hello
i have another question..
we all know that an EC is twice as likely to hit one time as it is to hit twice in a row and for it to hit two times in a row it is twice as likely to this than land three times in a row etc..
my question is
would we get the same results from 100 spins that we would get from 1000 spins or maybe more?
would we get a bigger difference (maybe 2 in a row is 3 times more likely to hit than 3 in a row) from a smaller sample or a bigger sample?
how far can results deviate from 'twice as likely'?
if i used the same amount of spins an investigated them would you more than likely get the same result all the time?
is there anywhere i can find out the answers to these questions or somewhere where i could do further studies?
thank you in advance fro your replies :)
Hi frost,
Generally speaking, in smaller samples, you're going to get more 'out of balance' results, so in 100 spins the difference is likely to be larger than in 1000 spins (the law of large numbers).
How far can results deviate? I don't know, but it's a good question. I've seen sequences where you don't get a series of 2 (only singles and series > 2) for 50+ spins. There's a statistical test for randomness called the 'Runs test' which takes into account the standard deviations of 'runs', so that might be a good place to start doing some research.
I've actually been meaning to post it here but haven't got around to it yet. If I get time in the next couple of days I'll post the details.
thank you bayes.
:thumbsup:
Hello eveyone
What if we had determind values?
So we take line 5 for example
This line can show a minimum of 1 time and a maximum of 6 times in a row. if we were to look at what happens in say 2000 spins is there a formula could use to work out this problem?
Also what topic does this come under? looked at the runs test but in order to use that would need to have the results ready
Am correct in my thinking?
Quote from: Bayes on Feb 08, 02:44 AM 2012
Generally speaking, in smaller samples, you're going to get more 'out of balance' results, so in 100 spins the difference is likely to be larger than in 1000 spins (the law of large numbers).
And with the lesser-chance bets. The higher the win/lose-odds, the lower/higher the SD. That's why it's so much easier to fall into the "trap" of perceiving peculiar distributions from the inside play of roulette.
what is the name of the topic i should look at in order to answer my question?
i no its statistics but what?
frost,
i will try to post up some of the specific math here about these SD's in a day or two, if i can locate it from my drives. (It's quite-difficult to find the sort of stuff that you can "get your hands on" off the internet. Lots of complex theory, but little direct application.)
I returned to this site, tonight, to only clean up the wording in my previous post.
okay.
thank you garnabby
Quote from: frost on Feb 08, 11:12 AM 2012
So we take line 5 for example
This line can show a minimum of 1 time and a maximum of 6 times in a row. if we were to look at what happens in say 2000 spins is there a formula could use to work out this problem?
frost,
What exactly is the question you want an answer to? do you want to know the number of series of 1-in-a-row, 2-in-a-row etc for lines? In other words, you want some relationship between series for lines as there is for an even chance bet, right?
For an EC bet, each series is half as likely as the previous one, ie; a series of 3 is half as likely as a series of 2. For a line, each series (streak) is one sixth as likely (ignoring the zero) as the previous.
ie; for an EC, the probabilities are: 1/2, (1/2)2, (1/2)3, (1/2)4...
for a line, they are: 1/6, (1/6)2, (1/6)3, (1/6)4...
Quote from: Bayes on Feb 09, 04:56 AM 2012
frost,
What exactly is the question you want an answer to? do you want to know the number of series of 1-in-a-row, 2-in-a-row etc for lines? In other words, you want some relationship between series for lines as there is for an even chance bet, right?
not entierly but please continue with your idea, it might spark something off.
my idea was this. we will use RED for the example.
in a series of spins we will have results like this
red land once - 100
red land twice - 52
red land three times - 25
red land four times - 12
etc
what i expected is these results to roughly remain the same reguardless of our sample size. if we used 1000 or 10000 spins we will still see that for a red to land 2 times in a row it will still be half as likely as a single red to show.
now i wanted to know
- how far away from the expected value (half as much) can the results go?
- am i right in thinking that no matter the amount of spins used we will always get the same trend of results?
- if there a formula i can use to work out the above questions?
- what kind of statistics should i be looking at to get a better understanding of what i want to know?
- what is the meaning of life?
thank you
Ha!, you beat me to it Skakus, I was about to post the same reply. ;D (if you don't know what this means, look up 'The Hitch-Hiker's Guide to the Galaxy').
@ frost,
Ok, so as I said in a previous post, you definitely will NOT get a nice even distribution with small sample sizes. In 100 spins you may only get 30 singles instead of the 'required' 50.
You can calculate the standard deviation (z-score) using the standard formula if you're only interested in how ONE series compares with the remaining (higher series). This is because you're only concerned with 2 outcomes. ie; since there are as many series of 1 (singles) as series higher than 1 (including series of 2,3,4...) then the probability is 0.5 for there to be as many series of 1 as remaining series in any sample. This applies no matter how far up the series you start. e.g. suppose you ignore singles and want to compare the number of series of 2 with the number of series higher than 2 - the probability that there will be equal numbers is still 0.5. And so on for any particular series you're interested in.
So in my example above of 30 singles in 100 spins, this is the same (from a probability point of view) as getting 30 reds in sample of 100, which has a z-score of around -4.0 - pretty rare.
However, if you want to take all series together (not just comparing one series with all higher series), then it gets more complicated. The maths isn't simple, but look up 'Multinomial Distribution'. There's a calculator here (link:://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/multinomial.htm).
in 100 spins, you would expect to get around 25 singles wouldn't you?
25 singles = 25 spins
12 doubles = 24 spins
6 trebles = 18 spins
3 4's = 12 spins
2 5's = 10 spins
1 6 = 6 spins
there or thereabouts?
monaco, you're right. In my previous post I shouldn't have said 'spins' but total streaks, so the total singles is 50%, which of course isn't the same as accounting for 50 spins out of 100. :thumbsup:
Nevertheless, the principle holds good.
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:
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?
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:
ah, got it, thanks
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.
:)
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?
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:
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 (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 (link:://:.freeproxyserver.ca/index.php?btxmnercdeqt=aHR0cDovL3dpemFyZG9mdmVnYXMuY29tL2ZvcnVtL2dhbWJsaW5nL2NyYXBzLzcxMDAtaW1wb3NzaWJsZS04NDMtb2YtMzIwMC13aW5zLw%3D%3D) , about mathematically determining which results aren't fair, or at least expected.
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?
@ 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?
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:
doh! thanks again :thumbsup:
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 (link:s://acrobat.com/app.html#d=dUaZa*i6TYme*urC8pGwNw) ; and work your way out, as at link:://mathproblems.info/prob11s.htm (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.