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Messages - mustangsally

#1
Math & Statistics / Re: how many spins until ..
August 21, 2015, 05:06:28 PM
Quote from: HansHuckebein on December 24, 2014, 08:07:55 PMI don't know if this could be calculated at all, still I have a question:

playing 4 single numbers on a single zero wheel,

how many spins should I need/ expect
until I have reached the minus 2,7% as a more or less steady result?
the 1/37 House edge is there every time you win a bet

it is the amount, on average, you are shorted on a win
it is how the house "stacks the deck" against the player
so they have the edge, always

so every winning bet suffers the house edge of exactly 1/37

not happy with that answer?

you want to know how many trials to make until you converge to that -1/37 value?

my opinion is it is how close do you want to be?
then it can easily be calculated
*****************
still not happy, well
that sounds like a personal problem so i thank you for sharing your thoughts here
thank you
***********************
but
here are a few #s i have
no one else in the universe has these, btw

make 37 bets and win exactly 4 of them with your example
you net $32 4 times on a win = $128
you lost 33 times $4 = $132
your bankroll is now $4 less after wagering exactly 37 *$4 = $148
-$4 / $148 =  -1/37
exactly the house edge

now how about if you win 3 times and lose only 34 times?
DUE the math and see a player advantage of -27.03%

4 wins has the highest probability followed by only 3 wins
DUE the math and see

but you still are not happy with any answers i give

the house edge over many bets is just an average...

so how about this 37*37 bets, on average

have fun!
no worry about the HE
it is what it is
#2
Math & Statistics / Re: repeaters formula
August 20, 2015, 04:32:41 PM
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!
#3
Math & Statistics / Re: repeaters formula
August 20, 2015, 03:20:15 PM
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
#4
Math & Statistics / Re: repeaters formula
August 20, 2015, 02:49:48 PM
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