I think on the "right" side of my brain and working to the front side of my brain but definitely not on the "wrong" side of my brain.

Quote:ScepticusI prefer to bet with method rather than haphazardly so choosing to let past numbers dictate what I should bet is only a method.

What is the second step in my "Pigeonhole Principle Method " ?

After the first two spins I choose the line containing those two dozens and consider the next two numbers in that column of the nine.

I do not bet the next ( 3rd ) number .If it wins I do not bet the next (4th ).

If the 3rd number loses I bet the indicated 4th dozen but only the 6 numbers of my chosen colour.

The last two numbers then become the first two numbers in the next series of four- and I stop when I have a profit or at Break- Even.

The calculation is 1/3 x 1/2 = 1/6 and since I get paid 5/1 these are the correct odds excluding the zero( I play only on a 37 number table )

.You will no doubt argue that the appearances of the zero will " kill " me in the Long Run .I dispute that and argue that this Long Run argument should be labelled "The Mathematicians Fallacy " the belief that an Expectation is a Certainty.

It is highly overrated as it

Assumes that the bettor makes the same bet with the same stake and

Assumes that all bets are therefore subject to this same calculation and

Assumes that I will live long enough to succumb to this

" inevitable "eventuality

Quote:Scepticus1-1-1-2-2-2-3-3-3

1-2-3-1-2-3-1-2-3

2-3-1-3-1-2-1-2-3

2-3-1-1-2-3-3-1-2

Do you understand why everyone is politely snickering at this system?

Quote:CrapsGeniousThis thread is giving me a migraine.

I think on the "right" side of my brain and working to the front side of my brain but definitely not on the "wrong" side of my brain.

Oh my God! I just figured out why I spelled my username wrong. It was the alcohol that altered the wrong side of my brain into "Obviously" thinking it was "right" and it was the night before Christmas when I "Googled" Craps gambling flaws because I was "Curious" and the "Wizard of vegas" forum was at the top of the search making it "Famous" and I had discovered a $1.00 flaw in the "prop bets" that made me "Genius"

My brain was thinking while under the influence of "the good stuff" made my fingers do a funky dance on the keyboard and tricked me into thinking that the word "Genius" was spelled like "Genious" because the keyboard layout shows the U, I, O together. so the question is:

1) Was it my Finger? Did my "right" side of the brain think it was wrong while curious and made my fingers assume that the "o" is part of the word "Genious" or was my middle finger feeling a little "bigger than normal" while "Under the influence of the "good stuff" and pressed the "i,o" together to form my username as the "i,o" are the only two letters together (Side by Side) as comparison to the rest of the letters to form my username.

2) Was it my brain? thinking that the right was actually wrong.

So many combinations can be formed from many random occurrences and no real answer to justify its integrety of what is real, right, wrong or...

anyway,

Thank you sir.

The nine - column block guarantees one of the nine lines will have at least three correct results.

These " trebles " are

1-2-3

1-2-4

1-3-4 and

2-3-4

We cannot bet the first two because we would need to bet all three . So the first two spins are used as " indicators ".

Any one from 4 is a 3/1 shot but if we bet 124 or 134 or 234 as a "double "of the third and fourth spin we need to bet 2 chips- betting that the 123 won't happen.So this reduces to a 7/2 shot for a 1 in 4 chance. Factoring in the zero still gives us an edge.

Quote:soxfanA cat would want to run a negative progression deep enogh to cover a certain mathemtical expectation at the gaming table. Of course, sometimes the math goes to sleep, and that's why some bj card counters endure prolonged downturns. And there are other practical considerations, hey hey.

So, does that mean that the block is " worthless" ?

Quote:scepticusPigeonholeMethod (2 )

The nine - column block guarantees one of the nine lines will have at least three correct results.

These " trebles " are

1-2-3

1-2-4

1-3-4 and

2-3-4

We cannot bet the first two because we would need to bet all three . So the first two spins are used as " indicators ".

Any one from 4 is a 3/1 shot but if we bet 124 or 134 or 234 as a "double "of the third and fourth spin we need to bet 2 chips- betting that the 123 won't happen.So this reduces to a 7/2 shot for a 1 in 4 chance. Factoring in the zero still gives us an edge.

Question: how do you bet "124" if the first two spins already happened?

Maybe an example of this would help. Assume you are talking about betting dozens (first 12, second 12, third 12), and the first two spins are numbers 1 and 4 (or 7 and 10). What are your next two bets?

Something tells me you are making your bets based on the assumption that odds that were in effect at the start of an event remain during that event - equivalent to saying that the chance of tossing a fair coin heads 10 times in a row is 1/1024 at the start of the tosses, and remains 1/1024 even if (a) the first nine tosses were heads (so you should bet on tails, which should be 1023 times as likely to happen), or (b) any of the first nine tosses was tails, in which case the probability of 10 heads in a row is zero.

Quote:ThatDonGuyQuestion: how do you bet "124" if the first two spins already happened?

Maybe an example of this would help. Assume you are talking about betting dozens (first 12, second 12, third 12), and the first two spins are numbers 1 and 4 (or 7 and 10). What are your next two bets?

Something tells me you are making your bets based on the assumption that odds that were in effect at the start of an event remain during that event - equivalent to saying that the chance of tossing a fair coin heads 10 times in a row is 1/1024 at the start of the tosses, and remains 1/1024 even if (a) the first nine tosses were heads (so you should bet on tails, which should be 1023 times as likely to happen), or (b) any of the first nine tosses was tails, in which case the probability of 10 heads in a row is zero.

Yes, Dozens.

By betting that the 3rd after the1-2 loses and putting 1pt. on each of the other 2 Dozens. If one wins put the 3 chips on it's 4th.

If the 3rd wins game over so the 3rd and 4th after the 1-2 can never be bet which leaves only 8 double s available.And 2 from 8 means odds of.... ?

Work it out and you'll see what I mean .

Quote:scepticusYes, Dozens.

By betting that the 3rd after the1-2 loses and putting 1pt. on each of the other 2 Dozens. If one wins put the 3 chips on it's 4th.

If the 3rd wins game over so the 3rd and 4th after the 1-2 can never be bet which leaves only 8 double s available.And 2 from 8 means odds of.... ?

Work it out and you'll see what I mean .

Let me see if I have this right:

Suppose the first two spins are both in the first dozen. In your chart, the next number is 2 (1-1-2-2), so don't bet on the second dozen, but instead bet 1 on both the first and third dozens.

If, say, the first dozen wins, then bet 3 chips on the second dozen (which is the fourth number in 1-1-2-2), since "one group is guaranteed to have 3 numbers hit".

Problem: if the fourth spin lands in the first dozen, then the third group of 4 in your chart (1-3-1-1) has 3 hits, and if the fourth spin lands on the third dozen, then the seventh group (3-1-1-3) does. Your "guarantee" still holds, but you lost your fourth bet.

Here's how I work this out:

First bet (third spin) - 1 on first 12, 1 on third 12

(a) If the first 12 wins, you lose 1 on the third 12, but gain 2 on the first 12, so you are ahead 1 so you bet 3 on the second 12

(a1) If the second 12 wins (1/9 of the time - ignoring green numbers for the moment), you gain 6, so you are 7 ahead

(a2) If the first 12 (1/9) or third 12 (1/9) wins, you lose 3, so you are 2 behind

(b) If the third 12 wins, do the same as in (a); 1/9 of the time, you end up 7 ahead, and 2/9 of the time, you end up 2 behind

(c) If the second 12 wins (1/3 of the time), you lose both bets, so you are 2 behind, and you start again.

Final expected result (again, assuming 0 and 00 don't show up): 1/9 x (+7) + 2/9 x (-2) + 1/9 x (+7) + 2/9 x (-2) + 1/3 x (-2) = 0.

If I did not apply it properly, then please explain, if both of the first two spins land in the first 12 (the first two numbers in 1-1-2-2),

(a) what you bet for the third spin;

(b) if the third spin is in the first dozen, what you bet for the fourth spin;

(c) if the third spin is in the second dozen, what (if anything) you bet for the fourth spin;

(d) if the third spin is in the third dozen, what you bet for the fourth spin.

(When I say "what you bet", specify the amounts and the specific dozens - terms like "put the 3 chips on its fourth" are confusing.)

Quote:ThatDonGuyLet me see if I have this right:

Suppose the first two spins are both in the first dozen. In your chart, the next number is 2 (1-1-2-2), so don't bet on the second dozen, but instead bet 1 on both the first and third dozens.

If, say, the first dozen wins, then bet 3 chips on the second dozen (which is the fourth number in 1-1-2-2), since "one group is guaranteed to have 3 numbers hit".

Problem: if the fourth spin lands in the first dozen, then the third group of 4 in your chart (1-3-1-1) has 3 hits, and if the fourth spin lands on the third dozen, then the seventh group (3-1-1-3) does. Your "guarantee" still holds, but you lost your fourth bet.

Here's how I work this out:

First bet (third spin) - 1 on first 12, 1 on third 12

(a) If the first 12 wins, you lose 1 on the third 12, but gain 2 on the first 12, so you are ahead 1 so you bet 3 on the second 12

(a1) If the second 12 wins (1/9 of the time - ignoring green numbers for the moment), you gain 6, so you are 7 ahead

(a2) If the first 12 (1/9) or third 12 (1/9) wins, you lose 3, so you are 2 behind

(b) If the third 12 wins, do the same as in (a); 1/9 of the time, you end up 7 ahead, and 2/9 of the time, you end up 2 behind

(c) If the second 12 wins (1/3 of the time), you lose both bets, so you are 2 behind, and you start again.

Final expected result (again, assuming 0 and 00 don't show up): 1/9 x (+7) + 2/9 x (-2) + 1/9 x (+7) + 2/9 x (-2) + 1/3 x (-2) = 0.

Your calculations are correct but the third is likely to lose twice in three. If there is a guarantee of three correct then betting the 1-2-4 has a one in four chance which pays 7/2.

The completion of any 4 spins chosen in advance is an 80/1 shot no matter how you slice it so IS less likely to happen .

If I did not apply it properly, then please explain, if both of the first two spins land in the first 12 (the first two numbers in 1-1-2-2),

(a) what you bet for the third spin;

(b) if the third spin is in the first dozen, what you bet for the fourth spin;

(c) if the third spin is in the second dozen, what (if anything) you bet for the fourth spin;

(d) if the third spin is in the third dozen, what you bet for the fourth spin.

(When I say "what you bet", specify the amounts and the specific dozens - terms like "put the 3 chips on its fourth" are confusing.)