@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 (http://web2.0calc.com/).
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 (http://web2.0calc.com/).
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. :stress:
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. :stress:
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. :thumbsup:
QuoteLooked good to start.
Ha, don't they all. ::)
Yeah thanks Nick
Bayes i got one more question if you could help, say i have 30 samples of 37 spins and i get between 2 to 18 unique numbers before first repeat happen in sample,when i get the average of unique for my 30 samples i get 5.9 or 6.1 uniques something in that range and standart deviation 2.9 to 3.0,could somehow use standard deviation or z score to find what would be best betting time..thanks and i hope i asked right question
Quote from: Bayes on December 31, 2014, 05:05:03 PM
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.
Hi!
This is totally true at roulette, 100% absolutely wrong at baccarat.
as.
@ 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.
Quote from: Bayes on January 10, 2015, 09:35:43 AM
@ 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.
Here are 3 million spins worth from Hamburg.
This is for the first repeater. It does look like 7 is the 'sweet spot'
2 10,925
3 21,007
4 29,818
5 36,698
6 40,599
7 42,534
8 41,406
9 38,225
10 33,919
11 28,188
12 22,631
13 17,399
14 12,875
15 9,012
16 6,005
17 3,711
18 2,374
19 1,293
20 695
21 369
22 214
23 56
24 25
25 14
26 5
27 3
28 0
29 0
30 0
31 0
32 0
33 0
34 0
35 0
36 0
thanks for input people...@Bayes what i am trying to do is to get amount of spins before repeat as random variable and see if it hits certain deviation does it regress back to normal..what 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
@ 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.
thanks very much Bayes...oh and last question ,say probability for DBL street to repeat as you calculated it prevoiusly is 0.61 can this value be taken and plugged in formula for longest amount of spins for the event to "sleep"..i think i found formula in one of your posts...thaks very much
Hi maestro,
I think I know the formula you mean; yes you can use it to find a sleep length corresponding to a particular z-score. Just be aware that it will be referring to 5 spin windows not individual spins. :thumbsup:
Thanks Bayes...yes true that will be window of spins..
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 (http://web2.0calc.com/).
Thought you might find this shoe interesting, relevant to the 90.6% figure above, played last week.
I shall write out the results from left to right using rows of 8 (I approached this shoe using column of eight).
My prime bet option is irrelevant, a secondary "Birthday Paradox" bet option was used, along with other opportunistic options.
BPBBBPPP
PPBPBPPP
BPPPPPPP
PBBBBPPP ("BP" column failure)
PPPBPBPP
PBBPBBPP (2nd "BP" failure, also this 8 hand sequence is same as last but one group of 8, 256/1 !!)
BPPBPPBB (3rd shoe "BP" failure, also mirror image of prior 8 hand sequence??)
PBPPBBBP (4th "BP" failure and
extreme rare event of 3 consecutive none-matching pairs group of 8 hands??
PPP
Granted Eirescott catered for prevention of ambiguous no-bets situations (BP BB now bet B), but still bleak to say the least,
Here is what the shoe looked like when recorded normally;
B
P
BBB
PPPPP
B
P
B
PPP
B
PPPPPPPP
BBBB
PPPPPP
B
P
B
PPP
BB
P
BB
PP
B
PP
B
PP
BB
P
B
PP
BBB
PPPP
Fair to say, I initially enjoyed amazing success incorporating this approach into my game, now however, rapidly trying to wean myself from even tracking it.
this one is also saying 7. so seven is the new devils number? >:D
http://arxiv.org/pdf/1309.7608v1.pdf
Quote
Conclusion:
We have presented a strategy for playing roulette in which a player has a positive expected gain per spin in contrary to the popular belief that in the long run the House always has the edge. It is, however, to be noted that one may have to play a large number of spins in order to realize an overall gain.
Quote from: Bayes on December 30, 2014, 12:14:24 PM
<snip>
where N is the number of equally likely (they must be equally likely!) outcomes and r is the number of spins/decisions.
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.
what you forgot to mention is that the 12 equally likely outcomes must total the sample space
and at 0 or 00 roulette that is not so
your example results are flat out not correct for your example because of that fact shown
and can be seen by calculating the success repeat rate for 2 spins
(should be 1/12)
actual prob = 108/1369 = 0.078889701
abouts 1 in 12.67592593
that is over a 5% difference that will wipe out any attempt to overcome the 1/37 house edge...
nows
that answer to a repeat for any of the 12 streets in 5 spins is easy solved by hand
when one has the time to do this by hand
(i use Markov chains as shown in photo)
and that answer = 0.59225602 (for 0Roulette)
way far away from 0.61805556 you gave
without even rounding
(00Roulette = 0.567855122)
this IS GREAT math use
(my college math professor used to say this a lot)
to show a proper formula and mis-apply(?) it in an example
i will not even mention the Baccarat example because that is never a 50/50 prop
but with Baccarat, a fact,
if one accurately counts the removed cards and calculates the new chance of a bet winning from the remaining distribution of the shoe
that is powerful, imo and leads to great winning percentages higher than at the start of a new shoe
the math is advanced so that remains for another time and place too
*********************
in closing
some may say and do say
no wonder this world is so messed up
know-it-alls
just my opinion
Sally
Quote from: maestro on January 08, 2015, 01:33:20 AM
Bayes i got one more question if you could help, say i have 30 samples of 37 spins and i get between 2 to 18 unique numbers before first repeat happen in sample,
when i get the average of unique for my 30 samples i get 5.9 or 6.1 uniques something in that range
your sample size is very small
the mean and variance can be exactly calculated, btw
the mean is about 8.3 (8.306669466) spins to get that 1st repeat
Pete (like a Rose... pretty in red)
Quote from: maestro on January 08, 2015, 01:33:20 AMand standart deviation 2.9 to 3.0,could somehow use standard deviation or z score to find what would be best betting time..thanks and i hope i asked right question
no use SD as the distribution is NOT normal
not normal
use the actual distribution for your system
this has been discussed on many forums B4
nut new here
here is part of the distribution
math is easy and so is Ion Saliu
by spin X spins on spin X
0.027027027 2 0.027027027
0.079620161 3 0.052593134
0.154245553 4 0.074625392
0.245678466 5 0.091432913
0.347613809 6 0.101935342
0.453406164 7 0.105792355
0.556815809 8 0.103409645
0.652639418 9 0.095823609
0.737132532 10 0.084493115
0.808177794 11 0.071045262
0.865206017 12 0.057028223
0.908922985 13 0.043716967
0.940923017 14 0.032000032
0.96327647 15 0.022353453
0.978164388 16 0.014887918
0.987606815 17 0.009442427
0.993300981 18 0.005694166
0.996559963 19 0.003258982
0.998326469 20 0.001766505
0.99923108 21 0.000904612
the photo shows the distribution to 21 spins
i B lazy for all
sure fun playing and winning at Roulette
when using actual math results
now for the Angels to win tonight!
yes, i know
Sally
Quote from: marvin on August 07, 2015, 03:01:50 AM
this one is also saying 7. so seven is the new devils number? >:D
http://arxiv.org/pdf/1309.7608v1.pdf
that paper is a joke, mucho math is wrong, imo
but entertaining +1
"Since a straight-up bet pays 35 to 1, using this strategy one wins 29 chips in case
of a success but loses only 7 chips, otherwise. Since the probability of a success
is 0.544 and that of a failure is 0.456, the expected gain per spin is 29(0.544) –
7(0.456) = 12.584 chips."
the prob of a win = 7/38 (0.18421052631578947368421052631579)
and NOT .544 (another wrong value)
this guy has to be on meds
so this forum (like most all others)
is all about wrong math to make it LOOK like a winner is the long term result
cool of you can sell that for real $$$$
DUE the math
because it is DUE!
I show no math because no one cares, really
just an opinion
remember if 20 spins has not produced a repeat, you have lost lots
i mean lots of real $$$ chasing that 1st repeat
imo, real $$$ is in the actual distributions, not some hocus-pocus-wand-waving-
HERE YA GO DUDES!