ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5673
February 21st, 2014 at 7:00:22 PM permalink
Quote: scepticus

Your calculations are correct but the third is likely to lose twice in three.

The third what?

Quote: scepticus

If there is a guarantee of three correct then betting the 1-2-4 has a one in four chance which pays 7/2.

There is a problem with your "four trebles" theory. Suppose the first spin is in the first 12 and the second spin is in the second 12. There are two possibilities for treble 1-3-4; column 1-1-2-2, and column 1-3-1-1.
There are also two possibilities for treble 2-3-4: column 2-2-1-2, and column 3-2-2-1.
Also, where exactly does "pays 7/2" come from?

Once again, you can easily show us that your idea has merit by showing us a simple example.
If your system applies to dozens, and the first two spins are 7 and 10 (i.e. both in the first dozen, or 1-1), then fill in the blanks:

(a) The third bet is __ on the first dozen, __ on the second dozen, and __ on the third dozen

(b) If the third spin was 6 (i.e. in the first dozen), then the fourth bet is __ on the first dozen, __ on the second dozen, and __ on the third dozen

(c) If the third spin was 18 (i.e. in the second dozen), then the fourth bet is __ on the first dozen, __ on the second dozen, and __ on the third dozen

(d) If the third spin was 30 (i.e. in the third dozen), then the fourth bet is __ on the first dozen, __ on the second dozen, and __ on the third dozen
scepticus
Joined: Oct 16, 2013
• Posts: 57
February 21st, 2014 at 7:39:45 PM permalink
1) 1-1
2)-1-1
3) 1-3
4)-2-2
Bet on the "third " spin 1st Doz and 3rd Dozen. If either win put the resultant 3 chips on the 2nd dozen on the " fourth" spin.If win collect 9 chips for a 2chip bet = 7/2.
If the 2nd dozen wins the " third " spin game over so we can never bet( as a double ) the "third" and " fourth " spin of the 1 and 2 winners- in this cas 2 and 2.
And, how likely is it that all four would be correct ?
I chose then 1-2 -4 because the first two being correct I consider it more than likely the winning "treble" would be in that column as it has more chances than the others. In other words some things ARE more likely.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5673
February 21st, 2014 at 8:40:06 PM permalink
Quote: scepticus

1) 1-1
2)-1-1
3) 1-3
4)-2-2
Bet on the "third " spin 1st Doz and 3rd Dozen. If either win put the resultant 3 chips on the 2nd dozen on the " fourth" spin.If win collect 9 chips for a 2chip bet = 7/2.
If the 2nd dozen wins the " third " spin game over so we can never bet( as a double ) the "third" and " fourth " spin of the 1 and 2 winners- in this cas 2 and 2.
And, how likely is it that all four would be correct ?

There's your problem. There is a 1/9 chance that all four are correct if you start counting after the first two spins. In the example where the first two spins are both in the first column, the "indicator column" is 1-1-2-2, and there is a 1/9 chance that both the third and fourth spins are in the second column, in which case all four trebles "win".

If you ignore that column, you do get a 1/4 chance of winning 7 and a 3/4 chance of losing 2. However, there is only an 8/9 chance of this happening; the other 1/9, you lose 2 (since the third spin was in the second column). The EV is now 8/9 x (1/4 x 7 - 3/4 x 2) + 1/9 x (-2) = 8/9 x (7/4 - 6/4) - 2/9 = 2/9 - 2/9 = 0.
scepticus
Joined: Oct 16, 2013
• Posts: 57
February 22nd, 2014 at 7:48:19 AM permalink
Quote: ThatDonGuy

There's your problem. There is a 1/9 chance that all four are correct if you start counting after the first two spins. In the example where the first two spins are both in the first column, the "indicator column" is 1-1-2-2, and there is a 1/9 chance that both the third and fourth spins are in the second column, in which case all four trebles "win".

If you ignore that column, you do get a 1/4 chance of winning 7 and a 3/4 chance of losing 2. However, there is only an 8/9 chance of this happening; the other 1/9, you lose 2 (since the third spin was in the second column). The EV is now 8/9 x (1/4 x 7 - 3/4 x 2) + 1/9 x (-2) = 8/9 x (7/4 - 6/4) - 2/9 = 2/9 - 2/9 = 0.

Depends on how you look at it.
From your theory you are right but in actual practice it doesn't necessarily happen as in your theory.The 3rd and 4th of the first two don't occur as in your tram lines maths theory - random ensures that. As for the airy fairy Long Run I should live so long !
I live and gamble in the real world and accept that risk is involved .Don't you defy the HE in your betting ? And if you do and profit what makes you think I don't ?
What works works .
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5673
February 22nd, 2014 at 8:43:24 AM permalink
Quote: scepticus

From your theory you are right but in actual practice it doesn't necessarily happen as in your theory.The 3rd and 4th of the first two don't occur as in your tram lines maths theory - random ensures that.

Huh? The third and fourth numbers in any column happen 1/9 of the time regardless of the first two numbers - random pretty much ensures that. The wheel doesn't remember what the previous two numbers were.
Quote: scepticus

As for the airy fairy Long Run I should live so long !
I live and gamble in the real world and accept that risk is involved .Don't you defy the HE in your betting ? And if you do and profit what makes you think I don't ?
What works works .

By that reasoning, Martingale "works" if you throw enough money into it. With a maximum bet that is 1024 times the minimum bet, it only collapses every 1165 times. True, you lose 2047 times your original bet, which more than erases your 1164 other wins, but "you should live so long", right?
scepticus
Joined: Oct 16, 2013
• Posts: 57
February 22nd, 2014 at 4:50:09 PM permalink
Quote: ThatDonGuy

Huh? The third and fourth numbers in any column happen 1/9 of the time regardless of the first two numbers - random pretty much ensures that. The wheel doesn't remember what the previous two numbers were.

By that reasoning, Martingale "works" if you throw enough money into it. With a maximum bet that is 1024 times the minimum bet, it only collapses every 1165 times. True, you lose 2047 times your original bet, which more than erases your 1164 other wins, but "you should live so long", right?

The wheel is an inanimate object so it cannot have a memory so that's a rather idiotic statement to make and why it has currency is only because it sounds " knowledgeable".
If there is a guarantee of 3 from four it means that some things can happen while some others cannot so your belief that all the 3rd bets and 4th bets all have the same chance is misplaced.
Our main point of difference is that you believe that a disadvantage means that we will lose at some future, unspecified date while I dispute your certainty. I preach uncertainty not certainty which is why I admit that I guess. Claiming to know
the future with certainty is in the province of clairvoyants , not probability theory which only deals with likliehood.
And you haven't explained why you gamble knowing that you must lose "eventually " just as you claim that I will.
And I only make flat bets so why do you think I bet progressives/regressives ?
scepticus
Joined: Oct 16, 2013
• Posts: 57
February 22nd, 2014 at 4:50:10 PM permalink
Quote: ThatDonGuy

Huh? The third and fourth numbers in any column happen 1/9 of the time regardless of the first two numbers - random pretty much ensures that. The wheel doesn't remember what the previous two numbers were.

By that reasoning, Martingale "works" if you throw enough money into it. With a maximum bet that is 1024 times the minimum bet, it only collapses every 1165 times. True, you lose 2047 times your original bet, which more than erases your 1164 other wins, but "you should live so long", right?

The wheel is an inanimate object so it cannot have a memory so that's a rather idiotic statement to make and why it has currency is only because it sounds " knowledgeable".
If there is a guarantee of 3 from four it means that some things can happen while some others cannot so your belief that all the 3rd bets and 4th bets all have the same chance is misplaced.
Our main point of difference is that you believe that a disadvantage means that we will lose at some future, unspecified date while I dispute your certainty. I preach uncertainty not certainty which is why I admit that I guess. Claiming to know
the future with certainty is in the province of clairvoyants , not probability theory which only deals with likliehood.
And you haven't explained why you gamble knowing that you must lose "eventually " just as you claim that I will.
And I only make flat bets so why do you think I bet progressives/regressives ?
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5673
February 22nd, 2014 at 6:24:34 PM permalink
Quote: scepticus

The wheel is an inanimate object so it cannot have a memory so that's a rather idiotic statement to make and why it has currency is only because it sounds " knowledgeable".

You're the one with the strategy that depends on what the previous two spins were.

Quote: scepticus

If there is a guarantee of 3 from four it means that some things can happen while some others cannot so your belief that all the 3rd bets and 4th bets all have the same chance is misplaced.

The third spin has 1/3 chance of being in the first 12, 1/3 of being in the second 12, and 1/3 of being in the third 12.
The fourth spin has 1/3 chance of being in the first 12, 1/3 of being in the second 12, and 1/3 of being in the third 12.
(In both cases, ignoring 0 and 00.)
Which of those two statements is ever wrong, and under what conditions (other than an unfair wheel)?
scepticus
Joined: Oct 16, 2013
• Posts: 57
February 22nd, 2014 at 7:11:56 PM permalink
Quote: ThatDonGuy

You're the one with the strategy that depends on what the previous two spins were.

The third spin has 1/3 chance of being in the first 12, 1/3 of being in the second 12, and 1/3 of being in the third 12.
The fourth spin has 1/3 chance of being in the first 12, 1/3 of being in the second 12, and 1/3 of being in the third 12.
(In both cases, ignoring 0 and 00.)
Which of those two statements is ever wrong, and under what conditions (other than an unfair wheel)?

NONE is the answer to your question but they are not all linked to the ones before -IN THE BLOCK - which guarantees that 3 WILL be linked which means that some won't.
The heading of this thread is Some things are more likely than others.
Consider the 1st of the 4 spins.
If 111 is correct then 222333 are wrong which means that the 222333 columns have only ONE chance each of completing 3 correct.
Each of the 111 columns each have FOUR so is the winning 3 not MORE likely to come from the 111s ? That does NOT mean that they will - only more likely. Random with it's variance can be our enemy but it is also the casino's enemy.No one knows what a future outcome will be so no one should claim that they do.
If I bet, say, 5 numbers ,what proof has anyone that I cannot average 1 win in seven spins whether or not there is one or even two zeros ? NONE ! So why do people continue to parrot the nonsense that we cannot win in the long run ? The maths only support the likliehood - not the certainty.
Nothing personal but ditch the certainty.
AxiomOfChoice
Joined: Sep 12, 2012