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All things being Equal

Started by frost, Feb 07, 01:18 PM 2012

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frost

okay.


thank you bayes.


where can i find information on the z-scores and the standard formula?


thank you also skakus, i will now go fulfil many lives  :lol:

monaco

so to get only 15 singles in 100 spins, z score would be:   

n:=100 (spins)
m:=15 (hits)

p:=18/36=0.5 (probability for ease)   

zscore:=(m-n*p)/sqrt(n*p*(1-p))

zscore:=(15-100*0.5)/sqrt(100*0.5*(1-0.5))= 8.5   is that right?

Bayes

monaco,

No, that's not correct. You've actually made the mistake I did, which you just pointed out to me. If you have 100 spins you first need to count the total number of streaks, then you can use the formula.

In your example, there may be 15 singles, but how many streaks were there > 1? say there were 25 in total. This gives a total of 25 + 15 = 40 streaks altogether, so this is the value for n in the formula. You have to do it this way because you need to compare apples with apples, what you've tried to do is compare apples with oranges.

@ frost, see my thread in this section called "standard deviation and z-score in plain English".

By the way, in 99.7% of cases, the z-score should come out between -3 and +3, so if you're getting values a lot higher than this, you've probably made a mistake.  :thumbsup:
"The trouble isn't what we don't know, it's what we think we know that just ain't so!" - Mark Twain

monaco


frost

thank you bayes.


i have read through the thread but some of the equations have gone all crazy. i cant read them.


but now i know what direction im heading in roughly.


:)

frost

how big should a sample number be?


if i get a SD within the 68% wouldn't that change if the sample number was bigger?

Bayes

It doesn't matter what the sample is, there is no 'right' number for this, it depends what you want to know.

If the sample number was bigger and the number of wins was the same, then yes, the SD would change.  :thumbsup:
"The trouble isn't what we don't know, it's what we think we know that just ain't so!" - Mark Twain

GARNabby

Quote from: frost on Feb 09, 03:39 AM 2012
okay.

thank you garnabby
First, here's some reference math from the site, link:://en.wikipedia.org/wiki/Casino_game ; then, my own approximating functions of SD based on the win/lose-odds for baccarat, eg.  (I could find no web functions for these particular SD calculations, probably because like for the odds, the ones above were more than likely just exactly simulated, over a wide sample group of "actual" results.  Likely because no one has really overcome those odds enough to stop, and think, "What're the new SD's?")

"Standard deviation:  The luck factor in a casino game is quantified using standard deviations (SD). The standard deviation of a simple game like Roulette can be calculated using the binomial distribution. In the binomial distribution, SD = sqrt (npq ), where n = number of rounds played, p = probability of winning, and q = probability of losing. The binomial distribution assumes a result of 1 unit for a win, and 0 units for a loss, rather than -1 units for a loss, which doubles the range of possible outcomes. Furthermore, if we flat bet at 10 units per round instead of 1 unit, the range of possible outcomes increases 10 fold.

Therefore, SD for roulette, even-money bet = 2b sqrt(npq ), where b = flat bet per round, n = number of rounds, p = 18/38, and q = 20/38.For example, after 10 rounds at $1 per round, the standard deviation will be 2 x 1 x sqrt(10 x 18/38 x 20/38) = $3.16. After 10 rounds, the expected loss will be 10 x $1 x 5.26% = $0.53. As you can see, standard deviation is many times the magnitude of the expected loss.The range is six times the standard deviation: three above the mean, and three below. Therefore, after 10 rounds betting $1 per round, your result will be somewhere between -$0.53 - 3 x $3.16 and -$0.53 + 3 x $3.16, i.e., between -$10.01 and $8.95. (There is still a 0.1% chance that your result will exceed a $8.95 profit, and a 0.1% chance that you will lose more than $10.01.) This demonstrates how luck can be quantified; we know that if we walk into a casino and bet $5 per round for a whole night, we are not going to walk out with $500.

The standard deviation for Pai Gow poker is the lowest out of all the common casino-games. Many, particularly slots, have extremely high standard deviations. As the size of the potential payouts increase, so does the standard deviation.

As the number of rounds increases, eventually, the expected loss will exceed the standard deviation, many times over. From the formula, we can see the standard deviation is proportional to the square root of the number of rounds played, while the expected loss is proportional to the number of rounds played. As the number of rounds increases, the expected loss increases at a much faster rate. This is why it is impossible for a gambler to win in the long term. It is the high ratio of short-term standard deviation to expected loss that fools gamblers into thinking that they can win.

It is important for a casino to know both the house edge and variance for all of their games. The house edge tells them what kind of profit they will make as percentage of turnover, and the variance tells them how much they need in the way of cash reserves. The mathematicians and computer programmers that do this kind of work are called gaming mathematicians and gaming analysts. Casinos do not have in-house expertise in this field, so outsource their requirements to experts in the gaming analysis field, such as Mike Shackleford, the Wizard of Odds."

My method's general SD for baccarat = (1 +/- 0.0476) X (2 or 1.95) X ( 0.4523574(+) or 0.4999536(-)) where: the 0.0476 is from the chance of a tie, at 0.0952, divided by 2; 2 is for a player win/loss of 1, and 1.95 is for a banker win/loss of 0.95/1; and 0.45... = sqr of (p X q) with ties included, but 0.49... = that w/o ties included in the P-B odds.  Those results reduce to 0.9478, and 0.9241; and 0.9523, 0.9285, respectively for the two P/B pairs; and compare favorably with the text-book ones of about 0.95, and 0.93. e.g., the P's SD ranges from 0.9478 to 0.9523, depending upon how you want to view the occurrences of ties in relation to the decided games.  From there it would be easy to adjust the numbers to other SD's derived from any better, or worse, set of P/B odds.

I don't include the ties in the (true) P-B odds, because it's the decided, non-tie games which must be overcome to begin to pull ahead.  Also, by ignoring the ties in determining the P-B odds, over the P-B decided games, the higher pair of SD's would, e.g., serve as a guard against an unusually-low number of tie-decisions over a session of play.  (The false P-B odds would underestimate the SD's.  Which of course, given the much-higher variance of ties, could and often does occur.)

Here's a recent thread from the Wizard's board, link:://:.freeproxyserver.ca/index.php?btxmnercdeqt=aHR0cDovL3dpemFyZG9mdmVnYXMuY29tL2ZvcnVtL2dhbWJsaW5nL2NyYXBzLzcxMDAtaW1wb3NzaWJsZS04NDMtb2YtMzIwMC13aW5zLw%3D%3D , about mathematically determining which results aren't fair, or at least expected.

monaco

hi Bayes - if you've got time, could you just check this over?

I just played on SkyVegas (1c rng) to test it out & these were the series i got in 1 spell:

1s-28
2s-21
3s-9
4s-8
5s-3
7s-1
there were also 9 zeros

thats 160 spins & 70 streaks not including the zeros (do they counts as 1s?)

(not counting the zeros)
n:=70 (no. of streaks)
m:=28 (no. of singles)

p:=18/36=0.5 (probability for ease)   

zscore:=(m-n*p)/sqrt(n*p*(1-p))

zscore:=(28-70*0.5)/sqrt(70*0.5*(1-0.5))= -21/sqrt17.5 = -5.019 ?


Bearing in mind you said anything beyond +/- 3 is rare, would these results start alarm bells ringing, or is it too small a sample?


frost

@ garnabby


where did you get this information from? and dose that mean that if we did have a winning system subjecting it to continuous spins will mean it will lose. thus meaning hit and run isnt a fallacy?


right?

Bayes

Quote from: monaco on Feb 15, 11:11 AM 2012

zscore:=(28-70*0.5)/sqrt(70*0.5*(1-0.5))= -21/sqrt17.5 = -5.019 ?


It should be -1.67, for some reason you got -21 in the numerator but it should be -7 (28 â€" 35).  :thumbsup:
"The trouble isn't what we don't know, it's what we think we know that just ain't so!" - Mark Twain

monaco


GARNabby

Quote from: frost on Feb 15, 05:37 PM 2012
Does that mean that if we did have a winning system, subjecting it to continuous spins will mean it will lose... thus meaning hit and run isn't a fallacy?

First, let me make clear that the ties, where possible, are only one type of undecided game.  Sitting out on a few, or many, games to "chart", or specifically to "no bet", can be viewed as "ties" which aren't bet.  From that perspective, we can further re-adjust our general SD's.

As to understanding the overall picture, it's best to begin with the "random walk" theory at link:s://acrobat.com/app.html#d=dUaZa*i6TYme*urC8pGwNw ; and work your way out, as at link:://mathproblems.info/prob11s.htm .

Sorry for the delay in getting back to you, frost.  Busy solving the real mysteries of the universe: I call it "indeterminance theory", IT for short.  (Yes, it's indeterminacy, but i prefer my own neoism.)  Today's objective was to complete my understanding of mass, specifically the reason that symmetry shows up there.  As a bonus, it appears that the photon is a sort of inverse mass, with its simultaneity pointed outward instead of inward.

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