@ TheCaviarKid,

Thanks for the stats.

@ maestro,

Quotewhat i am not certain is say average of spins to get repeat is 10 then should i drop first spin and count second sample from spin two to 11 or discard 10 spins and get new 10 sins...thanks

I would discard the spins and start afresh, but you could be tracking multiple windows of different lengths looking for a strong deviation in any of them; you don't have to stick only to the sweet spot of 7 spins.

Don't know whether this is "old skool" as in the sense of:

QuoteThe term is commonly used to suggest a high regard for something that has been shown to have lasting value or quality.

But I'm all for free and open debate and keeping it simple. Personally, I would get rid of all the personal blogs, community votes etc, which just distract from the purpose of a forum IMO. This isn't a "community", it's just a venue for like-minded people to discuss gambling and related topics, so I don't think there should be any requirement that we all get along.

@ maestro,

There was a thread on the "sweet spot" for repeaters somewhere, but damned if I can find it. As I recall, the average was 7, but I'll get back to you on this.

@ Jimske,

Quote1. Baccarat has a pre-defined game already in the shoe

But it has no bearing on the independence or otherwise of outcomes. And also, the fact that the shoe is predefined says nothing about whether anyone knows the composition of it (they don't). Probability is in the mind - if you have more information about an event than someone else then your probability of the outcome will be different from theirs (and it should be!); there is no probability "out there" in the world.

BetVoyager's "randomness control" feature is a good example of this. The spins/cards are actually preselected before they are displayed (before the player makes his bet), so each series of spins/hands is predefined, but if no-one knows what they are, does it actually mean anything insofar as the probability is concerned? That's the function of the hash key; although the outcomes are predefined, they cannot be changed after the player has made a bet (if they are the hash will change, and that's how the player knows he isn't being cheated).

Quote2.  The mechanical aspect of the shuffle may not produce true random numbers

True: it could happen that if the cards haven't been shuffled thoroughly after the previous shoe, then outcomes may be dependent to a certain extent. But again, it depends on your knowledge; if someone has just joined the table then they will know nothing about the composition of the cards in the previous shoe, so the poor shuffle will be useless to them.

Hi Nick, thanks for taking the trouble. 

QuoteLooked good to start.

Ha, don't they all. 

Here's a quickie example of how to use the formula to create a system.

The chance of at least one repeat in four spins for a double-street is 72%, and the chance for at least 1 repeat in 6 spins for a street is 78%.

Alternate the following two series of bets:

1. Wait for 1 spin (you will bet for 3 spins). On the 2nd spin, bet whatever DS hit on the first spin. If a loss, bet the last 2 DS's on the third spin. On the 4th spin, bet the last 3 DS's which hit. If a win on the 2nd or 3rd spin, just repeat the bet, but not if the win comes on the 4th spin. If all 3 bets are lost, increase the stake as per D'Alembert (+1 after a loss, -1 after a win) and reset to 1 unit immediately a new high balance is achieved (i.e., don't finish playing the series, but start a new one).

2. Betting on streets follows a similar pattern. Regardless of a win or loss using the above, bet the last street hit, then the last 2 streets on the next spin, the last 3 on the next, etc, up to a maximum of 6 streets. Use the same progression as above after a series loss.

I haven't!

HTH

De nada.

 

Thanks for the vids, greenguy. But what would really make my day is a pic of a random patch of sky from your neck of the woods, perhaps taken on your way to work? 

Betvoyager has Baccarat (Punto Banco). The spread is 10 cents - 100 euro for the no-house-edge game and 10 cents - 200 euro for the standard game.

Quote from: Rashid on December 31, 2014, 11:22:25 PM
If you consider it just gamblers fallacy, then surely the 90.6% figure you posted above simply can't apply? 

It does apply, but only over the entire sequence (4 outcomes or 8 hands). The fallacy comes in when you try to be clever and think something like "ok, so we know there's a 90% chance of at least one repeat of a pair in the 8 hands, so let's wait until there have been 3 different outcomes in 6 hands, then we have a 90% chance that the final outcome will be a repeat of one of the previous 3!"  WRONG, because the past 3 outcomes are now history - there is no uncertainty associated with them and therefore no probability, only certainty.

It's no different in principle than waiting for a run of 10 reds and then betting black, thinking that black must surely be due. Either there are no blacks in the 10 spins or there is at least 1. The probability of no blacks in the 10 spins is (19/37)¹⁰ = 0.001275, so the probability of at least 1 black is 1 - 0.001275 = 99.87%, but that doesn't mean that this is the probability that the next spin will be black when you get to the 9^th red.

Quoteout of curiosity what is the percentage figure if using triplets as opposed to doubles (pairs)

Do you mean over the whole 8 triplets? if so, the chance is 99.76%.

Quote from: greenguy on December 30, 2014, 10:12:35 AM
I think esoito's heart was in the right place, and his intentions were noble and good, but he just took it to a level beyond the general level of enlightenment existing within the forum community. Kind of like a guitar teacher who wraps the knuckles of his grade 2 students because they can't grasp the beauty or display the skill to execute a grade 6 piece of music.

Nicely put. esoito is "old school" (at least grammar, maybe public).

QuoteWhat's up Tangram, with your comment??

A feeble attempt at humour, with a reference to greenguy's avatar (you might not remember Bill Bixby's hulk on TV).

Hi Rashid,

I'm not familiar with the details of eirescott's system, although I've heard about it a few times over the years (I think it was refuted at imspirit's site). I don't think the formula will change, because it just tells you the chance of at least one repeat, although maybe I'm misunderstanding you.

QuoteEither way you will lose more hand per shoe than win, despite the 90% expectation figure.

Yep, unfortunately we can't bet directly on the result of a sequence of outcomes, only on the next outcome, which as we know is always 50:50. And if we could, the casinos would adjust the payouts accordingly. 

Every time we try to force the next hand/spin to conform to a sequential probability (as we often do when creating systems) we end up assuming that trials are dependent, which they're not, so we have committed the gambler's fallacy.

Don't make him angry, you wouldn't like him when he's angry...

Hi maestro,

I believe this was the formula you were looking for:

[math]P(repeat)=\frac{N^r(N-r)!-N!}{N^r(N-r)!}[/math]

where N is the number of equally likely (they must be equally likely!) outcomes and r is the number of spins/decisions.

The (!) means that the number which comes before it is multiplied by numbers which are successively reduced by 1 until you reach 1. e.g.

3! = 3 x 2 x 1 = 6
4! = 4 x 3 x 2 x 1 = 24

etc. This is called factorial and any scientific calculator will have a button for it.

Example 1

What is the chance of at least one street repeating in 5 spins?

Here, N = 12 because there are 12 streets, all equally likely to hit.
r = 5 because we're interested in the repeats over 5 spins.

So plug the numbers into the formula:

[math]P(repeat)=\frac{12^5(12-5)!-12!}{12^5(12-5)!} = 0.618[/math]

An approximately 62% chance.

Example 2

In an even chance there are 4 equally likely outcomes taking 2 decisions at a time, i.e., BB, PP, PB, BP. What is the chance of at least one repeat in 4 decisions (note that this equates to 8 hands, not 4, because each outcome consists of two hands)

N = 4 and r = 4, so

[math]P(repeat)=\frac{4^4(4-4)!-4!}{4^4(4-4)!} = 0.906[/math]

or a little over 90%.

You can find a scientific calculator here.

Quote from: Turner on December 29, 2014, 09:47:59 AM
It won't be long until the forum is modified under his terms as this place now is.

Glad I'm not the only one who's noticed.