It is not possible to make money on a RR in the long run. The goal is always to lose as little money as possible when betting on an RR. The ideal wager is no wager at all. But sometimes it may be necessary to make wagers on RR’s for a variety of different reasons. In those cases you want to make the wagers that will lose you the least amount of money in the long run. The reason being that you can expect to lose money whenever you wager on an RR so try to lose as little as possible.

You can expect to lose 14 cents over time for each $10 come bet with $10 odds that you make on a RR. Whereas with a place bet of $12 on the 6 or 8 you can expect to lose 18 cents on each of those wagers over time. Would you rather lose 14 cents wagering $20 or 18 cents wagering $12? That is a rhetorical question because the answer is obvious.

Furthermore if you want to wager less money on the RR to reduce your variance then don’t place odds on your come bet. A $10 come bet without odds will expect to lose the same amount (14 cents) over time as a $10 come bet with $10 in odds. But if you are trying to look more like a regular player, most of them take odds, then the $10 in odds does not cost you anything because your expectation is to break even on the odds bet over time since it is an even money wager.

Pass/come line wagers have a house advantage of 1.4%. Place bets on the 6 or 8 have a house advantage of 1.5%.

While it is true you are in a better position to win a place bet on the 6 or 8 when the come bet goes to the 4,5,9 or 10 that does not take into consideration the other ways in which you can win your come bet. When you look at all the ways to win/lose the come bet vs. win/lose the place bet, the come bet is a better wager.

To prove this is not just some theoretical math stuff I will show how it works counting all the ways possible to win/lose each of those wagers. This is going to get cumbersome but it will demonstrate my point.

Let us look at the wins and losses when one makes 1980, $10 come bets with $10 in odds and 1980, $12 place bets on the 6.

In 1980 rolls an RR can expect to throw 55 twos, 110 threes, 165 fours, 220 fives, 275 sixes, 330 sevens, 275 eights, 220 nines, 165 tens, 110 elevens and 55 twelves.

The $10 come bet will be resolved on the first roll with a 2, 3, 7, 11 or 12. The player will win 440 and lose 220 of those wagers.
The $10 come bet will go to a 4 or 10, 330 times. The player will win 110 and lose 220 of the wagers that go to a 4 or 10. The odds bet pays 2-1 winning 220 units and losing 220 units.
The $10 come bet will go to a 5 or 9, 440 times. The player will win 176 and lose 264 of the wagers that go to a 5 or 9. The odds bet pays 3-2 winning 264 units and losing 264 units.
The $10 come bet will go to a 6 or 8, 550 times. The player will win 250 and lose 300 of the wagers that go to a 6 or 8. The odds bet pays 6-5 winning 300 units and losing 300 units.

Adding up the wins and losses yields, 976 wins vs. 1004 losses for the base $10 come bet.

The player wagers $10, 1980 times for a total of $19,800 wagered while winning $9,760 and losing $10,040. The player has a net loss of $280 for the come bet.

The player will make a $10 odds wager when the come bet goes to a point number 1320 times out of 1980. Thus the player will wager $13,200 in odds bets winning $7,850 and losing $7,850. A net loss of $0 for the odds bet.

Please note that the loss of $280 out of $19,800 wagered yields a HA of 280/19800 = 1.4%.
If we add in the odds the loss of $280 out of $19,800 + $13,200 = $33,000 wagered yields a HA of 280/33000 = 0.848% which is the HA for single odds.

Now let’s look at what happens when making a $12 place bet on the six, 1980 times. The player can expect to win this bet for a payout of $14, 5 times and lose the $12 wager 6 times yielding 900 wins and 1080 losses.

The player wagers $12, 1980 times for a total of $23,760 wagered while winning $12,600 and losing $12,960. The player has a net loss of $360 for the place bet.

Please note that the loss of $360 out of $23,760 wagered yields a HA of 360/23760 = 1.5%.

Let’s recap what I have shown above.

A player making 1980, $10 come bets with $10 in odds can expect to wager $33,000 and lose $280.
A player making 1980, $10 come bets without taking odds can expect to wager $19,800 and lose $280.
A player making 1980, $12 place bets on the 6 or 8 can expect to wager $23,760 and lose $360.

In both situations, come bets, with or without odds will lose less money than making a place bet on the 6 or 8 over the long run when betting on an RR.

Taking odds or not is up to you as the player depending on your objective. If you want to look like a regular player take single odds. If you want to wager less money and lower your variance don’t take odds. But the come bet is the way to bet on an RR if you want to lose the least amount of money.

That is what the math says. But, in the infinite progression you cannot say what number will come up, only what could or maybe should. Depending where you enter and exit in the cycle, any variation of numbers can come up. I do not see how the theoretical makes you jump one way or the other. Theoretically in Video Poker you should get a royal in a little over 40000 hands. But, that is not guaranteed. It could take 80000 or even 200000 hands. Back to craps, the math says the come bet is better, I do not think you can guarantee that. Only hope.

If you tell me how much money and the number of pass/come bets and place bets you will wager on RR’s over your lifetime, I can give you a range of how much money you can expect to lose on both types of wagers with a 99.7% probability of certainty or greater. That is not 100% but it is pretty close.

Casinos build their business on the math of the game. The gaming industry invests and makes billions of dollars in a business that is built on the theoretical expectations in random games.

We use math all the time to try to gain an advantage over the house. A card counter uses math to tell him when the deck is rich with tens or not and adjusts his betting accordingly. There are no guarantees that a player will win when the deck is rich with tens and aces. But the math of blackjack tells us the player has an advantage when the deck is rich with tens and aces.

Back to craps, the math says the come bet is better, I can not guarantee that. But I can offer a lot more than only hope. If you make several thousand wagers on RR’s over your lifetime I can state with a very high probability of certainty, you will lose less money with come bets than with place bets.

WOW Skinny – You put it so well!

Phil – Like most players who have always used place bets, the concept of a comebet being a better bet is hard to swallow. It reminds me of the student who I write about in my book that just couldn’t believe that using the 5-count would save him money.

What he did was to put it to his own test. Now his sample was small – 15 sessions – but it was enough to convince him. What he did was put a $25 chip in his pocket every time the shooter didn’t reach the 5-count. He did this because he usually wagered on every shooter right out of the gate with a $10 pass line and $15 in odds.

I am saying that you should do the same thing, even if it is a small sample. Take 10 sessions and use the 5-count and then place your $12 6 and 8 for a total of $24 and record your wins and losses. Do the same thing with one comebet with single odds. I know you will find what others have found – comebetting after the 5-count will lose you less money

Dom

Dom-For an RR, I only put up one place bet for $12. It hurts to put come bet for $10 and $10 odds for a total of $20. I will try for a while and see what happens.

Don’t – just put out a comebet and NO ODDS – Or just $5 or $6 in odds – Try that Phil

Dom

The one very important fact is this. We, as Advantage Players, are playing for the long run. Anything can happen in the short run, one session, one trip, one year. But it is guaranteed that in the long run the house edge will prevail with a random roller. I want to make the bet with the lowest house edge. That bet is a come bet with odds. Yes, I have to bet a bit more than placing the 6 or 8, but over the long haul I will lose less.

A Come bet with single odds reduces the effective house edge of that Come bet to 0.8%. So I have decided to play the Come bets and not worry about what might have been with Place Bets because I know I am losing less.

NFF

Since I first met you NFF, the long haul has gotten shorter.

Very true words Phil. It was great seeing you in AC last weekend.

NFF

To put this in its proper perspective, when you make a $10 come bet with single odds on RR’s you will not be betting $20 every time. 1/3 of those wagers will be decided in one roll which are only $10 because the wager gets settled when the RR throws a 2,3,7,11 or 12. That happens 12 times out of every 36 rolls for an RR. The other 2/3 of the time the RR will establish a point number. It is only those times that you will be wagering $20 on the shooter.

So your average bet is only $16.67 when you plan on making a $10 come bet with $10 in odds on RR’s. 1/3*10 = 3.333; 2/3*20 = 13.333; 3.333+13.333 = 16.666 = 16.67

As I said before you can plan on losing 14 cents when you make a $10 come bet on an RR whether you take odds or not.

When you do not take odds you lose 14 cents with a bet of $10 so the House Advantage is .14/10 = .014 = 1.4%.
When you take single odds you still lose the same 14 cents. But now your average bet is increased to $16.67. The HA is .14/16.67 = 0.008 = 0.8%.

In both situations you can expect to lose an average of 14 cents. But when you take odds you lose a lower percentage of the total you wager so the HA is lower.

I am explaining all this to help you become a more informed player. If it makes your head spin, ignore the technicalities and just follow Dom’s advice. But for those that are interested I hope it gives you a bit more insight into how it all works.

Perfect Skinny!

Dom

I was trying to figure what Philham was thinking, where he was coming from.
There seems to be an apparent conflict between long term and short term.

MATH explains the LONG TERM.
The CASINOS figure on the long term and the math supports that
(not to mention that they also have a calculated edge in their games).
They are figuring on very many iterations, between an RR player throws,
the very many RR players, and the many years in the gaming business to flesh
out the validity of the statistics.

VARIANCE explains the SHORT TERM.
The RR PLAYER lives in the short term, and variance explains that.
He remembers only his limited number of throws, between his turn at the dice,
for a few turns in a session at a casino, and over only several casino visits.
He is looking at a low number of samples, not statistically large enough
to be calculated for the math to prove out.

What say you?

I

Thanks Skinny! Loved your "putting it in perspective" post. That was very helpful for me!

I am confused about something in one of your earlier posts though and I think this is an ongoing problem for me as far as word definition and how it relates to Craps.

Skinny, You said "If you want to look like a regular player take single odds. If you want to wager less money and lower your variance don’t take odds."

I would have thought the word should be volatility and not variance in this case. Skinny, Would you please give an easy definition of both variance and volatility as I continue to confuse these two. If you want to open another thread for this then that is fine with me. Anyone else have good and easy definitions to remember these two words! Thanks, I would be forever greatful! 😀

Here are the dictionary definitions of variance and volatility:

variance (vârˈē-əns, vărˈ-)►

n. The act of varying.
n. The state or quality of being variant or variable; a variation.
n. A difference between what is expected and what actually occurs.

Definition of volatility

:the quality or state of being volatile: such as
a :a tendency to change quickly and unpredictably-price volatility-the volatility of the stock market
b :a tendency to erupt in violence or anger-the volatility of the region-the volatility of his temper
c :the quality of being readily vaporizable at a fairly low temperature
-As each component of crude oil has a different relative volatility, they will evaporate at different temperatures. —Martin W. Stockel et al.

I know you asked for a simple definition and I didn’t give you that yet. But I will get there eventually, I hope 😀

If you notice the 3rd definition for variance states, "A difference between what is expected and what actually occurs."

In math that is what variance and volatility measure, just using different metrics, ie. different standards of measurement.

I don’t know of a definition for volatility in statistics. But in finance it is the square root of the variance which happens to be the standard deviation. The reason they do that is to convert the variance into a metric that is in the same units as the variable they are measuring. Let me give you an example which I hope will clarify what I am saying.

In probability theory and statistics, variance is the expectation of the squared deviation of a random variable from its mean. Informally, it measures how far a set of (random) numbers are spread out from their average value.

Let us say you do 5 practice sessions of 36 rolls each for a total of 180 rolls. You find you threw 3, 5, 7, 7 and 8 eights in those 5 sessions for a total of 30 eights in 5 sessions. The mean (average) number of eights you threw per session is 30/5 = 6. So you can expect to throw 6 eights in 36 rolls. To measure the variance for you in those 5 sessions we subtract the number you threw in each session from your average and square each of those numbers, add up the results and divide by the number of samples. Thus (3-6)(5-6)(7-6)(7-6) and (8-6) = -3, -1, 1, 1 and 2. So your actual throws were negative 3, 1 and positive 1, 1, 2 away from your mean (average number of eights expected). Note those numbers add up to zero. -3-1+1+1+2 = 0 That will always be the case since you are measuring the difference between what is expected (mean) and what actually occurs (what you threw). So in order to make the number meaningful we square each difference, add up differences squared, divide the result by the number of samples and that is the variance for the number of eights you threw in those 5 sessions. By squaring the numbers -3, -1, 1, 1, 2 we get 9, 1, 1, 1, 4 which adds up to 16. So the variance for the 5 samples is 16/5 = 3.2

But that number is not in the same units as the measurement we took (single units counting the number of rolls) so it is difficult to understand its meaning.

The standard deviation puts it back into the original units. The standard deviation is the square root of the variance. The square root of 3.2 = 1.79. Thus the standard deviation for these 5 sessions was 1.79. That is a relatively small number so we know you did not deviate much from what was expected by you in those sessions. One could say there was not much volatility or variance in the number of eights you threw in those 5 sessions.

Standard deviation (SD) is a widely used measurement of variability or diversity used in statistics and probability theory. It shows how much variation or "dispersion" there is from the "average" (mean, or expected/budgeted value). In statistics, the 68-95-99.7 rule, or three-sigma rule, or empirical rule, states that for a normal distribution, nearly all values lie within 3 standard deviations of the mean.

About 68.26% of the values lie within 1 standard deviation of the mean.
About 95.44% of the values lie within 2 standard deviations of the mean.
Nearly all, 99.73% of the values lie within 3 standard deviations of the mean.

That means one can expect 68% of the eights to be within 1 SD of the mean (6), 95% of the eights to be within 2 SD of the mean and 99.7% of the eights to be within 3 SD of the mean.
For the number of eights you threw each session, you can expect:
68% to be between 4.21 and 7.79 (1 SD of the mean).
95% to be between 2.42 and 9.58 (2 SD of the mean).
99.7% to be between 0.63 and 11.37 (3 SD of the mean).

Thus in mathematical terms variance and volatility are similar in that they both measure the difference between the actual and expected but using different metrics.

As an aside, I could also have measured your actual results against what is expected by an RR, calculating the variance and standard deviation against an RR. I could then use that to determine if you are controlling the dice or if your results are not that different from what could be expected of an RR. Of course, this is too small a sample to have any statistical significance. But with enough data that is how SmartCraps determines your ability.

i place on 5 and field it’s the lowest amount after i win 2 and get my money back i go up on 6 or8 another win place last number so if he loses your out nothing and i make money this way not whole lot usually but some and sometimes 40 to 80

Have you taken a class night rider?

Dom

Nightrider,

Here is how I see this. If you make a $5 field bet and place the $5 on each roll of a random roller here is how I see it:

1. I agree with Dominator, you either have not taken a GTC course or you were asleep during the lecture on betting on random rollers.
2. You have 20 ways to lose the field bet (5,6,7 and 8) and 16 ways to win the field bet (2,3,4,9,10,11,and 12)–a negative expectation in the 5% range. OK if the 2 and 12 pay double it comes up to 18 ways to win. Still a losing bet with a high negative expectation of a little over 2%.
3. Placing the 5 has a negative expectation of 4%.
4. By using this "hedge" bet you have two bets on the table with a negative expectation. Why would you want to increase the negative expectation with these two bets on each and every roll of a random roller’s roll of the dice? It is bad enough to make any one bet with a negative expectation on a random roller.
5. You are so much better off with a minimum bet on either the come or don’t come line, take or lay single odds and have a negative expectation of less than 1%.
6. Roll a 7 you lose two bets, roll a 6 or 8 you lose one bet. Roll a 5 you lose one bet, win one bet for a net win of $2. Is it worth it in the long run. I don’t think so.

I am sure that Skinny can give a much better math explanation.

Hope this will help you understand why your betting system is not to your advantage. Give this some thought.

Doc

I will make it as simple as possible.

Let’s compare what happens when you make a $5 come bet with $5 in odds vs. putting out $5 on the field and a place bet on the five for $5. Say the field bet pays double on the 2 or 12 to make it a better field bet. In both cases you are betting $10 so the bets seem even. Actually you will only average out betting $8.33 with the come bet but it will feel like $10 since that is what you put out when you take odds.

If you make these bets 1980 times here is what you can expect to lose on average:

$140 loss with the come bets.
$275 loss with the field bets.
$396 loss with the place bets on the five.

Would you rather lose $140 with come/odds bet or $691 with field/place bets?

To make it so that you bet $10 in both cases, you would have to make a $6 come bet with $6 in odds to average out to $10 with come bets. Your expected loss for 1980 bets would be $168, still a lot less than $691 you could expect to lose with field/place bets.

"Dr Crapology" wrote: Nightrider,

Here is how I see this. If you make a $5 field bet and place the $5 on each roll of a random roller here is how I see it:

1. I agree with Dominator, you either have not taken a GTC course or you were asleep during the lecture on betting on random rollers.
2. You have 20 ways to lose the field bet (5,6,7 and 8) and 16 ways to win the field bet (2,3,4,9,10,11,and 12)–a negative expectation in the 5% range. OK if the 2 and 12 pay double it comes up to 18 ways to win. Still a losing bet with a high negative expectation of a little over 2%.
3. Placing the 5 has a negative expectation of 4%.
4. By using this "hedge" bet you have two bets on the table with a negative expectation. Why would you want to increase the negative expectation with these two bets on each and every roll of a random roller’s roll of the dice? It is bad enough to make any one bet with a negative expectation on a random roller.
5. You are so much better off with a minimum bet on either the come or don’t come line, take or lay single odds and have a negative expectation of less than 1%.
6. Roll a 7 you lose two bets, roll a 6 or 8 you lose one bet. Roll a 5 you lose one bet, win one bet for a net win of $2. Is it worth it in the long run. I don’t think so.

I am sure that Skinny can give a much better math explanation.

Hope this will help you understand why your betting system is not to your advantage. Give this some thought.

Doc

ok i’ll try it tomorrow didn’t work out last weekend doing this thanks guys

"Skinny" wrote: I will make it as simple as possible.

Let’s compare what happens when you make a $5 come bet with $5 in odds vs. putting out $5 on the field and a place bet on the five for $5. Say the field bet pays double on the 2 or 12 to make it a better field bet. In both cases you are betting $10 so the bets seem even. Actually you will only average out betting $8.33 with the come bet but it will feel like $10 since that is what you put out when you take odds.

If you make these bets 1980 times here is what you can expect to lose on average:

$140 loss with the come bets.
$275 loss with the field bets.
$396 loss with the place bets on the five.

Would you rather lose $140 with come/odds bet or $691 with field/place bets?

To make it so that you bet $10 in both cases, you would have to make a $6 come bet with $6 in odds to average out to $10 with come bets. Your expected loss for 1980 bets would be $168, still a lot less than $691 you could expect to lose with field/place bets.

ok i’ll try it didn’t work out last weekend doing this thanks guys

"Dominator" wrote: Have you taken a class night rider?

Dom

yes dom my HERO remember I’LL DO IT TOMORROW MORNING

Here is one more way to look at it.

Your expected loss on a $5 come bet wit $5 in odds is 7 cents.
Your expected loss on a $5 field bet that pays double on 2 or 12 is 14 cents.
Your expected loss on a $5 Place bet on the five is 20 cents.

7 cents vs. 34 cents means you can expect to lose almost 5 times as much with a field/ place bet vs. come/odds bet.

$5 tables don’t exist here in mi 10 is lowest sometimes hard to find them 15 is the norm usually

Nightrider

Are you are saying it didn’t work out using comebets because you lost money? if so one session doesn’t mean anything. The 5-count will not be a positive betting method. I said that in class. There isn’t a betting method that you can use to make this game positive. The ONLY WAY to make this game positive for you is the three things I said in class

1. Learn the throw
2. calculate your edge
3. BET INTO YOUR EDGE

With come betting you are bringing the house edge down to a very small percentage, which means that YOUR edge doesn’t have to be that great to have a positive expectation. People have a tendency to want a magic bullet. 100’s of books have been written and so many you tube videos are out there claiming a magic bullet to beat craps. THERE ISN’T ANY!

Those of you that just don’t believe – I will say it again as I said in class – try it our way for 6 months. I know you will be better off. Remember all that we have tried it all. What we teach works and we are the proof.

Dom

"Nightrider" wrote: ok i’ll try it didn’t work out last weekend doing this thanks guys

That night was just all backwards as I had said. Smart bets lost and dumb bets won, it happens.