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Isn't it strange?

Started by Richard Meisel, Jun 08, 10:17 PM 2020

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Richard Meisel

Just a little humor for these days.
Isn’t it strange? You have $20 and you bet $20 on the 1st Dozen:
W12  L26
Win 12 x20 = 240 
Lose 26 x-20 = -520
Total = -280

What if you bet the $20 on 2 Dozens?
10 on 1st Dozen, 10 on 2nd Dozen
W24  L14
Win 24 x10 = 240
Lose 14 x-20 = -280
Total = -40

And what if you bet $20 on 3 Dozens! What?
1st Dozen = 12 x10 = 120
2nd Dozen = 12 x-5 = -60
3rd Dozen = 12 x-5 = -60
Loss on 0 & 00 = 2 x-20 = -40
Total = -20

You actually lose less betting on all 3 Dozens or all 3 Columns! Now I don’t bet this way and I don’t recommend this bet but I’ve never found on any betting forum or on the internet anyone mentioning this bet, except Saliu mentions it in passing. But isn’t it strange?

Bigbroben

Your calculations are wrong.

Please correct. 

First example should give 12x40 since payout is 2:1.  With a total of -40.

Third example has 10 on 1st doz and 5 on both 2 and 3?  If so, results are wrong.
Life is hard, and then you die.
Mes pensées sont le dernier retranchement de ma liberté.

Taotie

Isn't it strange, that I just wasted 3 minutes reading and replying to this nonsense.

Bigbroben

Quote from: Taotie on Jun 09, 04:47 AM 2020
Isn't it strange, that I just wasted 3 minutes reading and replying to this nonsense.

Not strange: you were probably hoping for an anomaly that would lead to a HG and put you on track to become a filthy-rich king of the world.

Deceptive.
Life is hard, and then you die.
Mes pensées sont le dernier retranchement de ma liberté.

Richard Meisel

Quote from: Bigbroben on Jun 09, 12:17 AM 2020
Your calculations are wrong.

Please correct. 

First example should give 12x40 since payout is 2:1.  With a total of -40.

Third example has 10 on 1st doz and 5 on both 2 and 3?  If so, results are wrong.
Yes, correction: 20 on 1st Dozen
W12  L26
Win 12 x40 = 480 
Lose 26 x-20 = -520
Total = -40

10 on 1st Dozen, 10 on 2nd Dozen
W24  L14
Win 24 x10 = 240
Lose 14 x-20 = -280
Total = -40

10 on 1st Dozen, 5 on 2nd Dozen, 5 on 3rd Dozen
W36  L2
Win on 1st Dozen = 12 x10 = 120
Win on 2nd Dozen = 12 x-5 = -60
Win on 3rd Dozen = 12 x-5 = -60
Loss on 0 & 00 = 2 x-20 = -40
Total = -20

Betting $20 on the 1st Dozen will lose $40 after 38 Spins. Betting $20 ($10 on 1st Dozen and $10 on 2nd Dozen) will lose $40 after 38 Spins. But Betting $20 ($10 on 1st Dozen and $5 on 2nd Dozen and $5 on 3rd Dozen) will only lose $20 after 38 Spins. Bad Bet. Whoever thought of Betting on all 3 Dozens at once or all 3 Columns at once? Saliu mentions it. But, still, isn't it strange? Some of you aren't very nice. It took a long time for me to be a nice guy.

plolp


What is strange is that you take a particular case to make it general

gizmotron2

Quote from: plolp on Jun 10, 01:31 AM 2020What is strange is that you take a particular case to make it general
Been watching your play at R-sim. Very interesting stuff. Do you think you have winning all figured out?
Reading Randomness is a single thread. It is backed up by a software instruction thread and software download threads. The Even Chance Pro 1.4 version is the best version to practice on.
gamblingforums dot com/threads/reading-randomness.14733/

Moxy

Quote from: Taotie on Jun 09, 04:47 AM 2020
Isn't it strange, that I just wasted 3 minutes reading and replying to this nonsense.

Meh.  I've seen stranger things.




I'll be here all week.

plolp


@ Gizmotron :

maybe not ... My last 21 games are all won by equal weight, in that I am happy.
then i improve this, for now i don't know.

Richard Meisel

Quote from: Taotie on Jun 09, 04:47 AM 2020
Isn't it strange, that I just wasted 3 minutes reading and replying to this nonsense.
Isn't it strange?
AMERICAN WHEEL
N   =   log(1 -DC)/log(1 -p)   When DC = 2/3 (66.6%) and   p = 1/38      
log(1 -DC)   =   log(1 -2/3)   =   log(1 - .66666)   =   log(.33334)   =   -0.4771
log(1 -p)   =   log(1 -1/38)   =   log(1 - .02631)   =   log(.97369)   =   -0.0115
N   =   -.47711/-.011579   =   41.2   =   42

EUROPEAN WHEEL
N   =   log(1 -DC)/log(1 -p)   When DC = 2/3 (66.6%) and   p = 1/37      
log(1 -DC)   =   log(1 -2/3)   =   log(1 - .66666)   =   log(.33334)   =   -0.47711
log(1 -p)   =   log(1 -1/37)   =   log(1 - .027027)   =   log(.97297)   =   -0.01189
N   =   -.47711/-.011899   =   40.09   =   40

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