Re: 30 numbers, 2 bets lost...

[Kon-Fu-Sed]

Hi all,

As I don't want to clutter Mr Chips' thread with this off-topic, I started this thread.
Mr Chip's thread, where it all began, is here: nolinks://vlsroulette.com/gambling-and-roulette-related/allocating-units-to-4-streets/


And at 11:35pm Richiechips write (I bold relevant parts):
---------------------------------------------
RAG Writes
I won't get into this with you there isn't enough server space here.
.....
RAG Writes
Instead of debating dependence vrs independence you should be trying what I suggested.  You might be pleasantly surprised and I am only trying to be helpful.
.....
RAG Writes
I don't agree at all but that is just me.  But I will repeat just try what I sent or did I waste my time?
However since I promised I'd send my math and you asked see below
but before you do I know you won't agree but that doesn't make you right.

So we have 30/37 covered which = 81.08% = .8108   This gives us an 81.08% probability of hitting on the first spin.  Here is how you compute that
1.  win   --   30/37 =  0.81081081081081081081081081081081 or  81.08%
2.  lose  --    7/37 =   0.18918918918918918918918918918919 or  18.92%
                                  100.00%
1.  win   --   81.08%  Total % of all possible wins is 81.08%
2.  lose  --   18.92%
        100.00%
This proves that you have a 81.08% chance on the 1st spin to win.
=======================================================================
However, we might miss on the first so we want to know the probability of hitting on the 2nd spin, assuming no spins were made.   This is how you compute the probability of 2 events.
1.  win, win   --   30/37 x 30/37 = 900/1369 = 0.65741417092768444119795471 or 65.74%
2.  win, lose  --   30/37 x   7/37 = 210/1369 = 0.15339663988312636961285609 or 15.34%
3.  lose, win  --     7/37 x 30/37 = 210/1369 = 0.15339663988312636961285609 or 15.34%
4.  lose, lose --     7/37 x  7/37 =   49/1369  = 0.03579254930606281957633308 or   3.58%
                                          100.00%
1.  win, win   --   65.74%
2.  win, lose  --  15.34%
3.  lose, win  --  15.34%  Total % of all possible wins is 96.42%
4.  lose, lose --    3.58%
      100.00%
This proves that you have a 96.42% chance by the 2nd  spin to win.
.....
Now please let's not debate this... neither one of us will agree...besides isn't it more important to find a decent system....I thought this was what I was trying to do.  If I overstepped my bounds I apologize. Just trying to help. I haven't debated since Collage.  I saw you were playing 4 streets and offered a suggestion.
------------------------------------------------
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Richiechips, I appreciate that you are trying to be helpful, but in fact you are not.
Believe me, Richiechips, I have tried! And I wasn't the least surprised...

But you don't want to debate this and I can see why - you are doing the EXACT mistake I said you did. Look here:

Your words:
1.  win, win   --   65.74%
2.  win, lose  --  15.34%
3.  lose, win  --  15.34%  Total % of all possible wins is 96.42%
4.  lose, lose --    3.58%
      100.00%
This proves that you have a 96.42% chance by the 2nd  spin to win.



NO! It is proof that you will win in 96% IF YOU STILL HAVE TWO SPINS IN FRONT OF YOU. Nothing else!

The SUM of win,win + win,lose + lose,win = 96.42% that is correct.
But then you also stipulate that it is still possible to win the first bet.
BUT if you know that the first spin was lost you CANNOT include the "win" alternatives for that spin, as in "win, win" and "win, lose".
The first bet WAS LOST. You KNOW THAT and therefore it cannot be calculated as a potential winner. It cannot be a winner, can it?

When the first bet is lost, the only possible combinations, after that fact, are "lose-win" and "lose-lose".
But you take into account all the four cases "win-win", "win-lose", "lose-win" and "lose-lose".
Why???

How can you say, that if the first bet was lost, a combination of "win-win" or "win-lose" is even remotely possible? You know it lost...
Please; you have to explain that! I get sooo KonFuSed

(And I don't think Victor will mind the bandwidth, if you CAN explain how a lost bet can be considered a potential winner - will you Victor?)

Also:
If you bet 30 different numbers (in any combination) you have a 30/37 chance to hit. That is 81.08% chance.
To get a 96.42% chance, you have to bet 36(!) different numbers as 36/37 = 97.3% (35 numbers = 94.6%)
But you actually claim that you, by betting only 30/37 numbers have a close to 36/37 chance to hit...
Is that what you do???


IN SHORT:
Each and every time you bet 30 numbers you have a 81% chance to win, regardless how many losses you have had before that.


And all this that I write, my dear Richiechips, is not an opinion or something I just make up to p* you.
It is the consensus among all who deal with probability, gambling theory and statistics.

Actually: The only times I read such mistakes as yours are:
* On gambling forums like this, written out of sheer ignorance of "Gambler's Fallacy"
* On roulette system seller's sites written out of sheer greed in hopes that "Gambler's Fallacy" theories sound appealing to the ignorant
(but then, of course, it's not by mistake...)

Yes, you are actually using "Gambler's Fallacy" arguments/theory to convince us: "The odds become better if you haven't won for a while!"
Ever heard that one before?
I think, hope, most of the people on this board are enlightened enough to not fall for that.

(A really heavy test of "wait before you bet" (GF) was performed by Grabb: nolinks://nolinks40.brinkster.com/grabb/ - check it out!)


KFS

[Kon-Fu-Sed]

Whoops!

I said I tried it but I forgot to post the results.
Here it is...

TEST #1:
--- I used a sample from the Hamburg casino
--- I started when there was two different 3-number bets
--- I always skipped the two LATEST showing

* First bet: Ten 3-number bets
- If won, I end the session and record it as a "1" session, because it won at the first bet
- If lost, I make a second bet

* Second bet: Ten 3-number bets
- If won, I end the session and record it as a "2" session, because it won at the second bet
- If lost, I make a third bet

* Third bet: Ten 3-number bets
- If won, I end the session and record it as a "3" session, because it won at the third bet
- If lost, I end the session and record it as a "0" session as it didn't win

I did a total of 3700 such sessions and the results were:
1-sessions: 3014
2-sessions:  552
3-sessions:  105
0-sessions:   29
----------------
Total:      3700

A perfect mathematical distribution would have been:
1-sessions: 3000 (30/37x3700)
2-sessions:  567.56757 (30/37x(3700-3000))
3-sessions:  107.37764 (30/37x(700-567.56757))
0-sessions:   25.05479 (the rest)

No surprises there... Quite close to what probability says it should be...



TEST #2:
--- I used another sample from the Hamburg casino (actually the data following what was used in the first test)
--- I started when there was two different 3-number bets
--- First bet I always skipped the two latest showing
--- Second bet I always bet the SAME ten
--- Third bet I always skipped the three latest showing
--- I make the bets and record the same way as in test #1

I did a total of 3700 such sessions and the results were:
1-sessions: 2991
2-sessions:  575
3-sessions:   99
0-sessions:   35
----------------
Total:      3700

A perfect mathematical distribution would have been:
1-sessions: 3000 (30/37x3700)
2-sessions:  567.56757 (30/37x(3700-3000))
3-sessions:   96.63988 (27/37x(700-567.56757))
0-sessions:   35.79255 (the rest)

Again, no real surprise there... Although a bit better than probability says on 2- and 3-sessions!



But what about the money?
You can insert the figures you like in the above two tables. I've made two examples for you:


In test #1 I bet 10u for a first bet 3700 times. That is -37,000u
On the first bet I won 3014 times giving me back 3014 x 12u = 36,168u. I'm now down 832u

For a second bet I bet 10u again, the 686 times I lost the first bet = 686 x 10u = -6860u
On the second bet I won 552 times giving me back 552 x 12u = 6624u - a loss of 236u. I'm now down 832 + 236 = 1068u

For a third bet I bet 10u again, the 134 times I lost the first two bets = 134 x 10u = -1340u
On the third bet I won 105 times giving me back 105 x 12u = 1260u - a loss of 80u. I'm now down 1068 + 80 = 1148u



In test #2 I bet 10u for a first bet 3700 times. That is -37,000u
On the first bet I won 2991 times giving me back 2991 x 12u = 35,892u. I'm now down 1108u

For a second bet I bet 20u - I double up on the same ten - the 709 times I lost the first bet = 709 x 20u = -14,180u
On the second bet I won 575 times giving me back 575 x 24u = 13,800u - a loss of 380u. I'm now down 1108 + 380 = 1488u

For a third bet I bet 36u - I quadruple on NINE - the 134 times I lost both the first two bets = 134 x 36u = -4824u
On the third bet I won 99 times giving me back 99 x 48u = 4752u - a loss of 72u. I'm now down 1488 + 72 = 1560u...


OK, I know this isn't exactly as Richiechips do it but I honestly didn't understand all of his description...


Regards,
KFS

(BTW: Why quadruple on the third bet when it hardly comes into play - only a little above 1/37 - when the first bet hits like 30/37...?)

[TwoCatSam]

KFS

I want to pose a question.  I am not good at math, I'll admit.  I can see RichieChips point of view, however.

Let's you and I flip a coin.  Every heads, I win.  I win the first flip.  What are my chances of winning the next flip?  It's the original chance (50%) X 50% for a 25% chance of winning two in a row.  Then it is 25% X 50% for a 12.5% chance of winning three in a row.  I think this is pretty cut-and-dried statistics as I have read them in about a dozen books, some having nothing to do with gambling.

Now, let's look at your chances of winning or my losing.  First flip we are equal, 50/50.  The next flip I have only a 25% chance of winning, so you have a 75% chance of winning.  The next flip I have only a 12.5% chance a winning and you have an 87.5% chance of winning.

Is the above not true?

So which is true?  On the third spin of our trot, do we still each have a 50/50 chance or does one of us have an 87.5% chance? 

Sam

By the way, I directed this to KFS, but anyone who wishes, please weigh in.

[Kon-Fu-Sed]

Hi TCS, and all interested,


Not that I'm a math wiz or something but I'll try my best to explain in a simple way what you have read... (And, as I understand it, misunderstood)
It will take a while but in the end I *think* you'll get it


Coin-flipping is OK! A true 50/50 possibility. (As long as it doesn't stand on its edge or disappears )
So you select heads each flip.

Let's say we are going to (note: "going to") do ONE and only one flip.
The only possible results are then Heads or Tails. Nothing more. So that's a 50% chance.
Two (and only two) possible outcomes makes it a 50/50 possibility. Correct?


So we decide that we are going to (note "going to") do TWO flips, no more no less.
The possible combinations here are four: H-H, H-T, T-H and H-H. Each a 25% possibility. OK?

Now we do the first flip: Result Heads. You WON!
As it was the first flip, it was a 50/50 possibility - as you correctly said.
As this result is now known, we also know that the only possible combinations for this two-flip game to end is H-H or H-T.
It cannot possibly end T-H or T-T because we've already seen the first result (H).
Do you agree on this? (I really hope so )

And two possible outcomes makes it a 50/50 possibility. Correct?


BUT suppose you LOST that first flip - it was Tails.
As this result is now known, we also know that the only possible combinations for this two-flip game to end is T-H or T-T.
It cannot possibly end H-H or H-T because we've already seen the first result (T).
Do you agree on this? (I really hope so )

And, again, two possible outcomes makes it a 50/50 possibility. Correct?


SO!
Before we flipped the first flip, we had four possible outcomes.
H-H, H-T, T-H and T-T, each having a 25% probability to come.
But after the first flip, we had only two left and so it is a 50/50 possibility.

And that is REGARDLESS of the first result - did you notice: If the first result was H, the total could be H-H or H-T (50/50) and if the first was T, the total could be T-H or T-T (also 50/50).

You always have to take ONLY future spins into account when dealing with probability ( = UNcertainty).
That is the opposite to what you do in statistics ( = certainty) where you use only PAST spins.
You just shouldn't mix them. Then it becomes Gambler's Fallacy.

The probability for an UNknown coinflip is 50% (or p=0.5) and for a known it is 100% (or p=1.0). The latter is simply 100% sure.

I mean: Something that HAVE certainly happened cannot suddenly have only a 50% chance to happen, can it?

Before any flip: 50% * 50% = 25% for each possible outcome (p = .5 * .5 = .25) - Four possibilities
After the first flip: 100% * 50% = 50% for each possible outcome (p = 1.0 * .5 = .5) - Two possibilities
After the second flip: 100% * 100% = 100% for each possible outcome ( p = 1.0 * 1.0 = 1.0) - Only one possiblity left
(The p I use is the common way to denote probability.)

<phew>

I hope I have explained it in an understandable way.
Otherwise I *can* expand it to three or four flips to show you that at the very last flip it's STILL a 50/50 chance for you to win that very last bet (or for me ) regardless of the starting results.

And please understand that this is not my own opinion or something I just make up for my own amusement or something.
If you compare what I've written to what you've read, you'll find that I'm correct. This is what any non-Gambler's Fallacy person will tell you.


Regards,
KFS

[TwoCatSam]

KFS

I totally agree with you and I have arrived at that same conclusion years ago.  I have even seen it explained that way.  I believe Marilyn vos Savant did it once in her PARADE column.

A coin comes off the mint and falls on the floor.  In its short life, it has never been flipped.  While it is cooling, I say to you, "I'm going to flip that coin three times.  What are the odds I will get three heads in a row?"  You should say 12.5%

Let's call the three flips an event--not three separate flips.  What are the chances for a three-headed event?  12.5%

On THE PRICE IS RIGHT, there was a game like this.  There are twelve squares before you on the floor, side by side.  When you started out, your chances of winning were tiny, you had to step on six correct squares in a row--no misses.  With each win, your chances of winning increased until you had two squares on the floor before you.  Now you had a 50/50 chance and what had past was of no consequence.

If I had bet on the girl to win before she stepped on the first square, she was a longshot.  When she was down to two, she was even money.

KFS, I can't disagree with you, but RichieChips makes a pretty good argument!!

Thanks for the response.

Sam

[Kon-Fu-Sed]

TCS,

YOU understand those basics, no doubt about that as you say: "Now you had a 50/50 chance and what had past was of no consequence".
And your conclusion
"If I had bet on the girl to win before she stepped on the first square, she was a longshot.  When she was down to two, she was even money."
Exactly. 

But these were R's arguments (I quote from his posts) that made me  :
"on your third spin playing the same 10 #'s even at quadruple you have a 99.32% chance of hitting."

I wouldn't have bothered if this was a mistake ("on your third spin..." instead of "in a total of three spins") but he goes on:
"I like to know I have a 99.32% chance on the 3 spin if I'm betting the same 10 streets."

And:
"1.  win, win   --   65.74%
2.  win, lose  --  15.34%
3.  lose, win  --  15.34%  Total % of all possible wins is 96.42%
4.  lose, lose --    3.58%
      100.00%
This proves that you have a 96.42% chance by the 2nd  spin to win."

Then he shows the same table for 3 spins and finishes that one with:
"This proves that you have a 99.32% chance by the 3rd spin to win"

It is all the way the same faulty way to use the correct numbers, that I explained above - he's mixing the past and the future. Simple Gambler's Fallacy. (Why? Haven't got a clue)
And maybe I wouldn't bother all that much (the top of this thread) if he hadn't been so... well, persistant and giving me replies like
"I won't get into this with you there isn't enough server space here" and
"Let's stay friends and not get into this either...." (this one as a reply to my statement "If the result is known, then for that result p=1.0")
to nearly all my questions and statements.

May I quote myself from above?
Actually: The only times I read such mistakes as yours are:
* On gambling forums like this, written out of sheer ignorance of "Gambler's Fallacy" (add: AKA "Forum Mathematics" )
* On roulette system seller's sites written out of sheer greed in hopes that "Gambler's Fallacy" theories sound appealing to the ignorant
(but then, of course, it's not by mistake...)


Regards,
KFS

[TwoCatSam]

KFS

Thank you for the discourse.  You are a gentleman and a scholar!

TCS

[RICHIECHIPS]

TWOCATSAM
Since you seem to have an open mind may I add to what I was attempting to say earlier......

The suggestion I made earlier about playing 10 streets by skipping the last 2 streets that came out and if you win a spin then you still skip the last 2 streets ....However, I haven't played this system for years and may have misspoke...I think it was Kon-Fu-Sed who said that he was confused and rightfully so.  What I should have said was after a win skip the last 3 streets and bet 9 streets.  Win or lose next spin bet 10 again skipping the last 2 streets that came out.  Sorry for the error...

Assuming 2 chips per street and a win on 10 streets and a win on 9 streets, your nets consecutively are 4 & 6 for a total of +10 chips in 2 spins and while covering 81% , you can often do it again & than you made 20 chips and I walk.

In addition, I see someone above tested that simple comment of mine for thousands of spins.  Allow me to be the first to admit I personally do not know of any system that you can just plug into a computer and it keeps winning.  I hope I did not give that impression.  This is not to say someone out there doesn't have one....I just don't, nor am I wasting my time looking for it. 

I am a firm believer one must bring their intelligence to the table.   The suggestion I was trying to make was simply to try 10 streets.  And using his own system he might find 10 streets better.  I really wasn't trying to explain an entire system.....The person I was talking to was using 4 streets and since I have had great luck with 10 streets that's all I was pointing him towards and to let him develop his own from there. 

However, it seems I opened Pandora's Box because a simple suggestion for him to build on was examined like it was a full blown system.  IT WAS NOT....It was a very small piece.   Remember. it is not a mechanical system that you can just plug into a computer.  I personally stopped looking for the "Holy Grail" many years ago and I just became satisfied to set goals + & - win a little and walk, sometimes lose & walk.  As long as there are more wins I'm satisfied. 

To further explain more about the 10/9 Street system above so you don't think I am a complete idiot, if I may, it is at its best as a hit and run system...walk with 30ish % between 4 & 20 spins.   Buy in for 60 chips walk with 78 to 80 chips.  Of course you can double everything.  But let's just stick with an opening bet of 2 chips on 10 Streets.   You need to choose your table carefully....only 1 or 2 repeat streets out of the last 16+.  The charting board is helpful and believe it or not with a little experience you can spot a good table from the board in seconds.   

On losses you immediately double & revert to 10 Streets again skipping the last 2 that hit, go for 1 spin on 10 Streets and if you won drop to 9 streets on the second spin.  This returns 20 of the 18 or 20 you lost.  Since 30 numbers covers 81+% of the table you should win 3 out of 4 spins so in 2  spins at double your net return is the 20.  Depending on your loss you could be 2 ahead.   You could also stick with 10 streets for 2 spins and net 16 chips.  This would depend on the table.  Now the odds leave you 1 more spin.  But covering between 27 & 30 #'s out of 37 should allow for a few hits right off.  All you need is 4 hits and you hit your win walk goal.  And if you lost during, you now know how to recover quickly with odds 81+% in your favor.

Betting to the Odds / Law of Averages actually gives you an additional edge.   I know what I just said will be faced with much disagreement.  Many people may get excited.  Isn't that good?  To open up dialogue so we can listen and learn!  I'm not suggesting I can teach but I know I'm about to receive plenty of criticism.  Maybe I will learn.  I travel a lot so don't expect too quick a response.

Now to try to explain a very controversial position I took earlier let me continue.  But first please let me say that what I'm about to say is truly meant to be helpful and possibly open up new directions for many of you who are a hell of a lot smarter than I.   PLEASE DON'T KILL THE MESSENGER.   Besides I would really try what I'm attempting to lay out here or you may eat your words.

Let's say you are betting 30 numbers.  Let's say you plan to bet the same numbers until one of your numbers hit.  The odds of hitting one or your 30 numbers on a single zero table on the first three (3) spins is (take my word for it).  Remember we haven't spun yet. 
1...81.01% on the first spin.
2...96.42% on the second spin.
3...99.32% on the third spin.
Remember these are the same 30 numbers.

The full blown math for the above is in a previous memo up top.

What Kon–Fu-Sed  said was absolutely right again, when he said that this is before you spin at all.

However, here is where I most respectfully differ with almost every Roulette player.  However, there a few real pros who have got it.  The principle is based on "Bayes Theorem".    Bayes Theorem is designed into your computers in an attempt to learn. 
 
Using the 10/9 system as an example that after a loss you revert back to a 10 street bet at double.  Remember the odds are high because you are covering the same 30 numbers in this case for all 3 spins. So before you spin your odds are 99.32% to hit one of your numbers.  I don't know about you but I really, really like those odds already.   

Now the popular theory is that after your first spin the odds change and revert to the same 81.01% chance.  And after the 2nd spin the odds change again and you only have 81.01% chance of hitting and the same by the 3rd spin.  Well I respectfully don't agree.  But give me a chance to explain.  I know, I know you don't agree...So what's new?  Well it is my humble opinion that the odds are the odds, are the odds, are the odds, are the odds, and they never change.  What made the odds change from 81.01% back to 81.01% after you spun once?  You are thinking in individual spins and you are right each spin is independent of each other but not independent as a group of spins.  The odds do not change just because you spun once or twice.  Maybe for the number you hit but not the other 29.  The odds of this same exact 30 number group hitting once in 3 spins is 99.32% and that never changes.   What changes is the anomalies or your luck or trends or whatever you prefer to call it.  Sometimes the odds will follow the odds and sometimes they don't but the odds never change.  You are betting a large group in a small fixed sample, a wheel with 37 numbers on it, not infinity.  It is my opinion that a whole new world of possibilities opens up to you if you just believe in playing to the "Odds" or "The Law of Averages".   If people are ever going to beat Roulette the best way to do it is play to the odds.  If you don't believe in playing to the odds I wouldn't play.  I'm not suggesting 1000% strictly...bring you intelligence and gut feelings to the table also.   Playing to the odds is simply playing to the best bet.  I think that makes sense.  Let me explain why many maybe all of you don't believe me and I don't blame you.  First you believe the most popular of the beliefs maybe because of what you read.  "Every spin is independent".  They are to each other but not taken in a group of spins.  Sure play 1 number and it could take 500 spins to hit.....that's the anomaly not the odds.  The odds are 1 in 37 chances.  That's  a 2.07% chance of hitting.  The house is stacked against you.  However, let's increase the #'s covered and look at this again.  Let's cover 36 out of 37 numbers.  Did your odds change?  Did they get better?  What is your chances of hitting one of your 36 numbers in the next 36 spins.  Let's cover 1 thru 36 and spin 36 times do you think the single zero will come out 36 times in a row?  Do you think you have good odds to hit one of your 36 numbers within 36 spins?  Or do all of your 36 numbers only have 1 chance in 37 and the odds are not any better if the 0 came out 30 times in a row.   No, the odds are better the more you cover and the more chances you give the entire group to hit when designing a system. 

The problem that blocks this fact from catching on is players look at it from playing only 1 number or the even bets like Red or Black.  The problem here is that there are so many house numbers or colored pockets that sometimes the house numbers or colors defy the odds.  An anomaly takes place.   It is much easier for 1 number to defy the odds & throw off anomalies and cloud the facts like miss for 500 spins than it is for 36 numbers to defy the odds.   The odds are what I'm talking about and this is what will open up new systems that work well enough to keep you ahead.  Remember not for a 1000 spins in a row but pick the right table and hit it with a solid plan...and win walk with 30%.  Get better or get beaten.

[Kon-Fu-Sed]

Richiechips,

Will you never give it up?
If I could understand why you are saying those things that just are not true?

"Remember we haven't spun yet."   <--- Your figures below are correct in THIS EXACT CONTEXT. But not your words...

"1...81.01% on the first spin."   <--- This is correct - absolutely 100% correct... if you had written 81.08% but OK)

"2...96.42% on the second spin."   <--- This is WRONG! You have a TOTAL of 96.42% chance to hit on FIRST, SECOND OR BOTH SPINS!
That is VERY different from ON the second spin

"3...99.32% on the third spin."   <--- This is WRONG! You have a TOTAL of 99.32% chance to hit on FIRST, SECOND, THIRD, ON ANY TWO OR ALL THREE SPINS!
That is VERY different from ON the third spin.

You reference your own math above and so do I: Add all figures except the one for LOSING ALL three spins and get a TOTAL of 99.32. It includes every other combination except losing all three. It's the TOTAL for ALL three spins - NOT just "on the third".

And you write:
"The odds of this same exact 30 number group hitting once in 3 spins is 99.32% and that never changes."
NO! That's also not correct. (You are correct that the odds never changes...)
The chance of a 30-numbers group hitting ONCE, TWICE or ALL THREE TIMES is 99.32% - Not just once, as you say.

The chance for it to hit once in three spins is only 8.71% as it's very hard to avoid hitting more than once when you cover 30 numbers. (And now I'm talking about a three-spins group - as you like it - NOT three independent spins.)

The chance to hit more than only once is 90.62%, actually. And to lose all three is 0.67% chance.
(The three possibilties added together = 100% as it should be... p = .0871 + .9062 + .0067 = 1.0 Correct)


Why do you twist the numbers like this?
If I have a 99.32% chance to win ON the third spin, like you say, why should I even bother betting the two first that have much worse odds?


Please read the post I directed to "TCS and all interested" above, as it seems like you are.
KFS


Edit:
Maybe a better (and absolutely true) table would be:
1. You have a 18.92% chance to lose one spin. Always.
2. You have a 3.58% chance to lose two spins if you play them all
3. You have a 0.67% chance to lose three spins if you play them all
4. You have a 0.13% chance to lose four spins if you play them all
5. You have a 0.02% chance to lose five spins if you play them all
6. You have a 0.004% chance to lose six spins if you play them all

No matter what has happened before the first spin.
And no matter which 30 numbers you select or when (changing them mid-way doesn't matter), the above holds true all the way.

[Kon-Fu-Sed]

Richiechips,

Maybe you (and others who don't understand what I say) understand if I explain it this way:

We have three spins IN FRONT OF US where we bet 30 numbers.
The chance to win one single spin is 30/37 or 81.08% - we agree on that, I believe.

To calculate the % for a Win-Win combination we use ".8108 * .8108" which gives .6574 (rounded) = 65.74%
To calculate the % for a Win-Lose / Lose-Win combination we use ".8108 * .1892" which gives .1534 (rounded) = 15.34%
To calculate the % for a Lose-Lose combination we use ".1892 * .1892" which gives .0358 (rounded) = 3.58%
And for three-spins combinations we just multiply once more, by the Win- or Lose-per centage.
According to your figures, you do exactly this.


So then. The following combinations are possible - in a three spins set IN THE FUTURE - and their %:

W-W-W = 53.3% chance to happen
W-W-L = 12.44% chance to happen
W-L-W = 12.44% chance to happen
L-W-W = 12.44% chance to happen
W-L-L = 2.9% chance to happen
L-W-L = 2.9% chance to happen
L-L-W = 2.9% chance to happen
L-L-L = 0.68% chance to happen
------------------------------
Total = 100% chance to happen (no more combinations are possible) Correct.


Now let's look at the chance to win on the FIRST spin BEFORE any has been spun...
The following combinations win on the first spin:
W-W-W = 53.3%
W-W-L = 12.44%
W-L-W = 12.44%
W-L-L = 2.9%
Add the % and you get 81.08% and that's your figure as well, I believe.
If you add the % for the other four, you get 18.92%. Reality check: 81.08% + 18.92% = 100% = Correct!
And we both agree to this!


Now, let's look at the chance to win on the SECOND spin (your own words: "2...96.42% on the second spin") BEFORE any has been spun...
The following combinations win on the second spin:
W-W-W = 53.3%
W-W-L = 12.44%
L-W-W = 12.44%
L-W-L = 2.9%
Add the % and you get 81.08%...?
If you add the % for the other four, you get 18.92%. Reality check: 81.08% + 18.92% = 100% = Correct!
...96.42%, you said?


Now, let's look at the chance to win on the THIRD spin (your own words: "3...99.32% on the third spin") BEFORE any has been spun...
The following combinations win on the third spin:
W-W-W = 53.3%
W-L-W = 12.44%
L-W-W = 12.44%
L-L-W = 2.9%
Add the % and you get 81.08%...?
If you add the % for the other four, you get 18.92%. Reality check: 81.08% + 18.92% = 100% = Correct!

Question: Where did the 99.32% go???
Answer: It never existed ON the third spin.
It is the sum of all chances to win something in three spins if you actually bet on ALL THREE SPINS.


If you do not agree with my 81.08% chance on the third spin above, please show how you calculate. In the calculations you made above ("this proves that...") you also add combinations that do NOT win on the third spin so they DO NOT show the chance to win on the third spin.


And I repeat my question above:
If I have a 99.32% chance to win ON the third spin, like you say, why should I even bother betting the two first that have much worse odds?


Regards,
KFS

[TwoCatSam]

Richiechips

I certainly would not kill the messenger, nor would I ignore him.  Had I not seen some logic in your original post, I would never have reopened this topic.

There is a writer named Anthony Svoboda who posts a system using your math.  Well, several in fact.  One of his has a 99.32% chance of winning while betting ten streets.  He always considers the three bets an "event".  While he does break it down into separate bets, he does it just like you do---always considering all bets and not jumping in on the last one.  That won't work.

Long before I heard of you fellows, I studied this paradox.  I do feel it is a paradox, although some don't.  The problem with Svoboda's method is when you lose, it's $500 or so. 

He does not advocate staying after a win, which I like better.  If you would reduce this to rules, I will test it a while.

Was this it:  Bet nine streets and omit the last three that came.  If you lose, bet ten streets omitting the last two that came. If you win, stick until you make 20 chips.  Buy in with 60 chips.

You fellows have disagreed like gentlemen should and I commend you for it. 

Sam

[Kon-Fu-Sed]

May I continue my reasoning from above?
(a phrase out of courtesy - not aimed at you, Sam)


Now say we LOST the first bet. Does THAT change anything if we do bet two more?

We have actually seen the result of the first spin/bet and it was a LOSS.
What combinations are now possible to end our three-spins session? And their %?

First and foremost: What's the % chance for a result we HAVE ABSOLUTELY SEEN? Is there ANY possibility for it to be something ELSE or is it 100% sure?
I'd say it really is 100% sure - the result is still on the marquee so there is no doubt. It is now "statistics" (= 100% certainty).
So for the lost first spin, the % chance for it to be a lost spin is 100% - it can't be a winning spin in even .0001%, can it?

So... after a LOSS (100% sure) on the first spin, these combinations are possible:

L-W-W = 100% (the first, 100% surely lost, spin) * 81.08% * 81.08% = 65.74%
L-W-L = 100% * 81.08% * 18.92% = 15.34%
L-L-W = 100% * 18.92% * 81.08% = 15.34%
L-L-L = 100% * 18.92% * 18.92% = 3.58%

Reality check: 65.74% + 15.34% + 15.34% + 3.58% = 100% = Correct.


Now, let's look at the chance to win on the second spin, BEFORE it has been spun...
The following combinations win on the second spin:
L-W-W = 65.74%
L-W-L = 15.34%
Add the % and you get 81.08%
If you add the other two, you get 18.92%. Reality check: 81.08% + 18.92% = 100% = Correct.
Do you recognise that "81.08%" from somewhere...?


Now, let's look at the chance to win on the third spin, BEFORE the second has been spun...
The following combinations win on the third spin:
L-W-W = 65.74%
L-L-W = 15.34%
Add the % and you get 81.08%
If you add the other two, you get 18.92%. Reality check: 81.08% + 18.92% = 100% = Correct.
Do you recognise that "81.08%" from somewhere...?


And supposing we lose this second spin as well (a 3.58% chance but that's life...)
Now, the only possible combinations, BEFORE the third spin is spun, are:
L-L-W = 100% * 100% (the second, also 100% surely lost, spin) * 81.08% = 81.08%
L-L-L = 100% * 100% * 18.92% = 18.92%

Reality check: 81.08% + 18.92% = 100% = Correct.


And supposing we ALSO LOST THAT THIRD BET... (to lose all three spins is a 0.67% chance)
Now, the only possible combination is:
L-L-L = 100%
And it's calculated like "100% * 100% * 100% (the third, darn 100% surely lost, spin) = 100%"

As there is only one single combination left it HAS to be 100%.


These are the hard and true facts:
1. You have a 18.92% chance to lose one spin. Always.
2. You have a 3.58% chance to lose two spins if you play them all
3. You have a 0.67% chance to lose three spins if you play them all
4. You have a 0.13% chance to lose four spins if you play them all
5. You have a 0.02% chance to lose five spins if you play them all
6. You have a 0.004% chance to lose six spins if you play them all


So Richiechips, I wouldn't even bet that third bet as - according to your logics and the way you use figures and words - I will have a whopping 99.99984% chance to win on the 8th spin. 
What's 99.32% in comparison...?


Regards,
KFS


Nawh... "Forum Mathematics": It's not a 99.99984% chance to win ON the 8th spin - it's a 0.00016% chance to LOSE ALL EIGHT - quite a difference.

[RICHIECHIPS]

Hi KFS
I totally understand your position and respect it and it is the most popular understanding.  Please allow me to have an opinion of my own.  I didn't expect there would be many takers.  I only hoped some would try it out.  As they say the proof is in the pudding.

Allow me to give you my live casino experience with this theory.  I can either be having a very bad day or if I get greedy and want the challenge of drafting all my color to my side of the table after having played the system for hours and hours at a time, same table, same buyin.  Now I do not recommend it unless you can afford a total loss and unless you are very experienced in this specific system.  However, this is the true test.  I have often sat for hours without going back into my pocket and almost never experiencing 3 losses in a row.  Maybe 1 or at the very most 2 in hours of play.  And I really can't remember 2.  (Then again I'm getting old)  Putting math aside this speaks volumes for the odds / probability of 1 of our 30 #'s coming out within 3 spins.  That is why I suggested people try it rather than analyze it.  I'm sure you know systems open up to us the longer we play them.  I know you also know that we can be fooled by looks.  I always tell people that for hundreds of years better people then we have looked for a solid winning system.  Maybe it is time to think out of the so called "BOX".  I know this is controversial.    I also realize we all have our favorite methods / systems and I am the first to admit I have a pretty closed mind when looking at someone else's system.   And I know better but I can't help it. 
Quick story
Found a great system (so I thought) on high speed computer and played it for weekly for almost a year in a live casino.  It never won.  So I shelved it....came back to it a year later and could do no wrong.  This should teach us a lesson.  Gambling is a gamble. 
Back on point;
Stop and really think about this next sentence.... "If each spin is independent then how do things even out over time?" There either is a Law of Averages or there is no such thing.  A numbers ability to appear on average 1 in 37 / 38 spins requires that its appearance either speed up or slow down over time as all other numbers thru each numbers natural statistical pressure also influence the average distribution of all numbers. There is dependence. The dependence is on an unbiased wheel, and each number and time. The fact that we know that #'s appear on average 1 in 37/38 spins is something to build a system on. It's also called the "ODDS".

All numbers will even out over time in an unbiased atmosphere with an unbiased wheel.  If the wheel was biased things would not even out.  On the other hand we do not think what it's called is as important as what it produces.  What is more important is that an unbiased wheel, through the "Law of Averages", its numbers are influenced to even out over time.  This only proves in my humble opinion that all numbers are dependent on each other.

There are a very few people who agree the wheel has any kind dependence.    However, we have carried it a step further and this new school of thought is that YOU CAN ASSIGN WEIGHT TO PAST OUTCOMES TO PREDICT TO SOME DEGREE OF ACCURACY THE OUTCOME OF FUTURE SPINS.  Something very, very similar to card counting without the cards.  (Think about that.) As long as there is a Law of Averages than it simular to card counting.   What hasen't come out is due.  We don't know exactly when but the past helps.  We are not talking about being telepathic.  We are talking about playing to the odds.   The same as card counting.  Giving 6 decks of cards a good mix is no difference than a random wheel and the LOA.  An important fact and despite what so called experts say, you actually will find that in roulette, previous numbers (both the ones that have hit and especially the ones that have not) have an important bearing on what is going to happen on future spins. 

The self professed professionals, "ad nausea", carry this independence a step further and continue to state as a fact that a winning system is not a winning system if it cannot hold up to and profit by thousands of spins.  (Computer or otherwise).   I'm sorry but to me winning systems do not have to withstand a million spins and rake in 2 million dollars to be a winning system.  If you look at Systems that way you will never ever win.  However, change your expectations of their outcome and your opinion of systems will change.  There is nothing like walking up to a table, playing for 6 to 15 spins and walk away with 15 or 20 chips to change ones attitude about systems.  20 $1chips = $20, 20 $5 chips = $100, and 20 $25 chips = $500.  What is that wood? 

HOWEVER, NO STRATEGY SURVIVES COMPLETELY ITS CONTACT WITH A WHEEL.   EVERY TABLE IS TOTALLY INDEPENDENT AND EVERY TABLE ACTS ANY WAY IT WANTS. THEREFORE NO MECHANICALLY SET STRATEGY WE KNOW OF WILL WORK ALL THE TIME. What we often totally neglect and ignore is to take into account our own ability to make conscious decisions during play which can enhance our ability to win.  We simply watch the table.   

In addition, if you keep track of the outcomes, that information is helpful on your decisions in what numbers or groups of numbers to play next or more importantly not play next.  When you skip betting locations you not only save money but your net profit is higher also.  A human being playing roulette can use his or her intelligence & memory to observe and chart a "connection" between past and future spins that can be used to create a system that is applicable on roulette wheels to predict the likely hood of the next outcome.  Not the next # but possibly the next group.  An extended famine of high numbers 24 to 36 (for example) provides a solid betting opportunity to cover those numbers after that long absence, provided the wheel is not biased.  I know you don't agree with that either. That is ok.  I'm only giving my opinion.  To most it is worth nothing but I hope to a few it will lead to better things. 

At this time, we feel neither casinos nor most gamblers realize that there is a discernible, weighted, mathematical relationship between the numbers that have hit and the numbers that will hit in the future. It is called the "Law of Averages".  They do not know how to rely upon the mechanical perfection of the device (wheel) or to be able to anticipate with some degree of success balanced statistics.  R.D. Ellison in his book, says it better than us, "... events with probabilities will move toward parity with the others.... There is an ongoing statistical pressure on every playable number to appear within a given time frame. As events accumulate you will see the Law of Averaging inclination to even things out."

This educated guessing is the art of success and the better you master it the more you win. Without this talent you will have to depend strictly on the mechanics of your system and you will need luck & have to hit and run at best.  It's like expecting the expected, while being prepared for the unexpected which sometimes happens? PREDICTION WITH DECENT ACCURACY is a key to allowing you to stay at a table with a good system or two and continue to win, win, win, and win.  Because of our specific choices which come from experience and knowledge our brain will tell us the choices to make and this when added to all the Bla, Bla, Bla is what can keep us ahead of the casino. Yes, you can continue to win without having to hit and run.  If you are looking for your "Holy Grail", the key to the "Holy Grail" we beleive is hidden in one's ability to concentrate and successfully predict with DECENT ENOUGH SUCCESS what is due or not due. 

We penned it "The Limited Law of Likely Events", (LLLE) statistical pressure, weighted numbers, the odds and much better known as "The Law of Averages."(LOA)  Webster defines the LOA as "the common sense observation that probability influences everyday life so that over the long term the possible outcomes of a repeated event occur with specific frequencies" the key here is a repeated event will occur with specific frequency.  Keep in mind that here we have a very, very small target.....the wheel.  It could be worse it could be infinity.

KEEP IN MIND THAT IT ONLY WORKS BECAUSE WE ARE DEALING WITH A SPECIFIC GROUP OF NUMBERS AND NOT INFINITY.  THE SMALLER THE ENTIRE SAMPLE GROUP THE SOONER THINGS WILL EVEN OUT AND THE EASIER IT WILL BE TO PREDICT.  There goes that word again.  Law of Averages isn't an exact science.  Law of Probability is a probability not a guarantee.  SOMETIMES THE WHEEL WILL RESPOND IN TOTAL DEFIANCE OF MATHEMATICAL PROBABILITIES.  CHALK IT OFF TO JUST ONE OF THOSE DAYS AND LEAVE.  DON'T STAY AND PUSH IT.  YOU'LL LOSE, WE HAVE, MANY TIMES.  THIS IS PROBABLY THE BEST ADVICE. 

Back in the 1700's there was a Reverend by the name of Thomas Bayes (1702–1761).  He developed many theories' many of which are in use today.  In fact Microsoft uses them in our computers to make intelligent decisions because some are based on odds of one answer being right over others.  Believe it or not here we are in the 21st century and they are just starting to recognize all the aspects of his most important theorem.  Bayes' theorem is a result in probability theory, which relates the conditional and marginal probability distributions of random variables. In some interpretations of probability, Bayes' theorem tells how to update or revise beliefs in light of new evidence.  In other words watch what comes out and make decisions based on this new information.

I know I have convinced no one but in spite of not believing do yourself a favor "Think out of the Box".  It's more than likely, like I, you are always looking for a better way.  The only one we won't see in this Great Forum is that guy who has that mechanical system that keeps winning over 1,000,000 spins.  He's in a live casino somewhere.
Color in and the best to you
RICHIECHIPS

[Kon-Fu-Sed]

Dear Richiechips,

So many words. Unfortunately I had to stop reading here:
"Please allow me to have an opinion of my own."

OPINION?


I didn't know there were different opinions on multiplication and division... 2 * 3 = 6. That's not an opinion.

What do you mean?
Does my calculator express its opinon when I enter "100 * 30 / 37" and it responds "81.081081081"
It should come with a sticker then: "Other calculators may have different opinions and give different answers"...
Maybe I shall send it to you, so you can persuade it to share your opinion?
What is YOUR calculator's opinion on "100 * 30 / 37"? Does it differ from my calculator's opinion?
If not; haven't you told it your opinion that it differs from the first to the third spin?


The plain and simple truth is, my friend: You make all the correct calculations, you just make the wrong conclusions.
That's all.


Opinion? ? ? On multiplication? ? ? Give me a break!

Now I'm really
Kon-Fu-Sed

[RICHIECHIPS]

Sorry it was too long for you to read...I will take the easy out also. 
The Best to you & Stay
kon-fu-sed

[RICHIECHIPS]

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