You are assuming a random distribution of 7’son the point cycle. By reducing the number on 7’s you change the math of the PL + odds.

I will let those better versed in the math dispute your claims.

BTW, I used to be a place bettor, but has since converted to come bets with odds. My BR thanks me.

Sevenout,

Reducing the amount of sevens during a point cycle and being able to hit points are two different skill sets.

The fact still remains that if one uses their skill to avoid the seven on the come out roll then they are significantly increasing the house edge on their pass line bet.

While the Pass Line bets are comparatively good bets, regarding player advantage, the "advantage" is in the long run. That does not make it attractive to me, which is why I don’t try to roll a Seven on the Come Out roll. Such a small return over time, even with my Player Advantage, is more akin to Black Jack and card counting – what a grind! Hours of play and concentration just to eke out a 2.6% advantage – in the long run. Shoot for Sevens on the Come Out? Nah. I’m ready to get past that and on to the real money.

I much prefer Craps over BlackJack because of the opportunity I have to multiply my buy-in amount by several times in less than 30 minutes. There is no grind to that kind of game.

As to Odds bets versus Place bets, I think we all know that there is little advantage to either in Vegas on a 3x 4x 5x table. Just use whichever makes you more comfortable. We have discussed that numerous times in our weekend seminars. However, the fact that a Place Bet on 5/9 pay only 7/5 instead of 3/2 annoys me. In fact, the lower payout on all Place Bets annoy me, but Vegas 3x4x5x makes Come Bets and Double-Dipping too distracting at times.

On tables with 20x to 100x odds, forget about Place Bets. My preference for Odds Bets is because the math is much simpler, so I have less to calculate to check the Dealer when I get paid. And I find Odds much simpler to press than Place Bets. Having to double check the Dealer distracts me from concentrating on my shot.

But even in Vegas, I will still use the Come Bet to pace my exposure to loss, early on. My roll does not favor any particular number, regardless of the math. If I Place Bet on the 6/8, I can go 8 rolls before hitting either one. But a Come Bet is a number I have thrown, and will throw again. I’m not guessing where I am in the game with a Come Bet. But starting out with a Place Bet? I don’t know when I’ll hit that number. It’s an unnecessary exposure for loss, to me.

Now this all relates to the early part of a roll. Once a roll is established, a Come Out 7 will bring down my Odds bets. Every time I recall that happening to me, it was well into the roll, and I was well into profit. At those times, I considered going back up with either a Place Bet or Put Bet with Odds, but I’ve never actually done that. I felt good having been forced to capture my winnings. And the few rolls it took to get them back up never bothered me – particularly since I seem to favor 2 numbers in any one roll.

But if you have the dice and want to set for Come Out Sevens, you’ll get no complaint from me, or anyone I know in Golden Touch. Go for it. And if you want to make Place Bets only, I’m wondering how many working bets you have early on? Just the 6/8, all inside, or 6 bets across? Just be careful of your bankroll, with regard to short rolls.

Lucifer,

You devil. 😆
I always wondered on this topic. And wandered, too. Thanks for bringing it up.
Sometimes an argument is based on ‘in the long run’ but I also think
we’ll be all dead in the run, and we have to deal with the variances in the short run
in our real everyday life at the craps table.

I respect Skinny’s posts on this subject and I also like the Wizardofodds website.
Hoping there will be a lively discussion on this forum.

7×7.

"Loucifer" wrote: Here is the BLUF:

Lets looks at what happens to the house edge if we eliminate those six ways to make a seven on the come out (i.e. controlled shooter avoiding the seven on the come out).

The overall probability of winning is 2/36 + 9648/35640 = 17568/35640 = 244/495
The overall probability of winning is 1980/35640 + 9648/35640 = 11628/35640 = ~161/495
The probability of losing is obviously 1-(161/495) = 334/495
The player’s edge is thus (161/495)×(+1) + (334/495)×(-1) = -173/495 ≈ -34.9494%.

I see how you would think what you do given the way you calculated the edge for a controlled shooter. However, you are missing a few key points in your math which changes the outcome significantly.

The first issue is that you are assuming pr(7)=0. That is virtually impossible. One would have to have an SRR of infinity in order for the probability of throwing a seven to be zero. The longest no sevens roll I once saw was 77 rolls with only the final roll being a 7. I know that for a fact because I was come betting this roll from the start and never had my come bets brought down by a come out 7. This shooter would have an SRR of 77, a phenomenal number. But even then the pr(7) = 1.3%. A very low probability when compared to the pr(7) = 16.67% for a random roller. Nevertheless it is greater than 0.

Another concern is what sevenout pointed out. By reducing the number of sevens you are increasing the number of box numbers thrown over that of a random shooter.

These both need to be taken into consideration in the calculation of house edge.

Granted a good controlled shooter is able to reduce the number of sevens thrown to much fewer than that of a random roller. But a controlled shooter still does throw 7 when not trying to do so. Let us take a reasonable good shooter with an SRR of 8. That is probably higher than most folks but I am willing to work with that.

For a random roller one would expect the distribution 1,2,3,4,5,6,5,4,3,2,1 for the 36 possible outcomes of two die resulting in the numbers 2,3,4,5,6,7,8,9,10,11,12.

For a shooter with an SRR of 8, one could expect 4.5 sevens to be thrown in 36 rolls. With the hardways set this shooter would also reduce the appearance of a 1 or 6 on either die, so one could expect a reduced number of 2,3,11 & 12’s being thrown. The sevens are reduced from random by 25%. By making the same assumption on the garbage numbers, we could expect .75,1.5,1.5,.75 of the numbers 2,3,11,12 in 36 rolls. This adds up to 4.5 garbage numbers. 4.5 sevens + 4.5 garbage numbers adds up to 9 rolls, leaving 27 rolls for the box numbers vs. 24 rolls for a random shooter. If we assume a proportional distribution of the 3 extra rolls we can expect the distribution for the 4,5,6,8,9,10 to equal 3.375,4.5,5.625,5.625,4.5,3.375.

In order to work with whole numbers we need to look at 288 rolls. With an SRR of 8, one could expect the following distribution of numbers over 288 rolls:

6,12,27,36,45,36,45,36,27,12,6 for the numbers 2,3,4,5,6,7,8,9,10,11,12.

With this distribution we can now plug those numbers into the formula to calculate the player’s edge. I will start with the formula for a random roller and then plug in the distribution above to determine it for a player with an SRR of 8.

"Loucifer" wrote:
Pass/Come
The probability of winning on the come out roll is pr(7)+pr(11) = 6/36 + 2/36 = 8/36.

The probability of establishing a point and then winning is pr(4)×pr(4 before 7) + pr(5)×pr(5 before 7) + pr(6)×pr(6 before 7) + pr(8)×pr(8 before 7) + pr(9)×pr(9 before 7) + pr(10)×pr(10 before 7) =

(3/36)×(3/9) + (4/36)×(4/10) + (5/36)×(5/11) + (5/36)×(5/11) + (4/36)×(4/10) + (3/36)×(3/9) =
(2/36) × (9/9 + 16/10 + 25/11) =
(2/36) × (990/990 + 1584/990 + 2250/990) =
(2/36) × (4824/990) = 9648/35640
The overall probability of winning is 8/36 + 9648/35640 = 17568/35640 = 244/495
The probability of losing is obviously 1-(244/495) = 251/495
The player’s edge is thus (244/495)×(+1) + (251/495)×(-1) = -7/495 ≈ -1.414%.

Pass/Come
The probability of winning on the come out roll is pr(7)+pr(11) = 36/288 + 12/288 = 48/288.

The probability of establishing a point and then winning is:
pr(4)×pr(4 before 7) + pr(5)×pr(5 before 7) + pr(6)×pr(6 before 7) + pr(8)×pr(8 before 7) + pr(9)×pr(9 before 7) + pr(10)×pr(10 before 7) =

(27/288)×(27/63) + (36/288)×(36/72) + (45/288)×(45/81) + (45/288)×(45/81)+(36/288)×(36/72)+(27/288)×(27/63) =
(2/288) × (729/63 + 1296/72 + 2025/81) =
(1/144) × (52488/4536 + 81648/4536 + 113400/4536) =
(1/144) × (247536/4536) = 1719/4536
The overall probability of winning is 48/288 + 1719/4536 = 3024/18144 + 6876/18144 = 9900/18144
The probability of losing is obviously 1-(9900/18144) = 8244/18144
The player’s edge is thus (9900/18144)×(+1) + (8244/18144)×(-1) = +1656/18144 ≈ +9.127%.

While it is true you will not have as many come out winners as a random shooter because you are throwing fewer sevens and elevens on the come out roll. You can expect to have only 9 come out winners in 36 rolls vs. a random roller expecting 12 come out winners in 36 rolls. But you more than make up the difference because you make far more points than the random roller because you are throwing more box numbers than random and fewer sevens in the point cycle than random.

You can increase your player edge even more if you take odds. I leave it to you to do the math but at 3-4-5X odds a controlled shooter with an SRR of 8 can expect to have a positive player edge of +20.587%

Even at 2X odds a controlled shooter with an SRR of 8 can expect to have a positive player edge of +18.492%

If you can afford 5X odds a controlled shooter with an SRR of 8 can expect to have a positive player edge of +21.450%

I will explain how and why the odds bet is mandatory for a controlled shooter in a later post. A controlled shooter can make the most money taking the most in odds that one can afford and feel comfortable taking.

In my new book, using comebets is stressed almost entirely. It is the recommended way to play this game even if you are not a controlled shooter with an edge. Bring the house edge down to a very workable edge is the way to make money.

Remember everyone – "The way to win more money is to LEARN to lose less money"
You do this by making comebets.

Re-read the edge you get from Skinny’s response. Think about that. Here is is again:
3-4-5X odds a controlled shooter with an SRR of 8 can expect to have a positive player edge of +20.587%

As I say in class you have to be an idiot not to be a come bettor

Merry Christmas and Happy Holidays to all

Dom

So Dom do you think that your new book will convince me to be a come better ? I will love to add it to my collection .
Also it is not easy to get a SSR of 8 . Just saying .
Good Rolling. 🙂 😉

Finisher,

A controlled shooter with an SRR of 6.5 can expect to have a positive player edge of +1.830%.

At 2X odds a controlled shooter with an SRR of 6.5 can expect to have a positive player edge of +4.601%.

At 3-4-5X odds a controlled shooter with an SRR of 6.5 can expect to have a positive player edge of +5.467%.

If you can afford 5X odds a controlled shooter with an SRR of 6.5 can expect to have a positive player edge of +5.750%.

A controlled shooter with an SRR of 7 can expect to have a positive player edge of +3.886%.

At 2X odds a controlled shooter with an SRR of 7 can expect to have a positive player edge of +9.521%.

At 3-4-5X odds a controlled shooter with an SRR of 7 can expect to have a positive player edge of +10.889%.

If you can afford 5X odds a controlled shooter with an SRR of 7 can expect to have a positive player edge of +11.383%.

Skinny that sounds better . I sure do hate those good 7s that they mention in class . Also hate the point 7s out that take a big affect on the SRR thing . Just saying . It just seems that I can throw less 7s but not that many repeaters YET.
Good Rolling. 😀 🙂

Skinny,

According to the fundamental arithmetic of the game, much of the value in pass and come bets is in the come out roll, so skipping it carries a high house edge.

Do you agree or disagree with this statement?

Finisher,

Yes!!! You hit the nail right on the head. Avoiding the seven, and making points (repeaters) are two different skill sets!

This is another reason why I dislike come bets. In short, you have to hit a number twice to get paid once! Huge disadvantage in my eyes.

One of these days I will make a post about the 13 roll threshold, but essentially 95% of hands will be 3 to 13 rolls long. That is not a lot of opportunities for repeaters and yet another reason to shoot for seven on every come out. You have to respect the math of the game!

"Loucifer" wrote: Skinny,

According to the fundamental arithmetic of the game, much of the value in pass and come bets is in the come out roll, so skipping it carries a high house edge.

Do you agree or disagree with this statement?

Yes, I agree that a controlled shooter will reduce the value of the 7 on the come out roll. But I disagree strongly that a controlled shooter will carry a high house edge on pass/come bets. In fact the exact opposite is true.

I also agree that you have to respect the math of the game. That is why I went to the trouble to calculate the math of the game for a controlled shooter with an SRR of 8 as it relates to pass/come bets. My statements are backed up with math.

You can not just look at the come out roll and say that a pass/come bet is a bad bet for a controlled shooter because he is reducing the power of the 7 on the come out. You have to look at the pass/come bet in its entirety. You have to take into consideration both the chances of winning on the come out plus the chances of winning during the point cycle. Pass/come bets are contract bets which consist of both the come out roll and the point cycle.

I will repeat what I said in my prior post because it is critical to this topic,

"Skinny" wrote: While it is true you will not have as many come out winners as a random shooter because you are throwing fewer sevens and elevens on the come out roll. You can expect to have only 9 come out winners in 36 rolls vs. a random roller expecting 12 come out winners in 36 rolls. But you more than make up the difference because you make far more points than the random roller because you are throwing more box numbers than random and fewer sevens in the point cycle than random.

You state:

"Loucifer" wrote: The Pass/ Come are a blended average of both the come out (huge player advantage) and point (huge player disadvantage) cycles.

That is certainly true for a random shooter. A random shooter has a 2-1 advantage on the come out roll which occurs 1/3 of the time and an approximate 3-2 disadvantage during the point cycle which occurs 2/3 of the time.

Let us take a look at the math to see how the blended average works out for both a random shooter and a controlled shooter with an SRR of 8. I hope this will help to clarify why I controlled shooter has a big advantage with pass/come bets even though he will win those bets less than a random roller on the come out roll.

Looking at the formula you posted for a random roller,

"Loucifer" wrote: The overall probability of winning is 8/36 + 9648/35640

Thus a random roller can expect to win:
during the come out roll: 8/36 = .22222
during the point cycle: 9648/35640 = .27071

Hence a random roller can expect to win a pass/come bet 49.293% of the time. Conversely he can expect to lose a pass/come bet 1 – 49.293% = 50.707% of the time. This results in a negative player’s edge because 49.293% – 50.707% = -1.41%.

Looking at the formula I posted for a controlled shooter with an SRR = 8,

"Skinny" wrote: The overall probability of winning is 48/288 + 1719/4536

Thus a controlled shooter with an SRR = 8 can expect to win:
during the come out roll: 48/288 = .166666
during the point cycle: 1719/4536 = .378968

Hence a controlled shooter with an SRR = 8 can expect to win a pass/come bet 54.563% of the time. Conversely he can expect to lose a pass/come bet 1 – 54.5634% = 45.4366% of the time. This results in a positive player’s edge because 54.5634% – 45.4366% = +9.127%.

Thus we can see that while a controlled shooter with an SRR = 8 can expect to win a pass/come bet on the come out roll less often than a random roller (only 16.67% vs. 22.22%) he can expect to win far more often during the point cycle (37.8968% vs. 27.0707%). When we look at the blended average we see that reducing the sevens during the come out roll does not carry a high house edge for the controlled shooter.

Furthermore since the controlled shooter significantly increases his opportunity to win during the point cycle, it is even more imperative that a controlled shooter takes as much in odds as he can afford and tolerate to risk. I will explain this in more detail in another post where I will prove using the math of the game the huge advantage of the odds bet on pass/come bets.

One more thing about your post which I forgot to comment on in the above.

"Loucifer" wrote:
This is another reason why I dislike come bets. In short, you have to hit a number twice to get paid once! Huge disadvantage in my eyes.

As you stated in your own post, the formula for calculating the house edge on a pass/come bet takes hitting the number twice into account.

"Loucifer" wrote: The probability of establishing a point and then winning is pr(4)×pr(4 before 7) + pr(5)×pr(5 before 7) + pr(6)×pr(6 before 7) + pr(8)×pr(8 before 7) + pr(9)×pr(9 before 7) + pr(10)×pr(10 before 7) =

For those not familiar with the math expressions in the formula let me explain it in English for you:

pr(4) means the probability of rolling a 4 (the first time)
pr(4 before 7) means the probability of rolling a 4 (for the second time) before rolling a 7.

In other words, the way you win on the 4 during the point cycle is you must first roll a 4 and then roll a 4 a second time before rolling a 7.

To calculate the probability of winning a pass/come bet during the point cycle one calculates the probability of hitting each of the box numbers and then rolling that box number again before rolling a 7.

The formula is pr(4)×pr(4 before 7) + pr(5)×pr(5 before 7) + pr(6)×pr(6 before 7) + pr(8)×pr(8 before 7) + pr(9)×pr(9 before 7) + pr(10)×pr(10 before 7).

In English it means:

the probability of rolling a 4 and then repeating the 4 before rolling a 7 plus
the probability of rolling a 5 and then repeating the 5 before rolling a 7 plus
the probability of rolling a 6 and then repeating the 6 before rolling a 7 plus
the probability of rolling a 8 and then repeating the 8 before rolling a 7 plus
the probability of rolling a 9 and then repeating the 9 before rolling a 7 plus
the probability of rolling a 10 and then repeating the 10 before rolling a 7 = the probability of winning a pass/come bet during the point cycle.

Hence, you state right in the beginning that the formula considers the probability of establishing a point and then winning by hitting it again.

A controlled shooter with an SRR = 8 has a positive player’s edge of +9.127% and that takes into consideration that he must hit his number twice in order to win during the point cycle. The math proves that even though you have to hit the number twice controlled shooters still have a big advantage with pass/come bets.

Skinny,

I know that you are trying very hard and I greatly appreciate your effort.

However, I think we both can agree the arithmetic of the Pass/Come by design heavily favors the Come out cycle of the game.

I am not swayed by an SRR 8 in the point cycle, because my counter argument is this:

If a shooter can temporarily adjust his SRR 4 while shooting for sevens on the come out they will have an even greater edge.

The overall probability of winning is 10/36 + 9648/35640 = 17568/35640 = X
The overall probability of winning is 9900/35640 + 9648/35640 = 19548/35640 = ~272/495
The probability of losing is obviously 1-(161/495) = 223/495
The player’s edge is thus (272/495)×(+1) + (223/495)×(-1) = 49/495 ≈ 9.8989%.

And that is just assuming a "random" point cycle. Sadly it is far too difficult to grab your SRR 8 math and pair it with my come out SRR 4 math using cut and paste on my phone.

My point is this, there is no evidence to support using the Hardways set and avoiding the seven on the come out. Bypassing the come out costs the shooter a significant amount of positive edge.

The mathematically superior way to play the pass line is to use the All Sevens Set (6-1’s on axis) and shoot for sevens on the come out. Then during the point cycle switch to avoiding the seven.

Sincerely,
Lou

Lou,

At least you have dropped the claim that a player will do better by never taking odds. The +20.587% player advantage by taking odds, as detailed by Skinny, leaves no question that taking true odds is the way to make money in craps, even for a moderately successful controlled shooter.

As to setting for 7’s so as to increase winning Come Out bets, I don’t see anything exciting nor comparatively rewarding by making a successful flat bet, unless you’re pressing your Pass Line Bet. You press your Pass Line Bet after a win, right? I certainly press my true odds bets following a win on that point. If not, aren’t you spending a lot of time to just win small flat bets?

As to your complaint that it takes 2 rolls to collect on a Come Bet, that’s only true the first collection. After that, it should be an “on and off” collection. If taking 2 rolls to collect on a Come Bet w/odds is still such a problem, then just start with a Place Bet that converts to a Come Bet w/odds.

In order to compare apples to apples, what would the shooter edge be on a 6 or 8, with an SRR of 8? Now you can compare a $60 6/8 with a $10 come with 5x odds. Which one pays more?

If the place bet pays better, what about the 5/9 vs 4x odds? Or the 4/10 vs 3x odds? Where does the math shift,if at all?

Preacher,

One battle at a time! I still contend that "Free Odds" is an expensive illusion, however currently we are debating the finer points of bets (Pass and Come) that, by design, heavily favor a positive player edge on the come out cycle and heavily favor a negative edge on the point cycle.

I am attempting to figure how did it become sage advice to bypass the come out roll with the Hardways set?!

It is counter intuitive to the basic fundamental mathematics of those bets!

With regard to all of those huge positive edges with a subjugated SRR 8. I am sorry but those numbers mean very little to me. It is all just theoretical!

There is this nasty little thing called variance.

It is truly horrible!

It can easily amplify or erase all of those theoretical numbers. It is the true gatekeeper of this game.

Here is a little secret: It is what wiped out our friend the Doey-don’t better! That and the fact that he created a pass line bet that loses on the come out and added even more negative house edge to the already horrible house edge on the point cycle!

Which brings me full circle to Free Odds. If you are buying in for $300 you have no business even thinking about Free Odds! Your buy in can not handle the variance in bankroll and the variance of the numbers.

Think about it….

When is the last time you have stood at a table and seen the perfect distribution of numbers? One "2", two "3’s", three "4’s", four "5’s", etc, etc. Now ask your self what is your Free Odds based on? The ideal best case scenario?!

I would encourage everyone to record and observe actual real live rolls. What do you actually see?

Here is a spoiler alert:

Just the other night I saw the "10" disappear for 72 rolls! How is that possible? 1 in 12 rolls (3/36) is supposed to be a "10". Is that normal?

Should I have called a search party to look for the "10"?!

Welcome to the desert of the real, are you ready to see just how far the rabbit hole goes?

Vale,
Lou

Sevenout,

Please help me understand the point you are trying to make?

A placed 6 or 8 for $60 wins $70
A $10 come bet with 5x odds wins $70

They are both the same with regard to payout. However, in my opinion, the place bet is a better bet.

1) The place bet can be fully taken down at any time, the come bet is stuck like chuck.

2) A mid roll Come out seven has no effect on the place bet, where as it wipes out every single come bet. The best case scenario is that it takes 12 rolls to get set up again and collect on all six box numbers. 95% of all rolls end by 13 rolls!!!

3) Place bets win more then come bets that don’t have at least 5x odds. Place bets are far easier on the variance of your bank roll. Especially if you look for 3-5 score hits within 8 rolls and pull them down.

4) Every single active come bet eventually loses to the seven out. If you are constantly placing a come bet to get across the board and reach off and on status you have created a parallel system for the casino. That seven out wins you unit for the bet in the come box but loses you 6 units for all the bets you had on the box numbers. And if each of those bets had 5x odds on them… oy, that is one heck of a hit to the ol’ bankroll. Place bet? Pull them down, live to fight another day…

Etc, etc…

Sincerely,
Lou

People we can argue, religion, and politics, but you can’t argue Lou with math.

2 + 2 will always equal 4

Happy New Year!

Dom

Dom,

Happy New Year to you as well!

I 100% agree with you, you cannot argue with the math of the game.

My Dad used to tell me the story of how he would tell this new manager that 1+1 = 2. This snot nosed young buck had the audacity to retort back with… "Not necessarily…" My dad concluded that he was a dumb@$$.

However, as I grew up and moved away from the small town sticks of PA, I learned a valuable lesson.

My Dad was wrong.

You see it a binary system 1+1 = 0. My dad’s mistake was that he was applying the rules of a base 10 system to a base 2 system, simply because he did not know any better. In other words, he didn’t know what he didn’t know.

Theoretical edges derived from theoretical best case scenario numbers, are not an apples to apples comparison of what is happening in the real world.

The true test of one’s edge is analyzing one’s shot with slow motion video, knowing which sets produce which numbers, and having the historical data to have statistical confidence in the outcome.

Also, it is absolutely imperative that one understands the variance involved!

As example the mean combination of a pair of dice is 7, the standard deviation is 2.415. The means that 68% of all the combinations of the dice should total 5, 6, 7, 8, 9 in a completely random game.

However, sometimes this percentage will be off due to the variecne of the game.

Knowing that there should be 5 "6’s" in 36 rolls is completely different than knowing that it is perfectly normal for there to be 3 to 7 "6’s" in 36 rolls. And in extreme cases it normal for there to be zero 6’s in 36 rolls! [(31/36)^36 = .004593 or ~ 1 in 218. Assuming 100 rolls hour, once every ~78.5 hours]

Anyways, thank you for giving me the opportunity to generate discussion! The one thing we all have in common is that we love this game.

Sincerely,
Lou

A come out 7 only affects the flat portion of the come bet. The odds get returned to my rack.

On a short roll, I can win a come bet before I get up on all my numbers (in my case 3, PL and 2 come bets). Placing the 6 & 8 there is no "consolidation" prize (last come gets some).

If I throw a come out 7, many times it is my canary in the coal mine.Great time to get money off the table.

Lou said that if you bye in for 300.00 you should NOT EVEN THINK about free odds . Well for the last 10 yrs. or so that is what I bye in for . I do take odds with different numbers . Some with 1xs some at 2xs .But that bye in does last for 6 or 7 hrs. play . When I come back after dinner I still bye in for 200.00 . I use to about 11 or so yrs. ago bye in for less but that was when the tables were 3.00 . This last trip I did bye in one session for 400.00 .I did not change my play much tho .I did do more come betting tho .
I told a story of one time I got on a table were there was a guy recording the numbers . He said that the 8 had not been rolled in 42 numbers . I said that I would like an 8 for 6.00 . Well 3 rolls later I told him that he was getting me worried about my big bet and that I was starting to sweet . WE all laughed and the 8 was rolled after that .
I made my 7.00 and took my bet down .
You just never know . But I did think the odds were in my favor even if the math was not . I think the math was in for a 7 period at that time .
Good Rolling. 😀 🙂

I am glad Finisher that you re becoming more of a come player. One thing though, and this is common for most players to do, – your comment about how much odds you take depending on the number —– Most people think this way. They will take 1X odds on the 4/10 and 2X odds on the 6/8 …… WRONG! Do me a favor the next time you go out to play – take the same amount of odds on any number that your comebet goes to. Do this for maybe 5 – 10 sessions. I know you will make more money or lose less money than you have

Dom

Rose and Doc always keep one thing in mind when we take the odds—the odds portion of the your come bet is never at risk. Crazy, perhaps to some. BUT remember since the odds portion of your come bet has no house edge (it is always paid at the true odds) the worst that can happen is that in the long run you will break even—even on random rollers.

Just make sure you are playing within your bankroll as anything can happen in the short run.

Just our $.02 cents worth.

A lot has changed since your GTC class in 2004, And Dom has more hair than when you last saw him (I understand he’s mellowed out a lot, too.) You should take a Refresher course to catch up.

Now, now – 😆 😆 😆

A lot has changed because as we get older we get smarter. Remember the old proverb, "Change is the only consistent in this world" or something like that!

Dom

Here is another another way to say that the odds portion of your PL / Come is not as risk that we got from Goldfinger several years ago–since the odds bet is an even bet with with casino but simply a temporary hit against your bank roll. The worse that can happen is that you will break even in the long fun even on random rollers.