If a coin is tossed 3 times and all three tosses land on heads, the PROBABILTY (odds) of the next toss coming up heads is only 1 in 16.
The "CHANCE", on the other hand, never changes and is 50% on EVERY toss. (Its all in the wording I guess)
Ken
The probability of the next toss is still 50/50.
But if you mean the actual string of events, then you just multiple them together like 2 x 2 x 2 x 2 = 16
Quote from: Mr J on July 20, 2014, 02:18:09 PM
If a coin is tossed 3 times and all three tosses land on heads, the PROBABILTY (odds) of the next toss coming up heads is only 1 in 16.
The "CHANCE", on the other hand, never changes and is 50% on EVERY toss. (Its all in the wording I guess)
Ken
Hmmm. If that's the case then how about that for a system?
You walk up to a roulette and decide in advance to bet on black. With a twist. First you wait until you lose 3 times in a row. Meaning RRR. (Of course you don't lose actual money. Only in the virtual loss mode). If the probability of the 4th spin to bring yet another red is 1/16, then the probability of black should be 15/16. Overwhelmingly in your favor. The question is this. With 50:50 chances, betting randomly, is it possible to keep on losing spin after spin after spin? How many times? And is there a limit after which, a 50:50 chance finally comes your way?
I use a similar system with great success. I bet randomly 3 corner bets (12 numbers). And change them around after each spin. However I wait till I lose the first 3 bets virtually and then bet the next 3 with actual money. With a very small progression. Most of the time it works. If I lose the 3 actual bets, I start another cycle. I've been testing it for 4 years and about 85% of the time, 3 random corner bets win within 6 consecutive spins, and 95% of the time within 10 spins.
Quote by palestis *****You walk up to a roulette and decide in advance to bet on black. With a twist. First you wait until you lose 3 times in a row. Meaning RRR. (Of course you don't lose actual money. Only in the virtual loss mode). If the probability of the 4th spin to bring yet another red is 1/16, then the probability of black should be 15/16. Overwhelmingly in your favor. The question is this. With 50:50 chances, betting randomly, is it possible to keep on losing spin after spin after spin? How many times? And is there a limit after which, a 50:50 chance finally comes your way? ********
Excellent EC method based on longterm odds . Begin with the first real bet at spin # 4 and a seasoned player familiar with this method knows the precise moment when to stop betting.
This method requires patience and reduces the exposure to the casino`s advantage. To select a table with the le partage rule is a must.
Nathan Detroit
Quote from: Nathan Detroit on July 22, 2014, 10:31:14 AM
Quote by palestis *****You walk up to a roulette and decide in advance to bet on black. With a twist. First you wait until you lose 3 times in a row. Meaning RRR. (Of course you don't lose actual money. Only in the virtual loss mode). If the probability of the 4th spin to bring yet another red is 1/16, then the probability of black should be 15/16. Overwhelmingly in your favor. The question is this. With 50:50 chances, betting randomly, is it possible to keep on losing spin after spin after spin? How many times? And is there a limit after which, a 50:50 chance finally comes your way? ********
Excellent EC method based on longterm odds . Begin with the first real bet at spin # 4 and a seasoned player familiar with this method knows the precise moment when to stop betting.
This method requires patience and reduces the exposure to the casino`s advantage. To select a table with the le partage rule is a must.
Nathan Detroit
Patience is the greatest virtue for the roulette player, if he knows what he's waiting to develop. The logic behind this system is that it is quite rare when you see 3 consecutive EC's to go on and become 5or 6 or 7 or more. Cycle, after cycle, after cycle. Maybe once, but I doubt if you will see it often doing the same routine 3 or more cycles in a row in the same roulette.
But I found that it doesn't work well if you use "ready made situations". (Meaning scanning with your eyes several roulette boards to see a ready made 3 EC's in a row. And walk up starting the bets immediately. You have to be there at the roulette and determine in advance what you will play. Then wait for 3 opposite EC's before you start on the 4th spin. When you see a ready made scenario it's a done deal. Therefore your 50:50 chance did not play a role. When you walk up to a roulette and decide in advance what you will play, it's only then when your 50/50 luck starts counting.
Palestis
Sir,
You do not know the full details of this system.. Your reply is based on assumption only . .
HAPPY WINNINGS!!!
ND
QuoteIf the probability of the 4th spin to bring yet another red is 1/16, then the probability of black should be 15/16.
No the probability is still 50/50 (excluding 0). The 1/16 refers to the particular string of red/black. Remember there's also the chance that it will be RRRB. Can just as easily be RRRR or RBRB or RRBB etc.
When I am walking up to a table seeing 3 REDS in a row I don`t look at them as " virtual bets" but as an indication to commence betting a predetermined number of spins for the change .
This method is not recommended playing for a streak ..
.
Quote from: Nathan Detroit on July 24, 2014, 11:20:40 AM
When I am walking up to a table seeing 3 REDS in a row I don`t look at them as " virtual bets" but as an indication to commence betting a predetermined number of spins for the change .
This method is not recommended playing for a streak ..
.
That's what I was doing in my first roulette days. Looking around for streaks of the same EC, (3 in a row and up), and then played for the change doubling ea. time if no hit. Well, it worked most of the time, but when the streak of 3 became 6,7,8, and up it was disastrous. I find it much more effective to start counting after you approach the table. Choose in advance what to play and wait for a streak of 3 opposites. And then bet for the change only for the next 2 spins. The chances for a hit are high and the progression stays low. Easily recoverable if there is a loss. The power of the 50% chance for the player cannot be underestimated. The probability to lose all 5 spins (3 virtual and 2 real), is 1/2x1/2x1/2x1/2x1/2 about 3.125%. It will happen sometimes for sure. But I doubt if a player will fall in the 3.125% exception each time for the same scenario. Instead of a longer progression, the bet amounts can be increased the second time a new cycle begins. I doubt that a player will run into the negative 3.125% twice in a row.
The above post * 8 is well written and to the point within the context of this subject Well done palestis.
ND
Quote from: Steve on July 22, 2014, 08:05:03 PM
No the probability is still 50/50 (excluding 0). The 1/16 refers to the particular string of red/black. Remember there's also the chance that it will be RRRB. Can just as easily be RRRR or RBRB or RRBB etc.
HERE IS A POST THAT I ENTERED IN ANOTHER FORUM AFTER I HAD A LENGTHY ARGUMENT WITH ANOTHER MEMBER WHO CLAIMS THAT BETTING ON AN EC IS ALWAYS 50% NO MATTER WHAT THE BETTING CIRCUMSTANCES ARE. ANY COMMENTS?
QuoteMY QUESTION. Show me the probability of hitting black in any of 5 consecutive spins
Palestis
HIS ANSWER. There's about a 97% chance of hitting. (1) - (18/38 x 18/38 x 18/38 x 18/38 x 18/38 ) = 97.275
REAL
AND HERE IS MY ANALYSISYou admit that commencing a 5 bet series on BLACK there is a 97% probability that you will hit it, at any time during those 5 bets.
And 3% that you will lose. With a Martingale progression the bets will look like this (provided you go all the way to the 5th spin).
$5 $10 $20 $40 $80. Total risk $155 to win $5.
JUST KEEP IN MIND THAT THE PROBABILITY TO WIN WAS DETERMINED TO BE 97%
Question #1: Is there any reason that, if you lost the first 2 spins, the ESTABLISHED probability will change in any way for the 3 spins remaining? NO. A series bet PROBABILITY does not guarantee that you will hit the target in the first 2 or 3 spins. You have to go thru all 5 spins.
QUESTION #2: What is the probability that you will hit a black in the remaining 3 spins? It still is 97%. Anything else will be contradictory to the principle of probability of series
Now let's say that your betting series won't be like 5 10 20 40 80. Instead it will look like $0.01- $0.01 - $5- $10- $20. Total risk $35.02.
WITH A STIPULATION: That the first 2 bets have to be losing bets (enter VIRTUAL LOSS).
Again will the probability change for the 3 remaining just because we lost the first 2 bets? NO NO NO Still is 97%.
The only thing that changed is the $ RISK which was reduced from $155 to $35. And some added waiting period waiting for 2 REDS.
ACTUALLY YOUR PROBABILITY OF 97% IS WRONG.
If it was correct, if you played 100 times you will win $485. (97x$5).
and you will lose $465 (3x$155). So you will always be ahead $20
To makes things simple it's not 97%. Accurately figured, if you play in this fashion (5 10 20 40 80) in the long run you'll will be losing an amount equal to the HE (house edge).
However I just proved to you that if you play with the first 2 spins in the virtual loss mode you will always be winning in the long term.
Even if I downgrade the probability of the series from 97% to 90% win rate, if you play the system 100 times (with the first 2 spins in the virtual loss mode) you will be winning $450 (90x$5) and you will be losing $350 (10x$35). A constant profit of $100. Simply by switching the first 2 spins in the virtual loss mode. And if you changed it to 3 virtual losses then the profit will be even more impressive.
The downside is the waiting period. But if I know that I will always be winning, waiting doesn't bother me at all.
We wait from 9 to 5 to make a living. Y not wait when we work with roulettes.
Ps: Any argument that after 2 reds the probability for a B (
in a pre-determined series) is 50%. is invalid.
Lets not forget that I made a decision in advance to bet Black 5 times. Therefore I REMAIN LOCKED IN THE PROBABILITY OF SERIES. NOT THE PROBABILITY OF EQUAL CHANCES TAKEN 1 SPIN AT A TIME.
which YOU ADMITTED it was around 97%. Meaning that I have a probability of 97% to hit it IN ANY OF THE NEXT 5 SPINS. LOSING THE FIRST 2 OR 3 OR 4 SPINS SHOULD NOT HAVE ANY EFFECT IN THE ALREADY ESTABLISHED PROBABILITY OF SERIES.
It still remains in effect for the entire duration of the series.
The next EC bet is 48.6% to win and 51.3% to lose
If you predict a series of say "2 wins in the next 7" thats Binomial distribution (probability) and is a %
At the start of your pre-determined march you have a probability of success in %. The next spin is always 48.6% to succeed.
Thats clear to everyone, surely?
Quote from: Turner on August 10, 2014, 07:23:12 AM
The next EC bet is 48.6% to win and 51.3% to lose
If you predict a series of say "2 wins in the next 7" thats Binomial distribution (probability) and is a %
At the start of your pre-determined march you have a probability of success in %. The next spin is always 48.6% to succeed.
Thats clear to everyone, surely?
The assumption that every spin has a 48.6% probability (for EC bets), must be wrong.
At least for practical purposes. Maybe it's true in theory. But does it really happen in real life situations? Like roulette? After seeing 7 reds in a row, can you recall how many times, those 7 reds turn to 12 reds in a row? I can count in my fingers the times I saw that happen over the years and hundreds of thousands of spins. The overwhelming majority of times it turned to black within the next 3-4 spins, if not immediately right after. Time after time after time. When you observe a roulette table, it won't be long before you see 4-5 black or red come up in a row. Anywhere from 5 to 10 minutes. However after seeing 5 black y is it so hard to see yet another 5? You can sit there for hours and hours and you might never see it. If it's so easy to see 5 black in a row, it should be just as easy to see another 5, after the already 5 came up. Yet it's not easy at all. What happened? For the first 5 black the roulette seems to have NO MEMORY? But when it comes to another 5, the roulette starts remembering things?
Maybe the picture below (taken from a basic probability book) can explain this phenomenon.
Palestis
The maths is real. What you said is illusion....
You have to break it down into what it actually is. Its consecutive, totally unrelated random numbers.
The fact they are on a carpet and painted on a wheel different colours, and stacked in a time line on the marque is where the illusion comes.
Think about this.
Palestis and Turner go to the casino and Palestis gets to the Table first. He sits and plays, and is following series and notices 5 reds, and plays Black. Turner overheard palestis at the bar saying "The assumption that every spin has a 48.6% probability (for EC bets), must be wrong"
Turner comes to the table, doesnt look at the Marque, just places chips on Black also.
The odds of Turner winning are 18/37 or 48.6%. The odds of Turner losing are 19/37 or 51.3%
Surely the odds arnt different for Palestis, just because he has looked at the marque and sat around for a while??
Quote from: Turner on August 12, 2014, 04:57:00 PM
Palestis
The maths is real. What you said is illusion....
You have to break it down into what it actually is. Its consecutive, totally unrelated random numbers.
The fact they are on a carpet and painted on a wheel different colours, and stacked in a time line on the marque is where the illusion comes.
Think about this.
Palestis and Turner go to the casino and Palestis gets to the Table first. He sits and plays, and is following series and notices 5 reds, and plays Black. Turner overheard palestis at the bar saying "The assumption that every spin has a 48.6% probability (for EC bets), must be wrong"
Turner comes to the table, doesnt look at the Marque, just places chips on Black also.
The odds of Turner winning are 18/37 or 48.6%. The odds of Turner losing are 19/37 or 51.3%
Surely the odds arnt different for Palestis, just because he has looked at the marque and sat around for a while??
Sitting around is not the case. If I decide in advance to play 5 black in a row, I'm locked in a probability of series event. VIRTUAL LOSS is an actual bet but it doesn't involve money. It's free. I don't just sit there observing numbers. I actually BET. That is I'm using my 48.6% chance to win. So by having 3 reds come up, means that I used my 48.6% chance to win 3 times and I lost all 3 times. The question is at 48.6% chance to win, how many times in a row can I lose? 10? Impossible. It never happened to me in the last 10 years. Betting 10 times a random EC and be wrong all 10 times.
If you are at the bar and run up and start betting on black, after 5 reds on the score board, it means nothing to you. Because you had no input (meaning tried your luck), in those 5 reds. The probability countdown starts the minute you commence betting. Whether it's virtual or actual.
Now I agree...as I said in my first post....setting a goal, say at 3 wins in the next 7 would be 28% chance...but that is worked out in binomial distribution, and the formula uses the odds of winning and losing...see?
7C3 x 18/37^3 x 19/37^4
=28.02%
This is "how many 3's in 7" x win probability x loss probability
No special odds for each spin. Just the odds of winning (18/37) and losing (19/37)
You can clearly see them in the formula. Win 18/37 and loss 19/37
do you see my point?
My point being, yes, a prediction has a probability which changes depending on the prediction, but each spin is same probability.
Thanks Turner, much appreciated !!
Ken
playing roulette will really challenge your patience. So good luck!
This system playing for a change ( e.g. after 3 Reds bet on Black a certain number of times) is being recognized by many seasoned gamblers as being what card counting is in Black Jack.
This method is highly recommended by 2 of the top US professional gamblers. J.P and F,S.
AMATEURS don`t need to apply.
.
MrJ thanks again for having started this thread.
Nathan Detroit
'Posted August 14th, 2014 1700 hrs EDST ( Filed)
Quote from: Nathan Detroit on August 14, 2014, 06:01:51 PM
This system playing for a change ( e.g. after 3 Reds bet on Black a certain number of times) is being recognized by many seasoned gamblers as being what card counting is in Black Jack.
This method is highly recommended by 2 of the top US professional gamblers. J.P and F,S.
AMATEURS don`t need to apply.
.
Amazing. The best systems turn out to be so simple and less complicated.
I agree with this system too. Though I play a different version. (I don't like progressions that shoot up too fast).
Instead of black or red missing for 3 spins, I Play a dozen missing for 6 or more spins (whenever I see it on a score board). And I play it for about 3 or 4 times, sometimes 5. And stop. The logic behind it is this. If a dozen is missing 6 times, will it always go on and become missing 10 times? Cycle after cycle after cycle? I doubt it. Sometimes you will see missing 20 + times, but since you only play it 4 times, you can't lose your shirt. However 6 misses turning out to become 10 misses each and every time, it's very unlikely.
Just as if red was missing for 3 spins, turning out missing for 6. Every time.
If you were playing at a 0/00 wheel then you could disregard the fixed table layout for the dozens by creating your own dozens.
Just check on google WASHOO2 2 dozen roulette system.
You find it also in the archives at one of Steve`s boards.
8/16/2014 Post on File
Summary (if you are interested :laugh:)
Q: Why are the odds of getting at least 1 black in 5 spins around 97%?
1) The odds of getting 1 and 1 only black in 5 spins
how many 1's in 5 = 5C1 = 5!/1!/4! =5 (! is factorial. 5! is 5x4x3x2x1)
win = 18/38 and loss = 20/38 or 0.472 and 0.526
5C1 x 0.472 x 0.526 x 0.526 x 0.525 x 0.526 (or 0.526^4)
=0.1806 = 18%
so the odds of getting just 1 black in 5 spins is 18%
2) odds of getting 2 blacks only
5C2 x 0.472^2 x 0.526^3
= 10 x 0.03242 = 0.3242 =32.42%
3) odds of getting 3 blacks only
=29%
4) odds of getting 4 blacks only
=13%
5) odds of getting 5 blacks
=0.472^5 = 2.41%
So....18%+32.42%+29%+14%+2.41% = 96%
BUT.....for a very close estimate, forget zero and use win 0.5 and loss 0.5 = 97%
Mr Turner,
For simplification purposes:
Afeter 3 REDS bet the next spin on Black. If you win that one then the series is over.
If you had lost that first bet on nBlack you now bet once more on Black. If you win then the series is over.
I
If you lost that bet you bet the third time on Black. Win or lose the series is over..
The odds that a 6th Red could appear is 63 : 1.
Just play it at your live local casino.
Thanks Nate..
I just saw that stat.....97%
Well...I believe nothing unless I see it for my self.
So I broke it down