Reducing the House Edge
Consider a 37 number table where the bettor receives half his bet back if zero occurs when he/she bets Red or Black.
The Probability of getting 3 correct in 3 consecutive spins is one in eight if we ignore the zero. But what to bet ?
By betting a particular chosen sequence of 3 or even 3
consecutive Blacks or Reds to be All Correct or All Wrong
we can reduce the House Edge by betting in 3s.
We let the first spin go without betting.This is our" Indicator"( trigger) of what to bet on the second spin of the three.
If the second chosen selection( our first bet) wins we then bet the resultant 2 points on the following spin .
If the first spin-which we don't bet- is wrong then we bet- as a double- the second and third spin to be wrong.
Thus we , in effect, bet the three to be correct or incorrect over the 3 spins.A 2in 8 chance (1 in 3) for which we get paid 3/1-correct odds ignoring the zero.
Now consider the chances of the zero.
If it falls on the first spin we don't lose as we haven't bet.
If it falls on the second spin we lose half a point.
If it falls on the third spin then either,
a) we lose nothing if our second bet loses as our double has lost. or,
b) we lose the win of the second spin but not our original stake so we lose -or win - nothing.
So the chance of LOSING to the zero is reduced.
Quote:Member since: Oct 16, 2013
instead of pouncing on him let's try to help him find who hacked into his account; he can't have been paying that little attention
surely a pure troll wouldnt have waited this long to get his jollies by tweaking the rest of us
edit: I just looked at the other threads he started
Quote: scepticusREDUCING THE HOUSE EDGE.
Consider a 37 number table where the bettor receives half his bet back if zero occurs when he/she bets Red or Black.
The Probability of getting 3 correct in 3 consecutive spins is one in eight if we ignore the zero. But what to bet ?
By betting a particular chosen sequence of 3 or even 3
consecutive Blacks or Reds to be All Correct or All Wrong
we can reduce the House Edge by betting in 3s.
We let the first spin go without betting.This is our" Indicator"( trigger) of what to bet on the second spin of the three.
If the second chosen selection( our first bet) wins we then bet the resultant 2 points on the following spin .
If the first spin-which we don't bet- is wrong then we bet- as a double- the second and third spin to be wrong.
Thus we , in effect, bet the three to be correct or incorrect over the 3 spins.A 2in 8 chance (1 in 3) for which we get paid 3/1-correct odds ignoring the zero.
Now consider the chances of the zero.
If it falls on the first spin we don't lose as we haven't bet.
If it falls on the second spin we lose half a point.
If it falls on the third spin then either,
a) we lose nothing if our second bet loses as our double has lost. or,
b) we lose the win of the second spin but not our original stake so we lose -or win - nothing.
So the chance of LOSING to the zero is reduced.
A system you say!
Quote: scepticusConsider a 37 number table where the bettor receives half his bet back if zero occurs when he/she bets Red or Black.
The Probability of getting 3 correct in 3 consecutive spins is one in eight if we ignore the zero. But what to bet ?
By betting a particular chosen sequence of 3 or even 3 consecutive Blacks or Reds to be All Correct or All Wrong we can reduce the House Edge by betting in 3s.
We let the first spin go without betting.This is our" Indicator"( trigger) of what to bet on the second spin of the three.
If the second chosen selection (our first bet) wins we then bet the resultant 2 points on the following spin.
If the first spin-which we don't bet- is wrong then we bet- as a double- the second and third spin to be wrong.
How can the first spin - the "indicator spin" - be "wrong"?
Do you mean that if your first bet loses, you bet 2 on the other color for each of two more spins? Or perhaps you bet 1 on the other color on the next spin, and then let the bet ride for another spin if it wins?
Quote: scepticusThus we, in effect, bet the three to be correct or incorrect over the 3 spins. A 2 in 8 chance (1 in 3)
The last time I looked, a 2 in 8 chance is 1 in 4.
Before I continue any further, let me make sure I understand your system.
Your first spin is a non-betting "indicator" spin.
Your second spin has a bet of 1 on the same color as the first spin.
If your second spin wins, your third spin has a bet of 2 on the same color.
If your second spin loses, your third spin has a bet of 1 on the opposite color, and if the third spin wins, your fourth spin is a bet of 2 on the same color.
ZCore13
OKQuote: scepticusConsider a 37 number table where the bettor receives half his bet back if zero occurs when he/she bets Red or Black.
Never ever ignore the zero!Quote: scepticusThe Probability of getting 3 correct in 3 consecutive spins is one in eight if we ignore the zero.
There are now 27 possible 3 spin sequences and not all have the same probability of happening
I know you know this to be true
here is my table for this
| index | pattern | zero | red | black | prob |
|---|---|---|---|---|---|
| 1 | 000 | 3 | 0 | 0 | 1.97422E-05 |
| 2 | 00R | 2 | 1 | 0 | 0.000355359 |
| 3 | 00B | 2 | 0 | 1 | 0.000355359 |
| 4 | 0R0 | 2 | 1 | 0 | 0.000355359 |
| 5 | 0RR | 1 | 2 | 0 | 0.006396462 |
| 6 | 0RB | 1 | 1 | 1 | 0.006396462 |
| 7 | 0B0 | 2 | 0 | 1 | 0.000355359 |
| 8 | 0BR | 1 | 1 | 1 | 0.006396462 |
| 9 | 0BB | 1 | 0 | 2 | 0.006396462 |
| 10 | R00 | 2 | 1 | 0 | 0.000355359 |
| 11 | R0R | 1 | 2 | 0 | 0.006396462 |
| 12 | R0B | 1 | 1 | 1 | 0.006396462 |
| 13 | RR0 | 1 | 2 | 0 | 0.006396462 |
| 14 | RRR | 0 | 3 | 0 | 0.11513632 |
| 15 | RRB | 0 | 2 | 1 | 0.11513632 |
| 16 | RB0 | 1 | 1 | 1 | 0.006396462 |
| 17 | RBR | 0 | 2 | 1 | 0.11513632 |
| 18 | RBB | 0 | 1 | 2 | 0.11513632 |
| 19 | B00 | 2 | 0 | 1 | 0.000355359 |
| 20 | B0R | 1 | 1 | 1 | 0.006396462 |
| 21 | B0B | 1 | 0 | 2 | 0.006396462 |
| 22 | BR0 | 1 | 1 | 1 | 0.006396462 |
| 23 | BRR | 0 | 2 | 1 | 0.11513632 |
| 24 | BRB | 0 | 1 | 2 | 0.11513632 |
| 25 | BB0 | 1 | 0 | 2 | 0.006396462 |
| 26 | BBR | 0 | 1 | 2 | 0.11513632 |
| 27 | BBB | 0 | 0 | 3 | 0.11513632 |
looks like BBB = RRR = RBR = BBR and so on
8 of the 27 patterns have the same chance to happen in any 3 spin set. This IS so cool ;)
Now you can easily calculate if you actually do change the house edge
Please submit your homework in a timely manner if you want the credit
fun fun
Sally
This leads me to ask scepticus
"Why do you want to reduce the house edge?"
Quote: ThatDonGuyHow can the first spin - the "indicator spin" - be "wrong"?
Do you mean that if your first bet loses, you bet 2 on the other color for each of two more spins? Or perhaps you bet 1 on the other color on the next spin, and then let the bet ride for another spin if it wins?
The last time I looked, a 2 in 8 chance is 1 in 4.
Before I continue any further, let me make sure I understand your system.
Your first spin is a non-betting "indicator" spin.
Your second spin has a bet of 1 on the same color as the first spin.
If your second spin wins, your third spin has a bet of 2 on the same color.
If your second spin loses, your third spin has a bet of 1 on the opposite color, and if the third spin wins, your fourth spin is a bet of 2 on the same color.
Well ,O.K. TDG I didn't check my post properly .It is 1 in 4 but it is still a 3/1 chance - ignoring the zero.
Once again, I may not have made myself clear.
Consider I choose to bet BOTH Red AND Black over 3 spins-the 3 to be either all reds or all black.Betting both red and black on the first spin is not advisable because (a) we won't profit and (b) a zero would incur a loss.So we avoid the first spin of the three.
If a red wins the first spin we bet 1point on red and if it wins we put 2 points on red again on spin 3. Similarly with black.
If, on the second spin black wins when I'm betting red the bet is lost and we don't bet the third spin.
In theory we are betting BOTH red and Black to win 3 in a row- a 3/1 shot ignoring the zero.
The bet is over a series of 3 spins ,there is no fourth .
Clearer now ? The question is does it reduce the whole HE over the series of three or has it been reduced ?
Quote: mustangsallyOK
Never ever ignore the zero!
There are now 27 possible 3 spin sequences and not all have the same probability of happening
I know you know this to be true
here is my table for this
index pattern zero red black prob 1 000 3 0 0 1.97422E-05 2 00R 2 1 0 0.000355359 3 00B 2 0 1 0.000355359 4 0R0 2 1 0 0.000355359 5 0RR 1 2 0 0.006396462 6 0RB 1 1 1 0.006396462 7 0B0 2 0 1 0.000355359 8 0BR 1 1 1 0.006396462 9 0BB 1 0 2 0.006396462 10 R00 2 1 0 0.000355359 11 R0R 1 2 0 0.006396462 12 R0B 1 1 1 0.006396462 13 RR0 1 2 0 0.006396462 14 RRR 0 3 0 0.11513632 15 RRB 0 2 1 0.11513632 16 RB0 1 1 1 0.006396462 17 RBR 0 2 1 0.11513632 18 RBB 0 1 2 0.11513632 19 B00 2 0 1 0.000355359 20 B0R 1 1 1 0.006396462 21 B0B 1 0 2 0.006396462 22 BR0 1 1 1 0.006396462 23 BRR 0 2 1 0.11513632 24 BRB 0 1 2 0.11513632 25 BB0 1 0 2 0.006396462 26 BBR 0 1 2 0.11513632 27 BBB 0 0 3 0.11513632
looks like BBB = RRR = RBR = BBR and so on
8 of the 27 patterns have the same chance to happen in any 3 spin set. This IS so cool ;)
Now you can easily calculate if you actually do change the house edge
Please submit your homework in a timely manner if you want the credit
fun fun
Sally
This leads me to ask scepticus
"Why do you want to reduce the house edge?"
Quote: mustangsallyOK
Never ever ignore the zero!
There are now 27 possible 3 spin sequences and not all have the same probability of happening
I know you know this to be true
here is my table for this
index pattern zero red black prob 1 000 3 0 0 1.97422E-05 2 00R 2 1 0 0.000355359 3 00B 2 0 1 0.000355359 4 0R0 2 1 0 0.000355359 5 0RR 1 2 0 0.006396462 6 0RB 1 1 1 0.006396462 7 0B0 2 0 1 0.000355359 8 0BR 1 1 1 0.006396462 9 0BB 1 0 2 0.006396462 10 R00 2 1 0 0.000355359 11 R0R 1 2 0 0.006396462 12 R0B 1 1 1 0.006396462 13 RR0 1 2 0 0.006396462 14 RRR 0 3 0 0.11513632 15 RRB 0 2 1 0.11513632 16 RB0 1 1 1 0.006396462 17 RBR 0 2 1 0.11513632 18 RBB 0 1 2 0.11513632 19 B00 2 0 1 0.000355359 20 B0R 1 1 1 0.006396462 21 B0B 1 0 2 0.006396462 22 BR0 1 1 1 0.006396462 23 BRR 0 2 1 0.11513632 24 BRB 0 1 2 0.11513632 25 BB0 1 0 2 0.006396462 26 BBR 0 1 2 0.11513632 27 BBB 0 0 3 0.11513632
looks like BBB = RRR = RBR = BBR and so on
8 of the 27 patterns have the same chance to happen in any 3 spin set. This IS so cool ;)
Now you can easily calculate if you actually do change the house edge
Please submit your homework in a timely manner if you want the credit
fun fun
Sally
This leads me to ask scepticus
"Why do you want to reduce the house edge?"
