@Bayes ...Bayes sorry being pain but you used to post formula for probability of repeat on the other forum which is long gone,can you please post it again..say i want to know if i have group of 18 numbers how many unique can be drawn before repeater,i know for 37 numbers are something 9 uniques..thanks

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: Bayes on December 30, 2014, 12:14:24 PM

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.

BB
PP
PB  lose 1 bet
P?   now what

Eirescott posted many years ago, to place the P bet after the PP, and PB outcomes, because if the result is P it then becomes ambiguous therefore a no-bet scenario for which to recoup the prior lost bet.

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

What is the formula if you include Eirescotts adaptation?

 

Was wondering when this would make it's appearance on this board.  Having played it, I did manage to accumulate £6000 profit inside 11 day and 11 sessions a few months ago.  Not a bad wedge for less than two weeks action, I must say.  Having played through the good and ugly aspects of it, Scotty will be proud his concept still lives on after all these years.     

thanks Bayes spot on as usual..

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.

Quote from: Jimske on December 31, 2014, 05:11:46 PM
Are you referring to EIRESCOTT's GRAIL FINAL EDITION  ??

Yes Eirescott's system, not necessarily his final Pre-fab offering...     

Quote from: Bayes on December 31, 2014, 05:05:03 PM
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.

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.

Correct, it was refuted at imspirits site. 

Would have to disagree regarding betting on the result of a sequence of outcomes, this is exactly what this "pair" bet section entails.  It is geared to return "hopefully" a win within a maximum of 4 bets per 8 hand sequence.

While I agree every hand is independent and yes anthropomorphizing prior results is a difficult habit to break, however "success" per block of 8 hands is by far outstrips "failures", yet there still exists the situation of losing more bets than winning bets, obviously due to the potential of taking a maximum four attempts to win a single bet.

If you consider it just gamblers fallacy, then surely the 90.6% figure you posted above simply can't apply? 

As maths seems your forte, out of curiosity what is the percentage figure if using triplets as opposed to doubles (pairs), I haven't been able to put together a viable method of implementing triplets (my very limited test results were poor), yet I feel doing so would be inherently stronger that this non-matching-pair" option, purely based on the maths.     

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%.

Very well explained, thank you.

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.

Quote from: Bayes on January 06, 2015, 08:40:20 AM
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.

Very good.  Can anybody run a sim and post one of those graphics for the above?

OK, here you go.

I did a quick 40,000 simulator spins using the DS strategy as above with the +1 on a loss and -1 on a win.

Looked good to start.

Cheers

Nick

Hi Nick, thanks for taking the trouble. 

QuoteLooked good to start.

Ha, don't they all. 

Yeah thanks Nick