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How to make a roulette progression…..
Posted: 31 May 2012 03:30 PM   [ Ignore ]   [ # 16 ]
RICHIECHIPS
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PROGRESSION BETTING
Progression betting is relative…it can be good and can be very bad….it all depends on your strategy….Let’s look at bad first…you cover 1 or 2 numbers and lose so you increase or double each bet trying to catch up on losses….very bad   it could take a few dozen spins before your 1 or 2 numbers hit.  You couldn’t afford to keep doubling until you hit.

Now let’s look at very good example of progressive betting…Let’s say you cover 30 numbers on a European Table …. covering 30#‘s has an 81.08% chance on the 1st spin to win. 

However, if you lose and you double your bets on the same 30 numbers….this progressive bet is not bad at all …..if you figure the math you have a 96.42% chance by the 2nd spin to win. 

Ok let’s say you ran over a black cat on the way to the casino and lost the second spin and you double again and bet all the same 30 #‘s for a third time. The odds are extremely high as long as you keep betting the same 30 numbers.  Covering the same 30 Numbers and doubling your bets to catch up is not a bad bet at all.  You now have a 99.32% chance by the 3rd spin to win.  I’ll take those odd any day.  So progressive betting is both good and bad and as you can see it simply depends on your betting strategy.

There will be those who will say “Yes but you can lose on the third spin because roulette is random or the wheel is independent or some other hypothetical comment.  Yes you can lose on the third spin but the odds are the odds or as has been said so many times on this forum the math is the math.

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Posted: 01 June 2012 05:40 AM   [ Ignore ]   [ # 17 ]
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RICHIECHIPS - 31 May 2012 03:30 PM

PROGRESSION BETTING
Progression betting is relative…it can be good and can be very bad….it all depends on your strategy….Let’s look at bad first…you cover 1 or 2 numbers and lose so you increase or double each bet trying to catch up on losses….very bad   it could take a few dozen spins before your 1 or 2 numbers hit.  You couldn’t afford to keep doubling until you hit.

Now let’s look at very good example of progressive betting…Let’s say you cover 30 numbers on a European Table …. covering 30#‘s has an 81.08% chance on the 1st spin to win. 

However, if you lose and you double your bets on the same 30 numbers….this progressive bet is not bad at all …..if you figure the math you have a 96.42% chance by the 2nd spin to win. 

Ok let’s say you ran over a black cat on the way to the casino and lost the second spin and you double again and bet all the same 30 #‘s for a third time. The odds are extremely high as long as you keep betting the same 30 numbers.  Covering the same 30 Numbers and doubling your bets to catch up is not a bad bet at all.  You now have a 99.32% chance by the 3rd spin to win.  I’ll take those odd any day.  So progressive betting is both good and bad and as you can see it simply depends on your betting strategy.

There will be those who will say “Yes but you can lose on the third spin because roulette is random or the wheel is independent or some other hypothetical comment.  Yes you can lose on the third spin but the odds are the odds or as has been said so many times on this forum the math is the math.

Every spin is a new event. Every number has the same chance of falling every spin.

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Posted: 01 June 2012 09:48 AM   [ Ignore ]   [ # 18 ]
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Every spin is a new event and every number has the same chance of failing every spin but that doesn’t change the odds / math above…betting the same 30 numbers on a 37 # roulette wheel ...on the 3rd spin you have a 99.32% chance of being successful.  A roulette wheel is random but not as independent as one might think.

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Posted: 12 June 2012 04:27 PM   [ Ignore ]   [ # 19 ]
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RICHIECHIPS - 31 May 2012 03:30 PM

Covering the same 30 Numbers and doubling your bets to catch up is not a bad bet at all.  You now have a 99.32% chance by the 3rd spin to win.

RICHIECHIPS - 01 June 2012 09:48 AM

...betting the same 30 numbers on a 37 # roulette wheel ...on the 3rd spin you have a 99.32% chance of being successful.

These two statements say different things!  To say you have a 99.32% chance of winning by the third spin means that you could win on the first spin, or the second spin if you lose on the first, or the third spin if you lose on the first two.  To say you have a 99.32% chance of winning on the third spin means that your chances of winning that spin depends on how you did on the first two spins.  By what mechanism do the results of spins 1 and 2 influence your chances on spin 3?  The answer, of course, is that they do not; betting on 30 of 37 numbers you have a 81.08% chance of winning on every spin regardless of what happened on previous spins, and regardless of whether or not you bet on any or all of those previous spins.

That said, a 99.32% chance of winning is pretty good—if all you are trying to do is win enough to pay for dinner that evening.  The problem is that 0.68% chance of losing, because when that happens you are out 210 units.

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Posted: 12 June 2012 05:52 PM   [ Ignore ]   [ # 20 ]
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You are correct that each spin independantly has 81.08% chance of hitting.  Pertty good as is.  However, what I suggest you consider are the odds of winning on one of your 30 #‘s by the third spin.  That is very different.  If you go up to a table and plan on playing the same 30 numbers for 3 spins, the odds of hitting one of your same 30 #‘s by the third spin is 99.32%.  I could show you the math but it takes up a lot of room.  This does not assume we will definitely win but your chance mathematically to hit one of your numbers by the 3rd spin is 99.32% as long as you plan and bet the same numbers for 3 spins.

I also don’t want you to assume this is a martingale or any other system….it is not, it is simple mathmatics….you are not out 210 chips as you suggest above….You could, for example, cover 30 #‘s with 5 chips placed on for example 5 line bets.  After 3 spins assuming 3 losses you are out 15 chips not 260.  I am not suggesting you play line bets but this was an example. 

When a player understands the importance of this dissimilarity new opportunities will open up to them.

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Posted: 13 June 2012 12:11 PM   [ Ignore ]   [ # 21 ]
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Use progression with care. The casinos special online make a lot of money, due to the habit of aggressive progressions.
Lets kill a myth; THE CASIONOS HAVE TABLE LIMIT TO PROTECT FROM MARTINGALE! Nothing is true here.

At the time you buy chipos for thousends, the casino got the money.  Every time you double, you do it with paid chips (WELCOME)

It you end up at a sucessful martingale and had 1024 chips on the table, the casino had little more than 50% chans to loose 1 CHIPS
and you the player a BANKROLL.

The limit is for they must have cash, if on some in special situation sombody PARLAY 50 times, and got it right, the casino must pay millions for a sold chip,
Thats may friend is another issue.

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Posted: 13 June 2012 01:04 PM   [ Ignore ]   [ # 22 ]
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RICHIECHIPS - 12 June 2012 05:52 PM

You are correct that each spin independantly has 81.08% chance of hitting.  Pertty good as is.  However, what I suggest you consider are the odds of winning on one of your 30 #‘s by the third spin.  That is very different.  If you go up to a table and plan on playing the same 30 numbers for 3 spins, the odds of hitting one of your same 30 #‘s by the third spin is 99.32%.  I could show you the math but it takes up a lot of room.

Actually the math is quite simple.  In order not to win a bet on 30 numbers within three spins you have to lose all three spins.  The probability of losing such a bet on any given spin is 7/37, so the probability of losing three spins in a row is (7/37)x(7/37)x(7/37).  The probability of not losing three spins in a row is therefore 1-((7/37)x(7/37)x(7/37)), or .9932, or 99.32%.

RICHIECHIPS - 12 June 2012 05:52 PM

This does not assume we will definitely win but your chance mathematically to hit one of your numbers by the 3rd spin is 99.32% as long as you plan and bet the same numbers for 3 spins.

Chosing the same 30 numbers is not a requirement.  You always have 7 ways to lose out of 37 possible results; hence you always have a 99.32% chance of winning within three spins if you chose 30 numbers.

RICHIECHIPS - 12 June 2012 05:52 PM

I also don’t want you to assume this is a martingale or any other system….it is not, it is simple mathmatics….you are not out 210 chips as you suggest above….You could, for example, cover 30 #‘s with 5 chips placed on for example 5 line bets.  After 3 spins assuming 3 losses you are out 15 chips not 260.  I am not suggesting you play line bets but this was an example.

To quote from your post on 31 May 2012 at 02:30 PM, “Now let’s look at very good example of progressive betting…Let’s say you cover 30 numbers on a European Table …. covering 30#‘s has an 81.08% chance on the 1st spin to win.

“However, if you lose and you double your bets on the same 30 numbers….this progressive bet is not bad at all …..if you figure the math you have a 96.42% chance by the 2nd spin to win.”  This seems to state explicitly that you were considering a Martingale progression.  You will also note that I did not say losing three spins in a row would cost 210 chips; I said it would cost you 210 units.  Regardless of how you divide your chips, if you are in effect betting the same amount on each number then you are betting one unit on each number.  A unit can be equal to, greater than, or less than one chip.  In your example, 5 chips on 5 lines, each chip is covering 6 numbers, so a unit would be 5/6 of a chip.

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Posted: 13 June 2012 02:43 PM   [ Ignore ]   [ # 23 ]
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To Midnight ....Thank you .....your comment below is the point I was trying to make and now at least we agree with the 99.32%.  I admit the way you proved my point mathematically was much shorter than I would have demonstrated it.  But we both got to the same answer.  We, I think agree, the 3rd bet has 1-((7/37)x(7/37)x(7/37)), or .9932, or 99.32%.  I don’t have to lose the 3rd spin to have these odds.
“Actually the math is quite simple.  In order not to win a bet on 30 numbers within three spins you have to lose all three spins.  The probability of losing such a bet on any given spin is 7/37, so the probability of losing three spins in a row is (7/37)x(7/37)x(7/37).  The probability of not losing three spins in a row is therefore 1-((7/37)x(7/37)x(7/37)), or .9932, or 99.32%.”

With all due respect I do not agree with this statement below:
“Chosing the same 30 numbers is not a requirement.  You always have 7 ways to lose out of 37 possible results; hence you always have a 99.32% chance of winning within three spins if you chose 30 numbers.”

Mathematically you are correct however once you cahnge your numbers you change your bet and once you change your bet you are not playing the same 30 numbers & therefore change you odds of hitting your number.  In order to have a 99.32% you have to pick and stick to the same 30 numbers for 3 spins.

In referance to your comments and my comments below
“To quote from your post on 31 May 2012 at 02:30 PM, “Now let’s look at very good example of progressive betting…Let’s say you cover 30 numbers on a European Table …. covering 30#‘s has an 81.08% chance on the 1st spin to win.

“However, if you lose and you double your bets on the same 30 numbers….this progressive bet is not bad at all …..if you figure the math you have a 96.42% chance by the 2nd spin to win.”  This seems to state explicitly that you were considering a Martingale progression.  You will also note that I did not say losing three spins in a row would cost 210 chips; I said it would cost you 210 units.  Regardless of how you divide your chips, if you are in effect betting the same amount on each number then you are betting one unit on each number.  A unit can be equal to, greater than, or less than one chip.  In your example, 5 chips on 5 lines, each chip is covering 6 numbers, so a unit would be 5/6 of a chip.”


In response to the above comments & again with due respect… you are not incorrect when you say I was implying a martingale type of progression bet.  That is true, however, in that instance were were talking about progressive betting and I was trying to point out when a progressive bet isn’t a curse.  Certainly I would never playing martingale on a color but playing Martingale on 30 pre planned numbers has a 99.32% chance.  This a much much better martingale progression than playing a color.  That was the simple point I was trying to make.
However, when we were discussing the 30 number concept having 99.32% I was simply talking math and I was trying to explain that in this case the 99.32% has nothing to do with any system, martingale or otherwise.  As far as chips or units allow me to clarify my point.  If you bet 5 chips ...1 on each of 5 line bets and lost and bet 5 chips and lost and bet 5 chips and lost you lost 15 chips.  I didn’t want anyone to thing you lost 210 chips.  This was my example for explaining the odds of 3 flat -30 chips bets.  Now if we are talking progression then I have no problem betting the martingale for 3 spins or 4 spins on the same 30 numbers to catch up when my odds of hitting is 99.32%.  I personally like those those odds.  I am not recommending this as system only explaining this specific progressive bet that even with using the martingale concept this has a much better chance then playing a color.  The point was to let people know some progressive bets are much better than others.  We shouldn’t throw the baby out with the bath water.  I am not a fan of progressive betting but certain bets with certain favorable odds I would consider doubling my bet.  Gambling with a 99.32% odds is a risk I would consider.  As you can see it is very difficult to explain a theory through a froum concept and I think when all is said and done I think we agree.

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Posted: 13 June 2012 05:59 PM   [ Ignore ]   [ # 24 ]
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RICHIECHIPS - 13 June 2012 02:43 PM

We, I think agree, the 3rd bet has 1-((7/37)x(7/37)x(7/37)), or .9932, or 99.32%.  I don’t have to lose the 3rd spin to have these odds.

I suspect we may not agree.  What we do definitely agree on is that we have a 99.32% chance of winning at least one of the next three spins.  This is not the same as saying that, having lost the first two spins, we now have a 99.32% chance of winning the third.  Why is it different?  Because our formula is no longer 1-((7/37)x(7/37)x(7/37)).  We know we lost spins 1 and 2, so the probability that we lost those spins is 1, not 7/37.  Consequently, the formula for winning at least one of the next three spins, given that we lost spins 1 and 2, is 1-((1)x(1)x(7/37)) = 81.08%.  (Now where have I seen that number before? oh oh )

RICHIECHIPS - 13 June 2012 02:43 PM

With all due respect I do not agree with this statement below:
“Chosing the same 30 numbers is not a requirement.  You always have 7 ways to lose out of 37 possible results; hence you always have a 99.32% chance of winning within three spins if you chose 30 numbers.”

Mathematically you are correct however once you cahnge your numbers you change your bet and once you change your bet you are not playing the same 30 numbers & therefore change you odds of hitting your number.  In order to have a 99.32% you have to pick and stick to the same 30 numbers for 3 spins.

Making all the standard assumptions (random game, independent trials, no dealer signature, unbiased wheel, no benevolent or evil spirits present, etc.) what difference does it make what numbers we choose?  They all have the same chance of showing on each and every spin.  Whichever set of 7 numbers I do not bet on, I have a 7/37 chance of losing.  Substituting one such set for another therefore does not change my probability of success at all.

RICHIECHIPS - 13 June 2012 02:43 PM

In referance to your comments and my comments below
“To quote from your post on 31 May 2012 at 02:30 PM, “Now let’s look at very good example of progressive betting…Let’s say you cover 30 numbers on a European Table …. covering 30#‘s has an 81.08% chance on the 1st spin to win.

“However, if you lose and you double your bets on the same 30 numbers….this progressive bet is not bad at all …..if you figure the math you have a 96.42% chance by the 2nd spin to win.”  This seems to state explicitly that you were considering a Martingale progression.  You will also note that I did not say losing three spins in a row would cost 210 chips; I said it would cost you 210 units.  Regardless of how you divide your chips, if you are in effect betting the same amount on each number then you are betting one unit on each number.  A unit can be equal to, greater than, or less than one chip.  In your example, 5 chips on 5 lines, each chip is covering 6 numbers, so a unit would be 5/6 of a chip.”


In response to the above comments & again with due respect… you are not incorrect when you say I was implying a martingale type of progression bet.  That is true, however, in that instance were were talking about progressive betting and I was trying to point out when a progressive bet isn’t a curse.  Certainly I would never playing martingale on a color but playing Martingale on 30 pre planned numbers has a 99.32% chance.  This a much much better martingale progression than playing a color.  That was the simple point I was trying to make.

My mistake. red face Returning to your 5 chips on 5 lines example, you net a win of 1 chip when you win but lose 5 chips when you lose.  Consequently, without some kind of progression it made little sense to me to quote a 99.32% chance of success when only the outcome of the first spin determines whether you are a winner or loser for the series.  I did not realize we were talking about two completely different situations.

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Posted: 13 June 2012 06:15 PM   [ Ignore ]   [ # 25 ]
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Before I respond & because I’m new to this forum I would appreciate if you helped explain to me how you get each persons response in their own boxes.  It makes it much easier
Thanks

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Posted: 14 June 2012 01:00 PM   [ Ignore ]   [ # 26 ]
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RICHIECHIPS - 13 June 2012 06:15 PM

Before I respond & because I’m new to this forum I would appreciate if you helped explain to me how you get each persons response in their own boxes.

This forum recognizes a form of Hyper Text Markup Language, or HTML for short.  This language allows posters to embed instructions to the display module about how to display the actual text of the message.  There are two methods for getting quoted text into its own boxes, and which one to use depends on whether all the quoted text is from only one or from more than one post.

In this case I quoting text from only one post, so all I had to do was use the “Quote” button under the post to which I am responding.  Your text was already inserted into the “Post a New Reply” screen I got, surrounded by the HTML tags indicating it was to be formatted as quoted text.  Had I wished to break up your text into separate boxes in order to respond to you point by point I could copy and paste the appropriate HTML tags, “quote author=...” and “/quote”, to break up your text and allow me to insert my own between the boxes.

Collecting text from multiple posts is a little trickier.  For each post from which you are collecting text you need to use the “Quote” button under it, select all of the quoted text that appears in the “Post a New Reply” screen, use your browser’s Edit->Copy function to get it onto the clipboard, then Paste it into a temporary area.  (Being on Windows I use Notepad.)  Repeat this process for each post you are quoting, appending the clipboard to the temporary area you are using.  When you have all the quoted text in your temporary area you can Edit->Select All and Edit->Copy to get it on your clipboard, then click “Post Reply” and paste into the “Post a New Reply” screen.

I heartily recommend using the “Preview Post” feature when manipulating quotes to any extent.  It saves a lot of editing, assuming you want your post to be readable.

HTH

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Posted: 23 June 2012 09:14 PM   [ Ignore ]   [ # 27 ]
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Hi Midnight Sorry it took so long to get back….

RICHIECHIPS - 13 June 2012 01:43 PM
We, I think agree, the 3rd bet has 1-((7/37)x(7/37)x(7/37)), or .9932, or 99.32%. I don’t have to lose the 3rd spin to have these odds.I suspect we may not agree. What we do definitely agree on is that we have a 99.32% chance of winning at least one of the next three spins. This is not the same as saying that, having lost the first two spins, we now have a 99.32% chance of winning the third. Why is it different? Because our formula is no longer 1-((7/37)x(7/37)x(7/37)). We know we lost spins 1 and 2, so the probability that we lost those spins is 1, not 7/37. Consequently, the formula for winning at least one of the next three spins, given that we lost spins 1 and 2, is 1-((1)x(1)x(7/37)) = 81.08%. (Now where have I seen that number before?  )  Independently your math is correct.  However, as a pre - planned group of betting the same 30 #’s for the next 3 spins your odds of 99.32% never change even if you lost the first 2 spins.  Only as it stands alone is it 81.08%.  RICHIECHIPS - 13 June 2012 01:43 PM
With all due respect I do not agree with this statement below:
“Chosing the same 30 numbers is not a requirement. You always have 7 ways to lose out of 37 possible results; hence you always have a 99.32% chance of winning within three spins if you chose 30 numbers.”  Here again we will have to agree to disagree….once you change even 1 number you have to go back to the beginning and you incur the 81.08%.  [/color[color=red][color=green]]Mathematically you are correct however once you change your numbers you change your bet and once you change your bet you are not playing the same 30 numbers & therefore change your odds of hitting your number.  In order to have a 99.32% you have to pick and stick to the same 30 numbers for 3 spins.[/color]Making all the standard assumptions (random game, independent trials, no dealer signature, unbiased wheel, no benevolent or evil spirits present,  LOL  etc.) what difference does it make what numbers we choose? They all have the same chance of showing on each and every spin. Whichever set of 7 numbers I do not bet on, I have a 7/37 chance of losing. Substituting one such set for another therefore does not change my probability of success at all. Once you change a number you expand the target group and that changes the odds.RICHIECHIPS - 13 June 2012 01:43 PM
In referance to your comments and my comments below
“To quote from your post on 31 May 2012 at 02:30 PM, “Now let’s look at very good example of progressive betting…Let’s say you cover 30 numbers on a European Table …. covering 30#‘s has an 81.08% chance on the 1st spin to win.
“However, if you lose and you double your bets on the same 30 numbers….this progressive bet is not bad at all …..if you figure the math you have a 96.42% chance by the 2nd spin to win.” This seems to state explicitly that you were considering a Martingale progression. [color=red]Yes you are absolutely right but that wasn’t the specific point I was trying to make.  What I was trying to point out to others, not you,  was that some progressive / Martingale type bets can be better than others.  [/color]You will also note that I did not say losing three spins in a row would cost 210 chips; I said it would cost you 210 units. Regardless of how you divide your chips, if you are in effect betting the same amount on each number then you are betting one unit on each number. A unit can be equal to, greater than, or less than one chip. In your example, 5 chips on 5 lines, each chip is covering 6 numbers, so a unit would be 5/6 of a chip.”  The bet that I had in my mind was a simple Martingale type bet.  1 chip of each of 5 line bets.  If I lost I lose 5 chips.  The second bet would be double or 10 Chip loss for an accumulated 15 chip loss after 2nd bet.  Assuming even with a 99.32% chance you still lost on the 3rd bet also you would lose 20 chips on the 3rd spin for an accumulated loss of 35 chips.  Please don’t assume from this that if I lost the first bet I wouls play this martingle for the next 2 spins.  It was just an example of a bet with very good odds of success.
In response to the above comments & again with due respect… you are not incorrect when you say I was implying a martingale type of progression bet. That is true, however, in that instance we were talking about progressive betting and I was trying to point out when a progressive bet isn’t a curse. Certainly I would never playing martingale on a color but playing Martingale on 30 pre-planned numbers has a 99.32% chance. This is a much much better martingale progression than playing a color. That was the simple point I was trying to make.My mistake.  Returning to your 5 chips on 5 lines example, you net a win of 1 chip when you win but lose 5 chips when you lose. Consequently, without some kind of progression it made little sense to me to quote a 99.32% chance of success when only the outcome of the first spin determines whether you are a winner or loser for the series. I did not realize we were talking about two completely different situations. I realize the confusion.  You were right to assume a martingale progression but as I said my point was some are better then others.  Too often, for many, progression type bets are taboo.  They have their place.  The point I was trying to make is there are very risky Martingale bets and much less risky Martingale bets & one should not dismiss doubling if you have great odds in your favor.  I think we beat that one to death….By the way sorry it took so long to answer but got busy at work and had to totally rebuild my computer.  That took forever.

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Posted: 24 June 2012 09:48 AM   [ Ignore ]   [ # 28 ]
The Midnight Skulker
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RICHIECHIPS - 23 June 2012 09:14 PM

... as a pre - planned group of betting the same 30 #’s for the next 3 spins your odds of 99.32% never change even if you lost the first 2 spins.  Only as it stands alone is it 81.08%.

I think I agree with this statement.  Betting on 30 numbers you always have a 99.32% chance of winning one of the next three spins.  After observing Spins #1 and #2 you can still say that you had a 99.32% chance of winning one of Spins #1-#3, but you now have a 100% chance of winning one of Spins #1-#3 if you won either or both of Spins #1-#2, or an 81.08% chance of winning Spins #1-#3 if you lost both of them.  You also have a 99.32% chance of winning at least one of Spins #3-#5.

RICHIECHIPS - 23 June 2012 09:14 PM

With all due respect I do not agree with this statement below:

“Chosing the same 30 numbers is not a requirement. You always have 7 ways to lose out of 37 possible results; hence you always have a 99.32% chance of winning within three spins if you chose 30 numbers.”

Here again we will have to agree to disagree….once you change even 1 number you have to go back to the beginning and you incur the 81.08%.  Mathematically you are correct however once you change your numbers you change your bet and once you change your bet you are not playing the same 30 numbers & therefore change your odds of hitting your number.  In order to have a 99.32% you have to pick and stick to the same 30 numbers for 3 spins.

Well, OK; I can agree to disagree, but I feel compelled to try to make one last attempt to show you that it does not make any difference which 30 numbers you choose.  Suppose instead we are dealing with coin flips, and we want to know the probability of getting two heads in two flips.  Does it make any difference whether or not we use the same coin to make the flips?  Is not the probability of getting two heads in two flips 25% regardless of whether we flip one coin twice or two coins simultaneously (assuming of course both coins are fair and are flipped fairly, and, oh yes, no spirits)?

My point here is that events having the same probability of occurring (flipping heads or spinning a particular number) can be freely substituted for each other when computing the joint probability of multiple occurrences.  Expressed symbolically, let the probability of Event A occurring be p(A), and the probability of Event B occurring be p(B).  The probability that Events A and B will both occur is then p(A) x p(B).  Your contention is that Event B is actually the same as Event A (i.e. we are betting on the same 30 numbers), and so the probability of both events occurring (i.e. Event A occurring twice in a row) is p(A) x p(A).  I agree, but my contention is that p(B) = p(A) (i.e. the probability of, in this case, losing a bet on 30 numbers is the same regardless of which 30 numbers are chosen) and so the joint probability of both events occurring is also p(A) x p(A).

RICHIECHIPS - 23 June 2012 09:14 PM

What I was trying to point out to others, not you,  was that some progressive / Martingale type bets can be better than others.

OK, that makes sense.  You have a better chance of winning a bet on 30 numbers than winning a bet on, say, 18 numbers (e.g. one of the even-money outside bets); therefore a losing progression on 30 numbers is safer than one on only 18 numbers.

RICHIECHIPS - 23 June 2012 09:14 PM

The bet that I had in my mind was a simple Martingale type bet.  1 chip of each of 5 line bets.  If I lost I lose 5 chips.  The second bet would be double or 10 Chip loss for an accumulated 15 chip loss after 2nd bet.  Assuming even with a 99.32% chance you still lost on the 3rd bet also you would lose 20 chips on the 3rd spin for an accumulated loss of 35 chips.  Please don’t assume from this that if I lost the first bet I wouls play this martingle for the next 2 spins.  It was just an example of a bet with very good odds of success.

Agreed, but I have to point out that if you lose the first spin you cannot come out ahead for the series even if you win one of the next two spins.  If you win Spin #2 your net win for that spin is 2 chips, which still leaves you down 3 chips.  Similarly, if you win Spin #3 after losing Spins #1 and #2 your net win is 4 chips, which still leaves you down 11 chips.  I do not think I will be trying this system!

RICHIECHIPS - 23 June 2012 09:14 PM

By the way sorry it took so long to answer but got busy at work and had to totally rebuild my computer.  That took forever.

I know the feeling.  I am now retired, but was a computer support analyst in a division of a company that performed facilities maintenance for its clients.  I would periodically visit project sites to upgrade their systems and applications, and just see how things were going.  Such visits were of the “I will get here when I get here, and I will leave when I am done” variety (i.e. my on-site work schedule was totally flexible).

At around 8:30pm on the last day of one of my trips, as I was basically walking out the door, I remembered our staff had asked me to define a new printer to the UNIX operating system they were using.  What should have been a trivial task for me turned into a major disaster when, after typing in the necessary mantras, the entire print spooling subsystem disappeared.  I couldn’t print anything anywhere!  Not good.

I screwed around for about a half hour to no avail and then called our UNIX guru at home.  We screwed around for another couple of hours, and it was looking like I was not going to be leaving the site the next day, but would instead be re-installing the UNIX operating system instead, a task I was really not looking forward to.  Suddenly, and without us doing anything that should have had any effect, the print spooler reappeared!  Needless to say I shut down, packed up, and double timed out of the area.  To this day I do not know what happened; truly a case of sometimes we’re so good we amaze even ourselves!

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Posted: 24 June 2012 09:09 PM   [ Ignore ]   [ # 29 ]
RICHIECHIPS
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RICHIECHIPS - 23 June 2012 08:14 PM
... as a pre - planned group of betting the same 30 #’s for the next 3 spins your odds of 99.32% never change even if you lost the first 2 spins. Only as it stands alone is it 81.08%.
I think I agree with this statement. Betting on 30 numbers you always have a 99.32% chance of winning one of the next three spins. After observing Spins #1 and #2 you can still say that you had a 99.32% chance of winning one of Spins #1-#3, but you now have a 100% chance of winning one of Spins #1-#3 if you won either or both of Spins #1-#2, or an 81.08% chance of winning Spins #1-#3 if you lost both of them. You also have a 99.32% chance of winning at least one of Spins #3-#5. 
[As long as you use the same target group your odds of hitting graduate from 81.08% on the first spin, 96.42% on the second spin and 99.32% on the third spin.  If you modify the group you affect the odds I guess if you pre planned for 3 spins you could also say the group has 99.32% chance].  RICHIECHIPS - 23 June 2012 08:14 PM
With all due respect I do not agree with this statement below:
“Choosing the same 30 numbers is not a requirement. You always have 7 ways to lose out of 37 possible results; hence you always have a 99.32% chance of winning within three spins if you chose 30 numbers.”
Here again we will have to agree to disagree….once you change even 1 number you have to go back to the beginning and you incur the 81.08%. Mathematically you are correct however once you change your numbers you change your bet and once you change your bet you are not playing the same 30 numbers & therefore change your odds of hitting your number. In order to have a 99.32% you have to pick and stick to the same 30 numbers for 3 spins.
Well, OK; I can agree to disagree, but I feel compelled to try to make one last attempt to show you that it does not make any difference which 30 numbers you choose. [ smile That is funny …Ok Go ahead ….you are just like me…you don’t give up too easily.  That is why I am sure you are a good Gambler].   Suppose instead we are dealing with coin flips, and we want to know the probability of getting two heads in two flips. Does it make any difference whether or not we use the same coin to make the flips?  Is not the probability of getting two heads in two flips 25% regardless of whether we flip one coin twice or two coins simultaneously (assuming of course both coins are fair and are flipped fairly, and, oh yes, no spirits)?  [:) That is funny….there had been a few times at the tables I would look over my shoulder…]
My point here is that events having the same probability of occurring (flipping heads or spinning a particular number) can be freely substituted for each other when computing the joint probability of multiple occurrences. Expressed symbolically, let the probability of Event A occurring be p(A), and the probability of Event B occurring be p(B). The probability that Events A and B will both occur is then p(A) x p(B). Your contention is that Event B is actually the same as Event A (i.e. we are betting on the same 30 numbers), and so the probability of both events occurring (i.e. Event A occurring twice in a row) is p(A) x p(A). I agree, but my contention is that p(B) = p(A) (i.e. the probability of, in this case, losing a bet on 30 numbers is the same regardless of which 30 numbers are chosen)  [Sorry this is where I have a problem…I’m still of the opinion that the target group must remain the same.  I could be wrong.  But don’t think so.  As soon as you introduce a new number or different # the odds change.  I’m having trouble explaining.  It’s akin to chasing your tail.  You can’t catch it because it is a moving target.  You have increased the size of the target.  Within my scenario we have the same 30 #’s.  Assuming we look at your suggestion I think the odds change because the target is still 30 #’s but we now have added an additional # outside of the group.  When you use letters in the example it is difficult to see the problem, because in theory based on a formula it looks the same.  If you substitute a new letter or new # then problem is easier to see.  However, I wouldn’t go “ALL IN” on me being right, but I think so.] and so the joint probability of both events occurring is also p(A) x p(A). 
RICHIECHIPS - 23 June 2012 08:14 PM
What I was trying to point out to others, not you, was that some progressive / Martingale type bets can be better than others.
OK, that makes sense. You have a better chance of winning a bet on 30 numbers than winning a bet on, say, 18 numbers (e.g. one of the even-money outside bets); therefore a losing progression on 30 numbers is safer than one on only 18 numbers. [ Yes you are right but it doesn’t have to be a losing progression bet].RICHIECHIPS - 23 June 2012 08:14 PM
The bet that I had in my mind was a simple Martingale type bet. I wasn’t suggesting it would recover losses. 1 chip of each of 5 line bets. If I lost I lose 5 chips. The second bet would be double or 10 Chip loss for an accumulated 15 chip loss after 2nd bet. Assuming even with a 99.32% chance you still lost on the 3rd bet also you would lose 20 chips on the 3rd spin for an accumulated loss of 35 chips. Please don’t assume from this that if I lost the first bet I would play this martingale for the next 2 spins. It was just an example of a bet with very good odds of success.
Agreed, but I have to point out that if you lose the first spin you cannot come out ahead for the series even if you win one of the next two spins. If you win Spin #2 your net win for that spin is 2 chips, which still leaves you down 3 chips. Similarly, if you win Spin #3 after losing Spins #1 and #2 your net win is 4 chips, which still leaves you down 11 chips. I do not think I will be trying this system!  [You are absolutely correct but that is why I qualified my previous comment to you with the following statement…I said ”Please don’t assume from this that if I lost the first bet I would play this martingale for the next 2 spins. It was just an example of a bet with very good odds of success not recovering losses.  However, there are Martingale / Progression type bets you could use to catch up.  Certainly you would agree the quicker you catch up on a loss the better.  If there is a very safe low risk money management progression that could recover loses in a couple spins why not use it, Right?]RICHIECHIPS - 23 June 2012 08:14 PM
By the way sorry it took so long to answer but got busy at work and had to totally rebuild my computer. That took forever.
I know the feeling. I am now retired, but was a computer support analyst in a division of a company that performed facilities maintenance for its clients. I would periodically visit project sites to upgrade their systems and applications, and just see how things were going. Such visits were of the “I will get here when I get here, and I will leave when I am done” variety (i.e. my on-site work schedule was totally flexible).
At around 8:30pm on the last day of one of my trips, as I was basically walking out the door, I remembered our staff had asked me to define a new printer to the UNIX operating system they were using. What should have been a trivial task for me turned into a major disaster when, after typing in the necessary mantras, the entire print spooling subsystem disappeared. I couldn’t print anything anywhere! Not good.  [Been there done that….I am jealous…I wish I was a computer tech…That was why I was lost for 4 days…doubled my capacity to a 500GB SSD, Windows 7, Office 10.  Which would have taken you a couple hours took me 4 days.  Now I have more room to add more software to screw up my computer more often now. I know you know exactly what I talking about.  I have more time now, as I too am just retired.  Just sold my company http://www.GraceIndustriesInc.com and only things left are a couple of law suits against the NY City & NY State.  So now I have to also become a lawyer. Well maybe my next life.  Actually after giving it a lot of thought I think I want to come back as a pigeon. ]I screwed around for about a half hour to no avail and then called our UNIX guru at home. We screwed around for another couple of hours, and it was looking like I was not going to be leaving the site the next day, but would instead be re-installing the UNIX operating system instead, a task I was really not looking forward to. Suddenly, and without us doing anything that should have had any effect, the print spooler reappeared!  [They have a habit of disappearing and reappearing…] Needless to say I shut down, packed up, and double timed out of the area. To this day I do not know what happened; [Like Roulette NEVER look back….][[/color] truly a case of sometimes we’re so good we amaze even ourselves! 
[ I swear I think they are teaching computers not only to think but feel.  Every time I am under pressure my computer gets stubborn, locks up.  It knows , Anytime I have a 50-50 chance of getting something right, there’s a 90% probability I’ll get it wrong.  ]

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Posted: 25 June 2012 08:39 AM   [ Ignore ]   [ # 30 ]
The Midnight Skulker
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RICHIECHIPS - 24 June 2012 09:09 PM
The Midnight Skulker - 24 June 2012 09:48 AM

My point here is that events having the same probability of occurring (flipping heads or spinning a particular number) can be freely substituted for each other when computing the joint probability of multiple occurrences.  Expressed symbolically, let the probability of Event A occurring be p(A), and the probability of Event B occurring be p(B).  The probability that Events A and B will both occur is then p(A) x p(B).  Your contention is that Event B is actually the same as Event A (i.e. we are betting on the same 30 numbers), and so the probability of both events occurring (i.e. Event A occurring twice in a row) is p(A) x p(A).  I agree, but my contention is that p(B) = p(A) (i.e. the probability of, in this case, losing a bet on 30 numbers is the same regardless of which 30 numbers are chosen) and so the joint probability of both events occurring is also p(A) x p(A).

Sorry this is where I have a problem…I’m still of the opinion that the target group must remain the same.  I could be wrong.  But don’t think so.  As soon as you introduce a new number or different # the odds change.  I’m having trouble explaining.  It’s akin to chasing your tail.  You can’t catch it because it is a moving target.  You have increased the size of the target.  Within my scenario we have the same 30 #’s.  Assuming we look at your suggestion I think the odds change because the target is still 30 #’s but we now have added an additional # outside of the group.  When you use letters in the example it is difficult to see the problem, because in theory based on a formula it looks the same.  If you substitute a new letter or new # then problem is easier to see.  However, I wouldn’t go “ALL IN” on me being right, but I think so.]

We have substituted one or more numbers from outside the group for the same number of numbers in the group.

Suppose we both start out betting on the numbers 7-36 with one chip each on what I call the half-dozens 7-12, 13-18, 19-24, 25-30, and 31-36.  Your plan is to bet on these same numbers for three spins.  You calculate the probability of winning at least one of those three spins to be 1-((7/37)x(7/37)x(7/37)) = .9932.  I, on the other hand, plan to switch a bet to the half-dozen that I was not covering if it hits, and to do this for three spins.  My bets might therefore look like
    Spin #1: 7-12, 13-18, 19-24, 25-30, and 31-36
    Spin #2: 1-6,  13-18, 19-24, 25-30, and 31-36
    Spin #3: 1-6,  7-12,  19-24, 25-30, and 31-36
How would you calculate my probability of winning at least one of those three spins?

RICHIECHIPS - 24 June 2012 09:09 PM

I swear I think they are teaching computers not only to think but feel.  Every time I am under pressure my computer gets stubborn, locks up.  It knows , Anytime I have a 50-50 chance of getting something right, there’s a 90% probability I’ll get it wrong.

Murphy’s Law, computer variation: “If an error can occur it will.  If an error cannot occur it probably will.”

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