### Author Topic: Mathematical questions  (Read 4671 times)

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#### BlueAngel

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##### Mathematical questions
« on: July 14, 2016, 07:52:00 PM »
Everyone is welcome to contribute in this topic, but I'd be particulary interested for "Bayes" response.
So here we go:

1) If we could win with 12 numbers VS 7 numbers which make us lose, would that ratio be better than 24 numbers to win VS  13 numbers to lose?
The payout for both situations remains the same, 2 units at risk with 1 unit profit.

2) About EC's, if we could avoid all losing streaks, from 2 loses in a row and more, in exchange for every 1st win, would that condition be an improvement?

Does it make any difference from the mathematical point of view, or it's about the same?

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#### BlueAngel

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##### Re: Mathematical questions
« Reply #1 on: July 14, 2016, 08:11:51 PM »
Second part of the mathematical questions:

If we assume that the minimum for any EC bet is 25 hits out of 100 outcomes, then would it be correct to assume that by adding 2 more EC bets, a total of 3 EC's simultaneously for 100 outcomes would reduce the deviation?

Example, 25 appearances in 100 results is 25%, deduct that from 48.65% which is the average expectation and we arrive to 23.65% deviation.
Round it up to 24% and divide it by 3 (3 EC's) and we arrive to 8%.
8% multiplied by 300 bets equals 24 more losing bets than the winning ones, in other words 138 wins VS 162 loses

If the above statement has no flaws, then it could be possible to create a progression based on this minimum expectation , thus enabling  us to be permanent winners even when the worst happens.

Is it valid according your point of view or not?

#### kav

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##### Re: Mathematical questions
« Reply #2 on: July 14, 2016, 09:57:00 PM »
Your way of thinking/calculating on the second post is wrong.

What I did find interesting is your first question
Quote
If we could win with 12 numbers VS 7 numbers which make us lose, would that ratio be better than 24 numbers to win VS  13 numbers to lose?
The payout for both situations remains the same, 2 units at risk with 1 unit profit.

This is a question I pondered while designing my new bet selection.
The answer is that at first sight 12 numbers vs 7 numbers is actually worse than 24 vs 13 numbers.
Let's calculate the expected value.
12 - (7x2) = -2    24 - (13x2) = -2
Both have an expected value of -2, but in the first case the disadvantage is 1 number in 19 numbers, or 5%, while in the second case it is 1 number in 37 numbers, or 2,7%.
However there is some benefit in the 12 vs 7 numbers case. If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.

So the 12 vs 7 numbers approach is worse in expected value but better in variance. To put it in another way, if two players played the 2 versions for 100 spins, then the one who played the 12 vs 7 has more chances to be in loss than the other player, but he also has more chances that his loss will be lower that the other player's loss.

#### BlueAngel

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##### Re: Mathematical questions
« Reply #3 on: July 14, 2016, 10:06:43 PM »

Quote
Your way of thinking/calculating on the second post is wrong.

Why?

Quote
If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.

If so, then it would be sound to devise a progression which puts you on the positive by aiming equal amount of wins VS loses.
Wouldn't it?

#### kav

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##### Re: Mathematical questions
« Reply #4 on: July 14, 2016, 10:18:26 PM »
Quote
Your way of thinking/calculating on the second post is wrong.
Why?
Because of this:
"divide it by 3 (3 EC's) and we arrive to 8%"
It doesn't work that way.

Quote
If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.
If so, then it would be sound to devise a progression which puts you on the positive by aiming equal amount of wins VS loses.
Wouldn't it?
I don't understand the connection between my quote and your conclusion.

#### BlueAngel

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##### Re: Mathematical questions
« Reply #5 on: July 14, 2016, 10:40:23 PM »
Quote
Your way of thinking/calculating on the second post is wrong.
Why?
Because of this:
"divide it by 3 (3 EC's) and we arrive to 8%"
It doesn't work that way.

Quote
If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.
If so, then it would be sound to devise a progression which puts you on the positive by aiming equal amount of wins VS loses.
Wouldn't it?
I don't understand the connection between my quote and your conclusion.

About the first leg of my question, you are of the opinion that the worst EC situation (25/100) could happen on all 3 EC's at the same  time?!
Sorry but it doesn't make sense if this is what you mean!

You said that 12 VS 7 ratio has lower variance  in comparison with 24 VS 13, right?
So a progression which aims in equal proportion of wins VS loses would stand better chance rather than a progression which tries to handle greater variance like 24 VS 13.

Especially when the aim of the progression is EQUAL proportion of wins VS loses while the winning situations are NOT EQUAL (12 VS 7) which means roughly 65% win chance against 35% losing chance.

What you have to realize is that is one thing to say 1 out of 2 and completely another 50 out of 100 possibilities, proportion may be the same BUT variance increases as the total does!

#### kav

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##### Re: Mathematical questions
« Reply #6 on: July 15, 2016, 12:17:34 AM »
Quote
What you have to realize is that is one thing to say 1 out of 2 and completely another 50 out of 100 possibilities, proportion may be the same BUT variance increases as the total does!

BA,
I think there is no point to continue the discussion because you lack some basic understanding.

#### BlueAngel

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##### Re: Mathematical questions
« Reply #7 on: July 15, 2016, 12:40:26 AM »
Quote
What you have to realize is that is one thing to say 1 out of 2 and completely another 50 out of 100 possibilities, proportion may be the same BUT variance increases as the total does!

BA,
I think there is no point to continue the discussion because you lack some basic understanding.

If you believe otherwise just let us hear it.

Personally, I'm reinforcing my objections with undeniable evidence, what do you have to deny my point?
Nothing, it's just your word against mine.

So why don't you enlighten me??
After all you could be correct and me wrong, BUT that has to be proved!
Where is the epicentre of your negativity towards my theory??

#### Bayes

##### Re: Mathematical questions
« Reply #8 on: July 15, 2016, 08:04:10 AM »
Hi BA,

I agree with Kav regarding the first part of your question, and as to the second part, your analysis is a little simplistic because variance is a non-linear function, so you can't just subtract and divide as you've done.

You could find the number of wins in 100 bets, given the same deviation and betting on all 3 EC's, like this:

First find the standard deviation for 25 wins in 100 bets on one EC, using the formula for the z-score:

Z = (w - np) / SQRT(npq)

w = no. of wins
n = no. of bets
p = probability
q = 1 - p

So for a single EC, Z = (25 - 0.4865 * 100) / SQRT(100 * 0.4865 * 0.5135) = -4.73 standard deviations below the mean

Now re-arrange the formula because we want the number of wins for the same Z score when betting on all 3 EC's (27 numbers). The probability for 27 numbers is 27/37 ~ 0.73

w = np + Z * SQRT(npq)

= 100 * 0.73 - 4.73 * SQRT(100 * 0.73 * 0.27) = 52.

So this would be the "worst case" number of wins if betting on all 3 EC's.

But as you know, there's always a trade-off between the number of wins and the amount you have to put down in order to recoup any losses, so betting more numbers doesn't really help, and it just increases your exposure to the house edge.

Quote
2) About EC's, if we could avoid all losing streaks, from 2 loses in a row and more, in exchange for every 1st win, would that condition be an improvement?

This is ambiguous, can you clarify? What do you mean by "in exchange for every first win"?
« Last Edit: July 15, 2016, 08:27:33 AM by Bayes »

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#### BlueAngel

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##### Re: Mathematical questions
« Reply #9 on: July 15, 2016, 11:33:38 AM »
Quote
This is ambiguous, can you clarify? What do you mean by "in exchange for every first win"?

I  mean to use stops after every loss and restarting after every 1st  virtual win.
« Last Edit: July 15, 2016, 11:38:33 AM by BlueAngel »

#### Bayes

##### Re: Mathematical questions
« Reply #10 on: July 15, 2016, 12:26:38 PM »
Ok, so does this mean that you stop after every loss and only restart on the first "virtual" win? I.e. betting on red:

R (vw) bet on R
R  W
R  W
B  L (wait for vw)
B     (no bet)
R       (vw. bet on R)
B L  (wait for vw)
R       (vw, bet on R)
B L

etc

#### BlueAngel

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##### Re: Mathematical questions
« Reply #11 on: July 15, 2016, 12:41:56 PM »
Ok, so does this mean that you stop after every loss and only restart on the first "virtual" win? I.e. betting on red:

R (vw) bet on R
R  W
R  W
B  L (wait for vw)
B     (no bet)
R       (vw. bet on R)
B L  (wait for vw)
R       (vw, bet on R)
B L

etc

yes

#### Bayes

##### Re: Mathematical questions
« Reply #12 on: July 15, 2016, 07:33:13 PM »
The problem with the scheme is that although you catch every streak of 2 or more, these only account for 50% of the total EC outcomes, the rest being singles, and that's where you get clobbered, because every choppy sequence will result in continuous losses.

Generally speaking you can't expect any virtual wins or losses to either give you extra wins or eliminate losses because what you gain on the swings you lose on the roundabouts (something that's usually overlooked by adherents). Like any other bet selection it can be worth trying for a while to mix things up, but all it really does is to redistribute your wins and losses to different patterns, it doesn't give you something for nothing (or at least, nearly nothing.  ).

#### BlueAngel

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##### Re: Mathematical questions
« Reply #13 on: July 15, 2016, 09:32:24 PM »
The problem with the scheme is that although you catch every streak of 2 or more, these only account for 50% of the total EC outcomes, the rest being singles, and that's where you get clobbered, because every choppy sequence will result in continuous losses.

Generally speaking you can't expect any virtual wins or losses to either give you extra wins or eliminate losses because what you gain on the swings you lose on the roundabouts (something that's usually overlooked by adherents). Like any other bet selection it can be worth trying for a while to mix things up, but all it really does is to redistribute your wins and losses to different patterns, it doesn't give you something for nothing (or at least, nearly nothing.  ).

I was expecting to hear this, in theory every sequence has the same probability but from my empirical viewpoint, please explain me why I'm under the impression that this sequence: R B R B R B R B R B R B is less frequent than this one: R R R R R R R R R R R R or that one: B B B B B B B B B B B B.

Is there any sensible explanation or it's just my idea that alternating EC's are more rare than series of the same EC ??
Another one of my empirical fallacies is that I've observed columns to  be more "choppy" than dozens which I consider them more streaky.
Again in theory it doesn't have any difference, 12 numbers one side vs 12 numbers the other, but does anybody else can confirm such observations?
Regarding the comparison, a fact is that 2 out of 3 dozens contain only "low" or "high" numbers, on the other side, none of the 3 columns contain only numbers from 1 EC without its opposing EC.

This could be an explanation, but why the same group of numbers (EC's) to be more streaky, i theory every number has the same chances, yet again we have witness again and again what we call unequal distribution, in other word some numbers to appear more than their probability while some others less.
I believe probability theory cannot satisfy all the questions completely, or perhaps an explanation regarding unequal distribution lays on the wheel's layout.
I'm not affirmative, just assuming that this could be an explanation.

Last but not least, please could you answer me the following: according my calculations I've concluded that 30 numbers can never be less than 72 appearances in 100 outcomes, could you confirm this?

#### scepticus

##### Re: Mathematical questions
« Reply #14 on: July 16, 2016, 04:33:55 AM »
You are right BA to think that “chops” are rarer than non- chops. Consider R/B s as Sets of three.
RB…RB….RB…RB
RB…RB….BR…BR
RB…BR….BR…RB
Your “CHOPS” occur in  only two of the possible 8  .
We still have the problem of making use of this “information” though .
Each of the 8 have an equal chance over 3 spins.
Which to choose ?
Murphy’s Law states that whichever one you choose it is the  wrong one !
You could even choose BOTH Streaks and Chops  !

« Last Edit: July 16, 2016, 04:36:05 AM by scepticus »