 Author Topic: Question about variance, lifetime session  (Read 36103 times)

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Joe

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Kav, I wish you would get the formatting fixed in this form, I've just tried to post an explanation of the discrete uniform distribution and the result is almost unreadable. COULD YOU PLEASE UPDATE THIS CRAPPY FORUM SOFTWARE SO THAT IT FORMATS CORRECTLY!
« Last Edit: August 07, 2019, 12:52:48 PM by Joe »

Joe

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The Discrete Uniform Distribution

This describes any set of outcomes having equal probabilities. The classic example is a dice throw; all outcomes have the same probability of 1/6. So of course it applies to roulette numbers too since each will hit with a probability of 1/37. Strictly speaking it doesn't apply to other bets on the table, (EC, dozs, lines, quads etc) because although each EC, doz, line etc has the same probability as any other in the same group, none of them includes the zero, so the sum of the probabilities doesn't quite add up to 1, as it should. However to a good approximation the DUD can be used on them too.

To use the DUD, each member of the group is coded with numbers starting at 1 and ending at n and the probability of a win is 1/n (eg. for roulette n = 37). So to code the singles code #0 = 1, #1 = 2, #3 = 4, up to #36 = 37.

To find the z-score of a DUD for some sample of outcomes, you need

*The average "score" of the sample, X
*The expectation, E
*The standard deviation STD

Then you plug those numbers into the formula for z, which is : Z = ( X - E ) / STD

Where E = ( n + 1 ) / 2

and STD = sqrt[ (n^2 - 1) / 12 ]

To find X, you need to get the average of the sample which involves adding up all the codes and dividing by the sample size.
Here's an example using lines (double streets).

First code each DS as 1-6 = 1, 7-12 = 2, 13-18 = 3, 19-24 = 4, 25-30 = 5, and 31-36 = 6

After 7 spins you have this sequence of 7 numbers which represent the DS's which have hit :

2,4,4,1,2,1,4

step 1 : To find X, add up all the numbers in the sequence and divide by 7:

(2 + 4 + 4 + 1 + 2 + 1 + 4 ) / 7 = 2.571

step 2 : calculate the expectation. There are 6 possible outcomes (ignoring zero) so n = 6 and
E = (6 + 1) / 2 = 3.5

step 3 : calculate the standard deviation. This is sqrt[( 6^2 - 1) / 12] = sqrt[ 35 / 12 ] = 1.708.

step 4 : calculate the z-score. We now have X, E, and STD so Z = ( 2.571 - 3.5 ) / 1.708 = -0.544

The z-score tells you how "skewed" the distribution is, so in this example the skew is towards the low numbers because z is negative.  A score close to zero means little or no skew in the outcomes.

Next up, the Negative Binomial Distribution.
« Last Edit: August 07, 2019, 02:50:21 PM by Joe »

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kav

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• Gender:  « Reply #107 on: August 07, 2019, 03:58:35 PM »
Joe, I have problem with the formating myself.
I do regular mentainance, but it seems that It may need a fresh install or something.
Anyway, I will try to look into it.
Thanks for your patience.

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scepticus

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" with a probability of 1/37. Strictly speaking it doesn't apply to other bets on the table, (EC, dozs, lines, quads etc) because although each EC, doz, line etc has the same probability as any other in the same group, none of them includes the zero, so the sum of the probabilities doesn't quite add up to 1,  "    Joe

Joe .would calculating x in 36 and then allow 1 bet in 37 as a loss allow the calcualation of Dozens , lines etc. ?
« Last Edit: August 07, 2019, 07:06:14 PM by scepticus »

Joe

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• Gender:  « Reply #109 on: August 08, 2019, 09:18:49 AM »
scep, what you're suggesting is more about the binomial model because it's defined in terms of binary outcomes : win/loss, success/failure etc. The uniform distribution is about how the set of outcomes, eg. dozen 1, dozen 2, dozen 3 is behaving as a group. The only set of outcomes which truly fits the discrete uniform are the single numbers 0-36, but you can use it on other groups as long as you accept the small error. And also, it wouldn't be appropriate to use it on all 6 even chances because they are not mutually exclusive (eg. red and odd together don't form a uniform distribution because they aren't mutually exclusive; only red and black, odd and even, high and low are).

You could use it on compound bets though. eg. if you take the patterns of 3 on an even chance there are 8 mutually exclusive outcomes which all have the same probability (1/8, ignoring zero):

1 RRR
2 RRB
3 RBR
4 RBB
5 BRR
6 BRB
7 BBR
8 BBB

So if you record outcomes in sets of 3 using the codes 1-8 you can calculate the z-score for all the sets taken as a group.
« Last Edit: August 08, 2019, 09:23:55 AM by Joe »

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scepticus

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