasue the random occurance of this happening is 1 time every other shoe.

And the fact that this happens on every table its not a statistical irregularity but contrived. I am trying to make a point about clumping and how you can effectively get someone who is oblivious to it to understand it not make a system or approach

at the Crescent Dealer's School

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Well, fine if that is what you think. To me, counters understand counts. But whatever approach you try is fine by me.

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They don't even understand they lose!!!There convinced they win as their bankroll diminishes waiting for the long term surprised

at the Crescent Dealer's School

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• 3 years later...

I love this! Perfect Ellis!!

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Well I'm not sure it does us much good to be dragging up these arguments and discussions from 2009. Today we understand the frequency of events from Baccarat. Since they are both 50/50 it makes no difference whether you are looking at Bank vs Player or high vs low. Streaky vs choppy is the same as looking at clumped vs random. But even that is beside the point.

On this BJ go around I'm trying to keep it plain and simple. In random cards you play 3rd base and favor basic strategy. In clumped cards you play first base and favor dealer strategy. You determine whether to hit or stand by what is running. That is far more accurate than basic strategy in clumped cards.

Sure, you could determine clumped vs random by knowing the normal frequency of events. But you don't need that. All you need do is look at your watch. If the cards are 2 hours old or more they are clumped. (First Base)If they are new, they are random. (3rd base)

Look at how crowded the casino is. The more crowded it is and the later in the day it is the more the cards will be clumped right up to the point of unplayable. The earlier it is, the newer the cards are, the fewer the players, the more it will favor random and third base.

Here is what you need to know: In random cards you have no idea of the next card so you play basic.

In clumped cards you have a good idea if the next card is high or low so you play accordingly instead of basic.

In random cards you have no idea of the dealer's hole card so you play basic.

In clumped cards you have a good idea of the dealer's hole card so you play accordingly.

Random cards you favor basic.

Clumped cards you favor dealer strategy. After all, that is what she is beating you with.

Practice enough and all this becomes automatic.

But know this: In spite of all the hupla and Hollywood movies, card counters have no chance of winning.

Practiced NBJ players win and we are the ONLY winners.

My films are not trying to teach you how to play. Not yet anyway. They are merely showing you the conditions under which you must learn to play and how those conditions were created.

The game is rigged. Learn to use their own rigging against them. THAT is NBJ.

Edited by Guest
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Well said Ellis! Now on to something very important (to me lol) God Damn it, it's my Birthday today!!

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