Baccarat Bet Selection Options

Yes

Yes again.

Correct. Ball 99 had 5 opportunities to duplicate, over twice as many as the 2 opportunities Ball 12 had to duplicate.

No delusion, your perspective is temporarily askew. It is easy to get off track dealing with odds within odds.

Hopefully, now that I'm zeroing in on the specifics of your question, it will come back into focus for you.

 

Damnit. That's not right.

Playing a 7 step marty against a repeating Column of 7 isn't going to make it anywhere close to 2340 shoes before it busts.

 

Ah, you really aren't going to like this.

Sticking with a 7 hand sequence. It isn't a 1 in 16,384 chance to see it duplicate. It is only 1 in 128. Yeah, you aren't gunna like that at all.
Recording outcomes into 7col. There is a 1 in 128 chance to get a repeat. If there are a total of eight 7Col in a six deck shoe, then you have a 1 in 7 divided by 128 chance to see a repeating 7Col. That's a 1 in 18.285 chance that Columns 1 and 2, or 2 and 3, or 3 and 4, or 4 and 5, or 5 and 6, or 6 and 7, or 7 and 8 are a match.

1 in every 18 six deck shoes will produce 2 identical 7Col adjacent to each other.

The calculations made against a 1 in 16,384 chance apply accurately to a 14 hand sequence, or a bag with 16,384 balls in it.

 

Correct, any ball in any order.

We already know the answer to that one, so can put it to bed. However it wasn't what I was thinking., incidently, would you happen to know the expected frequency of this event?

Nope, see my first answer above.

FYI, there no longer exists any 6 deck game in the UK that I am aware of thankfully.

8 deck games, approx 70 hands per shoe excluding ties, 8x 7 columns. Repeats occur a lot more frequent than that what you suggest..

This is the crux of it, which you didn't address!!

 

You shouldn't make too many assumptions, you do know me after all

I think we have passed that one, unless they occur back to back...

8 Columns in 8 deck shoes, sometimes you may get 9, but it doesn't matter, end of shoe etc.

 

This could turn out to be very valuable information for a few members, the bigger the gap distance, the odds of a repeat occurring are greatly increased, once you are aware of this, which I will admit I wasn't. Then you can tailor your bets accordingly, OR take more stringent action.

You might be bending me slowly, but I'm still not 100% sold on the logic,

FYI, feast your math skills on this.

128 Balls in a bag, drawn in the following sequence (yes I release in this particular case, there were more than 8 trials);

Pick 1 - 10 (number 10 is drawn)
Pick 2 - 120
Pick 3 - 90
Pick 4 - 90
Pick 5 - 50
Pick 6 - 24
Pick 7 - 5
Pick 8 - 10
Pick 9 - 10

I post this a few posts back
Shoe produced, odds of 16,384, then odds of 2097152 (128/1 x (128/1 x 128/1)

 

11 hour gambling session last night till the wee hours of the morning. Probably be sometime tomorrow afternoon before I can spend the proper time get back to this.

 

Oh yeah, real quick, I got something to keep you up at night.

You mentioned the Law of the Third. Applying it to a seven hand sequence suggests the possibility that 1/3 of 128 combinations will not manifest in 128 trials. If you are betting against a sequence that has already manifested, it may be a "hot" sequence!

 

I beat that by about 6 hours, but had the luxury of lying on my bed

Within 8 picks, doubtful...

 

Worked all damn weekend when I expected to be off. Ready for a bloody nap, but will be up all night if I do, so let's see if we can put this to bed at least/last.

I'll take responsibility for getting fouled up with 1 in 16,000 odds and switching between balls in a bag and a 7 hand sequence. I should have been more on target than I was.

The balls.....

A bag of 128 balls. Picking one, recording its number, returning it to the bag, and picking another one. Repeat this process 8 times, and we get 8 numbers. The odds of picking the same number back to back is 1 in 128. We have 7 opportunities to do this so the odds of picking the same number back to back is 7 in 128. Which is is 1 in 18. ( rounded down )

It is not 1 in 16,000 ( rounded down )

The first number picked is not a specific number, the odds of it appearing are guaranteed at 128 in 128.

First number is 99. Odds of choosing 99 again after returning it to the bag is 1 in 128.
Second number is 88. It did not match the first number. Odds of choosing 88 again after returning it to the bag is 1 in 128.
Third number is 77. It did not match etc etc, odds are always 1 in 128 to pull it out the second time.

The fact that there was a 1 in 128 chance to pull number 77 ( or any number ) isn't applicable because we didn't try to predict number 77. We are only concerned with if ANY RANDOM number repeats itself. That ANY RANDOM number has no odds, it is guaranteed. It is the first repeat that has any odds associated with it, and those odds are to be divided by the number of trials. In this example it is 7.

So this 1 in 2 million (rounded down) shot you are reeling from. 128x128x128 represents pulling ball 99, putting it back in the bag and pulling ball 99 again, putting it back in the bag and pulling 99 again again. Without predetermining that ball 99 was going to come out first, then it's odds of appearing at random the first time are guaranteed at 1. And the following two repeats carry the odds of 128x128. Which is 1 in 16,000 (rounded down)

Still considering pulling 8 balls out of the bag, seeing one ball being pulled out 3 times in a row only has 6 opportunities. 123, 234, 345, 456, 567, 678. So these odds would be 1 in 2730.667. Derived from the calculation of 128x128/6.

Now, ditching the balls for Baccarat

It is a 7 hand sequence that carries the odds of 1 in 128. I know you don't play a Marty, certainly not a 7 step one. But we, you, I, somebody, anybody can check the accuracy of our odds against a 7 step Martingale.

You propose that any occurrence of of a 7 hand sequence repeating itself anywhere withing eight recorded columns of 7 decisions represents the odds of 1 in 16,000. IF you were playing a Marty you would be risking 127 units to win 1. ( 1+2+4+8+16+32+64 ) According to your odds, this would be successful 16,000+ times before busting. This cannot be.

Waiting for a 7 hand sequence to develop, and then betting against it 7 times risks 127 units. The odds are 1 in 128 that you will lose seven bets in a row.
Every new 7 hand sequence continues to carry the same odds of 1 in 128 that it will repeat itself.

If a 7 hand sequence repeats itself, and you lose seven bets in a row betting against it, and then continue the Marty from step 7 on to step 14 you bets increase by 128, 256, 512, 1024, 2048, 4096, and finally 8192. For a total risk of 16,383 units. You will only win this bet 16,383 times. So the odds of a 7 hand sequence appearing back to back to back can't be 1 in 2 million.

 

Pick 1 has odds of 1 in 128. Nothing but balls in the bag, guaranteed to pick any ball.
Pick 2 has odds of 1 in 128, coulda been number ten, coulda been any number, all share the same odds.
Pick 3 has odds of 1 in 128
Pick 4 had the odds of 1 in 128, even tho it is the same as Pick 3, Pick 3 was random and had a 1 in 128 chance to repeat.
Pick 5 has odds of 1 in 128
Pick 6 has odds of 1 in 128
Pick 7 has odds of 1 in 128
Pick 8 has odds of 1 in 128
Pick 9 is a repeat of Pick 8 and still carries the odds of 1 in 128.

Now, if you took Pick 1, and bet against Number 10 for the next 8 picks. You had 1 in 16 chance to lose that bet. (8/128). Your odds of losing that bet twice within 8 bets would be 1 in 292. (8/128)(7/128) or 16 x 18.285. Your odds of losing that bet twice in a row somewhere within 8 bets would be 1 in 2048. (8/128)(1/128) or 16 x 128.

 

My point about Law of the Third was....

Waiting for a 7 hand sequence ( or any sequence for that matter ) to develop and betting against it may have you betting against odds 1/3 less than expected.

If 42 of the 128 possible combinations of 7 decisions are going to be a 'no show' for the next 128 Columns of 7 then you must get at least 42 repeats. Or worse, a hand full of combinations repeating 3 times or more. And the one you are betting against could be a hot one. No guarantee that the repeat is back to back, but the odds are increasing against you trying to avoid a repeat.

It is true, that it remains "doubtful". But if you consider the Law of the Third seriously, than you aren't playing against full odds. As long as you are betting against a repeating sequence that is.

And if you are always changing your 7 hand sequence to be the very last 7 hand sequence that developed then you are barreling towards a head on collision with any and all 7 hand sequences that repeat, three peat and more.

 

This is what I get for not taking a nap, and doing math.

Only 7 opportunities to lose that bet twice in a row. So it would be 1 in 2340. (7/128)(1/128) or 18.285 x 128.

 

I don't think so. (128/1 x (128/1 x 128/1)

(1 / 128) = 0.007813

0.007813 * (1 / 128) = 0.000061

0.000061 * (1 / 128) = 0

The floating point on my 64 bit computer math function will not process this deep of a small number, so it just points to 0.

That means there is zero chance according very powerful floating point computers.

So to double check this I did the simple arithmetic trick

0.007813 * 128 = 1.000064

That confirms the first odds for 1 in 128.

 

If a punter was lucky enough to pick three winning horses at odds of 128/1 during the course of a race meet, whether the races were consecutive or otherwise, I think most lay people would combine the odds of 128/1 x and produce whatever. the answer is.

This is now just becoming semantics, which really doesn't matter.

We are going to have to disagree on this as it it now just going around in circles. The odds of each pick is 1 in 128, no argument there, but using that same logic would also mean the odds of picking 8 different balls would be the same as picking the same ball 8 times, or even 2 balls 4 times apiece. While the maths point to this being the equally as likely, neither scenario is going to happen.

 

I suppose it is just semantics. You could say that there is a 100% chance that a thing will not happen, or you could say there is a 0% chance a thing will happen.

If you want a long shot thing to happen 3 times out of 10 tries then you would say a 3 in 10 chance. That would be written as (3 / 10) = .3

If you want the odds on that long shot thing happening then you divide the times it will happen by the number of choices it must chose from. So to get the odds for a long shot thing happening just once you would write (1 / the number of choices) In this case you would write (1 / 128). So you get that decimal number (1 / 128) = (.007813). So that decimal number could be said to be seven thousand eight hundred thirteen millionths.

So you take the 3 in 10 chance of .3 and multiply it times the seven thousand eight hundred thirteen millionths, .3 X .007813 and you get 0.0023439, or twenty three thousand four-hundred thirty-nine ten-millionths. Now that is both grammatically correct, semantically correct, and numerically correct. 100% X (100% X 100%) = 100%, or 1 So you would write it numerically as 128 / 128 = 1

(128 / 1) = 128,
So 128 X (128 X 128) = 2,097,152

If you are going to use odds to make bet selections then you might want to know what the odds actually are. I'm happy with the mathBoyz odds. 18 / 37 = .48648648648, could be said to be 48.6%.

Wow, google's online calculator has a monster sized floating point, one hundred billionths.

 

Sometimes it all becomes a bit too hard and it is wiser to put 256 balls into a bag and reduce the number of trials

 

What the fuck, over.

 

If the fuckers can't shuffle cards properly, then you have to do whatever is necessary.

 

Real shame Baccaritic has gone AWOL, as I believe his maths is misleading and can give a false sense of security, in the sense of picking the same ball from a bag is less likely back to back as opposed to two variable spots in within 8 trials, subscribing to that chain of thought can prove costly, much preferable to say, 128/1 where-ever.