Some things ARE more likely than others !
In roulette the three possible results 1-2-x becomes 1-2-3 for dozens or columns.The promoters were " helpful" and devised "plans" to take out some of the guesswork. One such was a "Nine Column Block" covering 4 matches - think spins.There were 9 of these blocks in total covering all possible 81 results.
Each of these blocks "guaranteed " that one of the 9 lines would have at least 3 correct no matter the results.I give an example below- not the original but I think it fills the bill. You will quickly find that you cannot bet the "first two spins" but can use them to assess what to bet on the next two spins so they become "Indicators ".
Logic dictates that if something is guaranteed then some things have indeed a better chance than others .Choice is still needed but all gambling carries risk and the casino is at risk of variance as well as we gamblers.
1-1-1-2-2-2-3-3-3
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
Have fun.
Quote: scepticusHave fun.
Thanks, you too.
Quote: scepticusEach of these blocks "guaranteed " that one of the 9 lines would have at least 3 correct no matter the results.
Step 1: pigeonhole principle.
Step 2: ???
Step 3: Profit
Yes, you are guaranteed to have 3 wins on one line, but, as you say, you can't bet them "in advance", and once the first two spins are done, the guarantees disappear. The guaranteed 3 wins include situations where the first two spins were correct, ones where one spin was correct, and ones where neither spin was correct.
Quote: ThatDonGuyThe problem is, if you treat the pools as a series of individual bets, then the one line with the 3 wins means you are +1 on that line; meanwhile, you break even on the lines with 2 wins, and are -1 on the lines with 1 win. The pools strategy "works" because they are pari-mutuel.
Yes, you are guaranteed to have 3 wins on one line, but, as you say, you can't bet them "in advance", and once the first two spins are done, the guarantees disappear. The guaranteed 3 wins include situations where the first two spins were correct, ones where one spin was correct, and ones where neither spin was correct.
Yes. it's true that the "trebles" disappear but the guarantee still stands .After the first two have been spun there is still the guarantee that some column/s will have one correct while one will have two correct in the next two spins - barring zero/s.
So the question boils down to "which one ? ".
Quote: ScepticusIn roulette the three possible results 1-2-x becomes 1-2-3 for dozens or columns.The promoters were " helpful" and devised "plans" to take out some of the guesswork. One such was a "Nine Column Block" covering 4 matches - think spins.There were 9 of these blocks in total covering all possible 81 results.
Each of these blocks "guaranteed " that one of the 9 lines would have at least 3 correct no matter the results.I give an example below- not the original but I think it fills the bill. You will quickly find that you cannot bet the "first two spins" but can use them to assess what to bet on the next two spins so they become "Indicators ".
Logic dictates that if something is guaranteed then some things have indeed a better chance than others .Choice is still needed but all gambling carries risk and the casino is at risk of variance as well as we gamblers.
1-1-1-2-2-2-3-3-3
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
Have fun.
What about the zero(s)? And why does it matter what hit on the previous spins?
Are you traveling forwards and backwards in time?
If you're going to write down the possible outcomes, then you should write down ALL of the possible outcomes. If you do so, then you'll find that there are more possible outcomes than what the odds actually pay.
Quote: KeyserWhat about the zero(s)? And why does it matter what hit on the previous spins?
Are you traveling forwards and backwards in time?
If you're going to write down the possible outcomes, then you should write down ALL of the possible outcomes. If you do so, then you'll find that there are more possible outcomes than what the odds actually pay.
I 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 .
Maths is useful only insofar as the basis on which it is grounded is solid.There are too many assumptions in the Long Run Theory for this .
Have you guys not learned lessons from the many disasters in the financial world where maths geeks had such faith in the certainty of outcome in their models that they did not even consider whether the basis for them was flawed ? One complained of a ten sigma event having occurred !
Your fundamentalist faith is touching and may be suitable for indulging in theory - but not in the real world of roulette gambling where that little white ball can make fools of us all . I understand the risks but don't tell me that you have a crystal ball that gives you the certainty you claim.
Will I win in the future ? I don't know - you don't know - nobody knows . The difference between us is that I know that I am guessing while you don't know that you are only guessing.
You are entitled to your opinion but you are not entitled to ridicule others who differ from you .
Left cheek: even
Right cheek: odd
Both: red
Neither: black
It works about as well as your method.
+1
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.)
Quote: scepticusYour calculations are correct but the third is likely to lose twice in three.
The third what?
Quote: scepticusIf 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
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.
Quote: scepticus1) 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.
Quote: ThatDonGuyThere'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 .
Quote: scepticusFrom 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: scepticusAs 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?
Quote: ThatDonGuyHuh? 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 ?
Quote: ThatDonGuyHuh? 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 ?
Quote: scepticusThe 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: scepticusIf 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)?
Quote: ThatDonGuyYou'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.
They gave him an IQ test and it came back negative!
It's obvious that he (or, for all I know, she) is sticking to his point no matter how much mathematics and statistics we dig up.
Quote: scepticusQuote: ThatDonGuyHuh? 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 ?
HEY! Don't be calling people idiotic, or otherwise trample their statements with nasty comments, when they are attempting to help you especially if their analogies would be apt to describe your own lack of mental acuity. I have read this entire thread. You are using a betting system with absolutely no merit. Your ideas to support it show a lack of understanding in mathematics. That is completely within your rights, and it is good for us who profit from gambling, because without people like you, casinos would be much more worried about us. However, you are the one who wished to discuss this matter and asked for input. If you are so sure about your theories, go get rich off your system.
Quote: AxiomOfChoiceIs this serious or a bad troll?
There are 81 ways of 4 being correct in 4spins of the wheel so it is less likely that one of my nine will be the winner even if the first three are correct .
I believe so - you guys don't.
Never the twain shall meet.
'bye !
Quote: scepticusI believe so - you guys don't.
Never the twain shall meet.
'bye !
Thank you OP, I hope you enjoyed your stay.
I nominate this thread for the Double-Talk, Mumbo Jumbo Hall of Fame.
Quote: gpac1377Thank you OP, I hope you enjoyed your stay.
I nominate this thread for the Double-Talk, Mumbo Jumbo Hall of Fame.
I am honored to second that nomination !
Quote: gpac1377Thank you OP, I hope you enjoyed your stay.
I nominate this thread for the Double-Talk, Mumbo Jumbo Hall of Fame.
It will have a hard time topping the crap in the baccarat forums.
I still don't even understand what he was saying. He correctly notes that there are 81 possible sequences and then somehow concludes that they are not all equally likely. I have no idea where this erroneous leap of faith came from.
Quote: AxiomOfChoiceI still don't even understand what he was saying.
For a while I thought maybe it was my lack of reading comprehension, but clearly there's no dialogue in this thread because the OP only talks, he doesn't listen.
Quote: AxiomOfChoiceI still don't even understand what he was saying. He correctly notes that there are 81 possible sequences and then somehow concludes that they are not all equally likely. I have no idea where this erroneous leap of faith came from.
Here's what I think he was saying:
If you look at his original list of 9 columns of 4 numbers, each row represents a spin, and each number represents 12 numbers (e.g. 1 is 1-12; 2 is 13-24; 3 is 25-36). One column is guaranteed to have at least three bets win. His strategy was:
(a) Wait for two spins, and note which dozens they were in; each pair of dozens appears in exactly one column. (For example, if both spins are in the first 12, use the first column, since it starts with "1-1"; if the first spin is in the second dozen and the second is in the third, use the sixth column, which starts with "2-3".)
(b) Bet 1 on the two dozens other than the one in the third row of the column in (a). (For example, if the first two spins are both in the first dozen, use column 1; the third number is 2, so bet 1 on the first dozen and 1 on the third dozen.)
(c) If the third spin is in the dozen you didn't bet on, then the first column is the one column with the three (or more) wins, so start over again.
(d) If the third spin is in one of the two dozens you didn't bet, then bet 3 on the fourth number in that column.
I think his reasoning was, there were four possible ways that the "guaranteed three wins" can happen:
(1) The first three spins all win;
(2) The first, second, and fourth spins all win;
(3) The first, third, and fourth spins all win;
(4) The second, third, and fourth spins all win.
If #2 is correct, then you win 1 chip on your first bet, and 6 more on your second bet
Otherwise, you will lose 2 chips (either losing 2 on your first bet, or winning 1 on your first bet and losing 3 on your second)
His assumption was that 1,2,3,4 were equally likely, so the EV was 1/4 x (+7) + 3/4 x (-2) = +1/4
In fact, they are - if you ignore the possibility that all four spins will win. If you wait until two spins have been done, then each of the four possibilities remains in two columns and "all four win," which is also a -2 result, is in the remaining column. When you include that possibility, the EV is 0. However, he insisted that "in reality" this would happen far less than 1/9 of the time. (Never mind that, later, he agreed that there are 81 possible combinations for the four spins, and there are 9 combinations in the original 9 columns, so obviously there's a 1/9 chance that one of them will have all four spins win.)
Thanks for at least trying to understand what I was about rather than the usual knee-jerk responses .
Yes, there is a 1 in 9 chance of 4 in-a-row but only on the third spin because if the 3rd in-a-row happens there can never be 4 -in -a-row because a"treble" has won so we restart after a "treble" has occurred.
Thanks again.
Quote: scepticusJust a quick note "That Don Guy. "
Thanks for at least trying to understand what I was about rather than the usual knee-jerk responses .
Yes, there is a 1 in 9 chance of 4 in-a-row but only on the third spin because if the 3rd in-a-row happens there can never be 4 -in -a-row because a"treble" has won so we restart after a "treble" has occurred.
That's true - but by ignoring the possibility of all 4 winning, your assumption that all 4 trebles are equally likely is false.
Let's go back to the "first two spins are 1 (i.e. the first dozen)" example.
Columns 5, 6, 8, and 9 can no longer have a treble.
If the next spin is 2 (the second dozen), then column 1 is a 1-2-3 (first spin, second spin, third spin) treble.
If the next spin is 1, then the third spin is either 1 (1-3-4 column 3 treble), 2 (1-2-4 column 1 treble), or 3 (2-3-4 column 7 treble).
If the next spin is 3, then the third spin is either 1 (2-3-4 column 4 treble), 2 (1-2-4 column 1 treble), or 3 (1-3-4 column 2 treble).
The 1-2-3 treble is 50% more likely than each of the others - and you lose 2 every time the 1-2-3 treble happens.
It appears as though Scepticus doesn't comprehend what you're trying to explain to him.
@Scepticus,
The wizardofodds.com has a great section on the gambler's fallacy that all new players should read.
Quote: tilt247Wow, I guess I just love to torture myself by reading these silly posts about progression systems and how people swear by them. I say, whatever works for you do it. Life is too short to care about what other people think about your beliefs, superstitions, eating habits etc. However, most of us don't want to hear how you have this system and it works and you win etc blah blah blah. On the other side, this part of the discussion board was created for a reason... Humor I think. Moving on...
I thought this forum was used to help each other in ways of making money. If that takes turning a persons beloved betting system on it's head then I think it should be done.
