ZCore13

But it does work and the system generally results in the side with the larger bankroll having a greater chance of success.

Usually the casino has the larger bankroll compared to the player.

Some big players could present a problem to a smaller casino if their betting limits were too aggressive with that player.

Odds are designed to cause players to bet more than they would normally bet so that the house can have an advantage against the player as a result of having a larger bankroll.

That is, in fact, an edge (having more money and using a system of encouraging another party to bet beyond their limits).

If the player wins, the casinos have all kinds of backup measures to get the money back anyway.

Once the house edge is non-zero, the argument become murky and there are math guys who argue about infinite bankrolls and so on.

But at a zero house edge, martingale, I believe works great to benefit the side with more money!

Quote:AhighSome people claim martingale is a betting system that doesn't work.

But it does work and the system generally results in the side with the larger bankroll having a greater chance of success.

Usually the casino has the larger bankroll compared to the player.

Some big players could present a problem to a smaller casino if their betting limits were too aggressive with that player.

Odds are designed to cause players to bet more than they would normally bet so that the house can have an advantage against the player as a result of having a larger bankroll.

That is, in fact, an edge (having more money and using a system of encouraging another party to bet beyond their limits).

If the player wins, the casinos have all kinds of backup measures to get the money back anyway.

Once the house edge is non-zero, the argument become murky and there are math guys who argue about infinite bankrolls and so on.

But at a zero house edge, martingale, I believe works great to benefit the side with more money!

Another incorrect post by ahigh.

Martingale would not work on a negative game, even with infinite time and bankroll and no betting limit. The expectation would still be negative.

Quote:sodawaterAnother incorrect post by ahigh.

Martingale would not work on a negative game, even with infinite time and bankroll and no betting limit. The expectation would still be negative.

You failed to understand that I was speaking specifically towards the bet with zero edge, not the entire game.

This gets debated frequently. Here's a good link from 6 years ago on the topic:

http://rec.gambling.craps.narkive.com/tNTphNtB/overbet-your-buy-in-with-free-odds-bust-out-lose-less-the-usual-suspects

ZCore13

Quote:Zcore13I have to appologize for Ahigh. Someone that is able to answer you with an inteligent and fact based answer should be around shortly. While Ahigh's posts are entertaining, they generally go on forever and never say anything.

LOL. As if you could even spell intelligent.

The short answer according to the math is that is doesn't change the expectations in the long run.

But for a single event, you're more likely to bust out with a limited bankroll.

Betting bigger makes it more likely for you to lose all of your money. The casinos don't have to work as hard to get it all basically.

This is like arguing Keynesian economics. The theory is different from the practice.

If you are truly looking for just the math answer, it doesn't matter and changes nothing in theory in the long run.

The FAQ answers the math.

But if you want to know the whys, it's not about the long run, but the short run.

As Keynes said, "in the long run we're all dead."

Quote:n708s8thI agree a limited bankroll of less than $2,000.00 means that taking odds will hurt a player and is not in his favor. But everyone says always take the odds

I want to add to this question that this is specifically what I told the Wizard TO HIS FACE that "rubs me the wrong way."

I don't think this is explained well enough to most folks myself.

Even if plenty of folks understand this 100% on this forum, folks who visit the forum can and do end up betting too much to try to "save money" ensure that they are getting the "best deal."

On average it doesn't have a negative effect. But there's a loser for every winner out there even just looking at the theoretically zero house edge bets that don't hurt you in the long run.

Quote:n708s8thStill the question is why put up 5% of your money to improve your odds less than 1% ?

The logic isn't to improve your odds, it's to win more money. The house advantage isn't lowered as the pass line and the odds are separate bets and of course you can take the odds wager behind the line down at any point while you can't take the pass line wager down.

Quote:NokTangThe logic isn't to improve your odds, it's to win more money. The house advantage isn't lowered as the pass line and the odds are separate bets and of course you can take the odds wager behind the line down at any point while you can't take the pass line wager down.

The advantage as a percentage is lowered, but the raw dollar cost (per roll) is not.

The minimum cost is a single line bet. Period. At least until you get to a $25 minimum bet table with buy bets that have vig on the win rounding $1.25 vig down to $1.00.

Adding more money doesn't lower the cost, only the fraction of the cost to your total amount wagered.

Quote:bigfoot66Perhaps a better way to phrase the advice is "Max out the odds before increasing your pass bet beyond the table min". No one is encouraging you to bet more than you are comfortable betting just because the odds are free.

I'm not sure that is good advice. And, in fact, the casino is encouraging you to bet more than you are comfortable/can afford. That keeps them in business so I was once told and observed. It isn't only about the house edge contrary to what you read on here.

Example: Someone is down $20,000.usd and only has a $20,000.usd credit line. He/She wants to get the $20,000. back so they go get $5,000.usd on their VISA Credit Card and start betting larger amounts.

Quote:n708s8thStill the question is why put up 5% of your money to improve your odds less than 1% ?

You do not make an Odds bet to improve the odds on the line bet.

Its simply to get more of your bankroll in action on a bet that has no House Edge and No Player Edge.

Most of your action should be in odds bets.

Quote:FleaStiffMost of your action should be in odds bets.

So I go to Sam's Town with $100 in my pocket on a 20x table and bet $5 on the pass line and a four rolls. I should be $95 on the odds and wish I had another $5?

Or I should always bet more than single odds?

Not sure what you mean "most of your action." Do you say you should max out your odds ALWAYS or just do single odds or better always?

Also, should you always work the comeout roll?

There's an underlying assumption that you want to gamble as much as the free odds will enable you to do.

I think a lot of people quickly get to a point where their limited bankroll is a bigger concern than whether they get enough free bets for their $0.02 per roll per $5 cost to play the game.

Rule #1 should always be money management and bank preservation.

As for the odds bet you are talking about, you are getting a better deal by betting full odds. But it is your job to have the bankroll to play the correct way, or dont play the game.

Either way, you will lose everything if you play long enough.

Quote:BozEither way, you will lose everything if you play long enough.

At $0.02 per roll and 100 rolls per hour, that's $2.00 per hour. $32 per 16-hour day. $160 per 5-day play week. $8320 per 52 week year or $499,200 for a 60-year career.

So if you're rich enough and bet low enough limits, you won't lose everything. We can go back to Keynes who says "in the long run we're all dead."

Mathematically, your right, though. If you play infinitely long, you will lose everything no matter how much money you have and how little you bet. And of course that's how most people on this forum define right and wrong is by equations and other such cliff notes as they push up their black glasses with white tape between the lenses.

You know how many players at most every casino have ever won over the long term? None. It doesn't take equation and other such cliff notes or glasses with tape. It's just real life. Casinos do not lose in gaming. They may screw things up in other ways, with expenses, bad investments, horrible losses in other departments, but slots and table games NEVER lose.

ZCore13

Mathematical work has been done on this. I asked the Wizard about it a long while back, but he thinks it is a canard. The contrary thinking is that whether you go broke or note does not have a lot of mathematical meaning, a player's practical problems with it notwithstanding.

No house advantage does not reflect on chance of winning, or odds of winning, or risk of ruin, or anything besides being a "math expression" that means the "odds bet" is paid at the same chance of winning. Besides the "math meaning" it has no other significance.

You can have 1000x odds on the point of 4 that takes the house edge down to a smidgeon and on the point of four you still have only one out of three chances of winning.

Quote:Zcore13There are are no long time winners to speak of.

You seem to forget, humans aka gamblers know they are going to die. Animals don't. So, the animal has an advantage in the long run. A gambler only has so long to win/lose and enjoy it.

So if you are comfortable betting $20 each time, then it is better to put $5 on the pass line and then put $15 odds behind it versus putting the whole $20 on the pass line right from the start.

Like Alan said, it doesn't increase your chance of winning the roll, but if you DO win you will get paid more on the $5/$15 pass line/odds combo versus just getting paid 1:1 if you had the whole $20 on the pass line from the start.

So if I am a $5 low-roller bettor (which I am), I'm not going to put up $95 in odds just to reduce the house edge. But if I'm a $100 bettor, I would think about putting $5 on the pass line and then putting up those $95 in odds (assuming the table odds limits allow it). It is the same $100 being risked, but I would get paid more when I win by utilizing the odds.

Am I way off in my thinking?

Quote:blount2000Correct me if I'm wrong, but I tend think of odds as a better way to allocate the "base" amount you are comfortable betting. As opposed to just adding more to the initial base bet in order to reduce the house advantage. Especially if adding more to your base gets you beyond comfort level.

So if you are comfortable betting $20 each time, then it is better to put $5 on the pass line and then put $15 odds behind it versus putting the whole $20 on the pass line right from the start.

Like Alan said, it doesn't increase your chance of winning the roll, but if you DO win you will get paid more on the $5/$15 pass line/odds combo versus just getting paid 1:1 if you had the whole $20 on the pass line from the start.

So if I am a $5 low-roller bettor (which I am), I'm not going to put up $95 in odds just to reduce the house edge. But if I'm a $100 bettor, I would think about putting $5 on the pass line and then putting up those $95 in odds (assuming the table odds limits allow it). It is the same $100 being risked, but I would get paid more when I win by utilizing the odds.

Am I way off in my thinking?

I think this is the correct way to think about it. Decide how much you want to bet per decision, then divide that money with the table minimum on the pass/don't pass and the rest on odds, up to the table maximum. If the amount you are wanting to bet per decision is higher than the table minimum plus the odds multiple (i.e. willing to bet $50 per decision at a $5 min with 2x odds table) then up your pass\don't pass bet accordingly.

To more fully understand the concepts associated with the OP's question, however, does involve several additional mathematical concepts than simply the house edge--including expected value, expected loss, variance and standard deviation. And when we include the issue of the size of the bankroll, we then must take into account further questions such as: is there a win goal and/or loss limit and will the amount of the bet remain constant or will the bets vary (for which you would need to do simulations).

Anyone who has studied even the very basic math of craps knows what the OP states is true: The house edge or house advantage for the Passline is 1.41%, and taking 2x odds reduces the house edge to 0.0606%. This is arrived at by taking the expected loss and dividing it by the total of the flat and odds bets.

This is obviously acceptable. However, goatcabin (Alan Shank), who has posted here a number of times including comments about this subject, thinks this is misleading and I agree with him. [If goatcabin is reading this, I hope he joins in the conversation.] The expected loss is always 1.41% of the flat bet, regardless of how much you take in odds. Since the expected value of the odds bet is zero, the expected loss of the total bet (including both flat and odds) does not change.

We can look at this in a specific example using your suggested Passline bet of $5.00 without and then with 2x odds. Let's say the point is 10. The chance of winning is 1/3. The expected value of the Passline bet without odds is (1/3) * $5.00 + (2/3) * -$5.00 = -$5/3 = -$1.67. When you take 2x odds, you are adding $10.00. What is the result? Your chance of winning is still 1/3. But now you have the chance of winning a total of $25.00, but a chance of losing $15.00. Let's again calculate the expected value of both the Passline and the odds bet combined: (1/3) * $25.00 + (2/3) * -$15.00 = -$5/3 = -$1.67. The same amount either way!

Taking odds does not affect the expected value or the expected loss. It only increases variance. You can also call this volatility. It is helpful to have more variance if you hope to reach a win goal with a relatively small bankroll. But variance also means that you are likely to lose that small bankroll. The variance or volatility works both ways.

I think this is the crux of the OP's question.

When you take odds, you are obviously risking more money. If the dice goes against you, you will lose money more quickly, so that with a short bankroll compared to your total bet size, you will more likely bust your session bankroll. Even though taking odds does not add to the expected loss (since there is no house edge on the odds bet), it does add variance. This increases your "risk of ruin" since you are betting more.

However, since variance works both ways, you are also more likely to reach your win goal with odds compared to playing without taking odds (i.e. you don't have to be "as lucky").

Quote:JimboHowever, since variance works both ways, you are also more likely to reach your win goal with odds compared to playing without taking odds (i.e. you don't have to be "as lucky").

This is only true assuming you are betting the don't pass or that you are betting more than just one bet.

Going back to the example of $95 odds on a four point, you need to be a lot more lucky to make that first win compared to betting the whole thing on the pass line.

The fact that odds bets normally lose is part of answer. You do need to be lucky to win an odds bet. And you need to be more lucky than you would need to win a passline bet. Even on a six or an eight, you need more luck to win odds on a 6 or an 8 than to win on the pass line.

This is a big part of what I think trips people up too. There's an underlying assumption that everyone is going to bet every chance they can on the passline and on the odds for more than a single bet. But that assumption needs to be clarified.

IE: there's nothing wrong with putting your whole damn bankroll on the passline! And Wizard goes into how to get the absolute best chance doing the don't and laying odds, but then more process to get an exact double and not much better chance to get the double.

Therefore all the confusion in so many ways!

What?Quote:AhighThe fact that odds bets normally lose is part of answer. You do need to be lucky to win an odds bet. And you need to be more lucky than you would need to win a passline bet. Even on a six or an eight, you need more luck to win odds on a 6 or an 8 than to win on the pass line.

You're saying that when you add odds to the Passline bet, your "luck" goes down as opposed to only betting the flat bet? And it is therefore better not to take odds since it requires more luck to win?

This makes no sense.

ZCore13

Quote:n708s8thLet me say I love your Web site I think it’s easily the best out there. But I have a question a bought the math on a universal given on Craps. Everyone knows that there is no house advantage on odd’s bets. The problem I have with that statement is that it assumes that your bank and the houses bank are the same. Let me explain, a pass line bet has a house advantage of 1.41%. So I put a $5.00 bet on the pass line and if I do nothing else over the long time the house only has that 1.41% advantage. Now I take 2x odds for $10.00. The house advantage drops to 0.606% due to there being no house advantage on the odds bet. But if you’re an average gambler and you come to the table with $200.00. You just put up 5% of your bank to lower the house advantage 0.606%. This seems to me to risk too much for not enough gain. It seems that you need between $1000.00 to $2000.00 dollars to start with to make the math work out. If this was an insurance policy and I could cover 98.59% (100%-1.41%)of my house for $5.00 why would I pay $15.00 to cover 99.394% (100%-0.606%)of my house ?Where is my mistake?

I think this is just depending on how a player looks at the Odds Bet. I view the Odds bet as a completely separate bet to a Line/Come bet. It is true that having a Line/Come Bet is a requirement to making an Odds Bet, (Unless someone doesn't take Odds and the House and/or that Player allows you to make the Odds Bet for him/her) but making a Line/Come bet does not require you to make an Odds Bet.

That having been stated, the House Edge on the Line/Come Bet will be 1.41% on the Pass Line and 1.40% on the Don't Pass. Whatever money you use in order to make that bet will always be subject to that House Edge whether you make an Odds Bet or not. The Odds Bet, as a separate bet, has no House Edge whatsoever and could be looked at as reducing the edge on the, "Overall play," but again, to me, they are two separate plays, or bets, so I don't look at it that way.

In this case, to answer the last question, you are simply having two different insurance policies. The first one covers 98.59% of your house and costs $5.00 while the second one covers 100% of your house and costs $15.00, but you must first buy the lesser insurance policy in order to buy the second one. I guess, technically speaking, you are also insuring two different houses...with all due respect...the metaphor doesn't really work that well with what I am trying to say.

I think it is also strange to look at the combination as one bet because the House Edge (assuming a point has already been established) on a Pass Line bet is significantly greater than 1.41%, and varies depending on the point. You'll notice that many houses will allow you to take down a Don't Bet (bad move) after a point has been established, but good luck trying that with a Pass Line bet!

Quote:JimboWhat?

You're saying that when you add odds to the Passline bet, your "luck" goes down as opposed to only betting the flat bet? And it is therefore better not to take odds since it requires more luck to win?

This makes no sense.

You have a better than 49 out of 100 chance of winning the pass line bet.

You have a 45.45% chance of winning the odds bet in the best case.

You need more luck to win an odds bet.

Quote:AhighYou have a better than 49 out of 100 chance of winning the pass line bet.

You have a 45.45% chance of winning the odds bet in the best case.

You need more luck to win an odds bet.

I think this is misleading. In the first case, you are addressing both "aspects" of the line bet, i.e. the comeout, which is much more likely to win than lose, and the "point-seven", which (for "Rightside") is more likely to lose than win. In the second case, you are ignoring the comeout part. This is similar to the logical error many players make when comparing place bets to come bets, arguing that the come bets pays even money on a point, while the place bets pays better.

In order to get to the odds bets there's a "prerequisite" that a point number was rolled on the comeout. Whether or not you take odds, the probability of winning a bet, given that a point has been established, is the same.

Cheers,

Alan Shank

I believe you are mistaken if you insist on looking at it in this way.Quote:AhighYou have a better than 49 out of 100 chance of winning the pass line bet.

You have a 45.45% chance of winning the odds bet in the best case.

You need more luck to win an odds bet.

The 0.4929 probability of making the Pass includes the chances of winning on the come out together with the probability of a point repeating before the 7.

Quote:Jimbo

Anyone who has studied even the very basic math of craps knows what the OP states is true: The house edge or house advantage for the Passline is 1.41%, and taking 2x odds reduces the house edge to 0.0606%. This is arrived at by taking the expected loss and dividing it by the total of the flat and odds bets.

This is obviously acceptable. However, goatcabin (Alan Shank), who has posted here a number of times including comments about this subject, thinks this is misleading and I agree with him. [If goatcabin is reading this, I hope he joins in the conversation.] The expected loss is always 1.41% of the flat bet, regardless of how much you take in odds. Since the expected value of the odds bet is zero, the expected loss of the total bet (including both flat and odds) does not change.

OK, I will weigh in here. I agree with Jimbo.

Quote:JimboWhen you take odds, you are obviously risking more money. If the dice goes against you, you will lose money more quickly, so that with a short bankroll compared to your total bet size, you will more likely bust your session bankroll. Even though taking odds does not add to the expected loss (since there is no house edge on the odds bet), it does add variance. This increases your "risk of ruin" since you are betting more.

However, since variance works both ways, you are also more likely to reach your win goal with odds compared to playing without taking odds (i.e. you don't have to be "as lucky").

We should distinguish between ADDING an odds bet to a flat bet and making an odds bets INSTEAD of some portion of the flat bet. IOW:

1) compare a given flat bet to the same flat bet plus some odds multiple

The result is exactly the same expected value (loss) plus increased volatility

For example, compare a 60-bet session of $5 pass bets to the same for $5 pass taking 3, 4, 5X odds. Instead of an average bet of $5, the average bet will be $18.89. For 60 bets, the calculated ev will be -$4.24 in both cases, but the standard deviations will be $38.73 versus $190.37. The player taking odds will take vastly better advantage of good luck but get slammed much harder by bad luck, and have a considerably higher probability of breaking even or better, assuming neither player busts.

OR

2) compare a given (more-than-minimum) flat bet to a smaller flat bet plus some odds multiple, resulting in roughly the same total bet handle for a session

The result is a LOWER expected loss and somewhat increased volatility.

For example, with a $5 pass bet taking 3, 4, 5X odds, the average bet is $18.89. For a session of 60 resolved bets, the calculated ev is -$4.24, the standard deviation $190.37. For a $20 pass bet, no odds, the ev is -$16.97, the SD $154.9. Assuming a bankroll large enough that busting before 60 bet resolutions is not an issue, the flat + odds player is considerably more likely to break even or better than the $20 pass player. We can calculate the point along the "continuum of luck" where the lower volatility means the flat-only player will lose less than the odds player.

16.97 - 4.24 = 12.73

190.37 - 154.9 = 35.47

12.73 / 35.47 = .359

What this means is that, if both players suffer bad luck to the tune of .359 of one standard deviation, they will both lose the same amount (roughly); any worse, and the odds-taker will lose more. Any better, and the odds-take will lose less/win more.

A word about these calculations: Obviously, one cannot lose $4.24 in a $5 game for 60 sessions. What this number represents is the mean outcome of a very large number of such sessions.

Cheers,

Alan Shank

Quote:AlanMendelsonTo respond to the original post:

No house advantage does not reflect on chance of winning, or odds of winning, or risk of ruin, or anything besides being a "math expression" that means the "odds bet" is paid at the same chance of winning.

Actually it is paid at the inversion of the chance of winning, i.e. 2-to-1 if the chance of winning is 1/2, etc.

Quote:AlanMendelsonBesides the "math meaning" it has no other significance.

The "math meaning" IS its significance, the same significance that a house advantage of 16.67% on "Any 7" has. It is derived from the expected value of a bet, which is a function of two factors: 1) the probability of winning the bet AND 2) the payoff should the bet win. For years, on rec.gambling.craps, you could never grasp the interaction of these two factors.

Quote:AlanMendelsonYou can have 1000x odds on the point of 4 that takes the house edge down to a smidgeon and on the point of four you still have only one out of three chances of winning.

The idea that taking 1000X odds changes the house edge is very misleading; the house edge on the flat bet does not change when odds are added, and the expected value, whether of just the flat part or both parts, does not change.

BTW, did you ever figure out how odds behind DP/DC bets do NOT have a player advantage, even though they have a better-than-50% chance to win?

Cheers,

Alan Shank

Quote:Beethoven9thI actually agree with Ahigh. Since an odds bet cannot be made on the come-out roll, then why wouldn't you ignore the come-out when specifically focusing on the winning percentages of the free odds bet? Once you put your money on the table to take odds, then you have either a 33.33%, 40%, or 45.45% chance of winning that bet.

You miss the point. The odds of winning a passline flat bet include the player-advantage comeout roll; after a point is established, the odds of winning the flat bet and any odds bet are the same. Ahigh is comparing apples with oranges.

Cheers,

Alan Shank

Quote:goatcabinYou miss the point. The odds of winning a passline flat bet include the player-advantage comeout roll; after a point is established, the odds of winning the flat bet and any odds bet are the same. Ahigh is comparing apples with oranges.

Cheers,

Alan Shank

Maybe I'm missing something, but isn't this the whole reason why Ahigh is saying that a PL bet has a higher winning percentage? For example, if you put $5 bets on the line, and I'm allowed to take odds on that bet whenever a point is established, then you will have a higher winning percentage than me at the end of the session because of the advantageous come-out.

Take the same example, but put it in reverse. Let's say that you are always betting $5 on the Don't on every come-out roll. And whenever a point is established, I lay the point (and pay the vig). At the end of the session, I will end up having a higher winning percentage than you do.

Quote:n708s8thLet me say I love your Web site I think it’s easily the best out there. But I have a question a bought the math on a universal given on Craps. Everyone knows that there is no house advantage on odd’s bets. The problem I have with that statement is that it assumes that your bank and the houses bank are the same. Let me explain, a pass line bet has a house advantage of 1.41%. So I put a $5.00 bet on the pass line and if I do nothing else over the long time the house only has that 1.41% advantage. Now I take 2x odds for $10.00. The house advantage drops to 0.606% due to there being no house advantage on the odds bet. But if you’re an average gambler and you come to the table with $200.00. You just put up 5% of your bank to lower the house advantage 0.606%. This seems to me to risk too much for not enough gain. It seems that you need between $1000.00 to $2000.00 dollars to start with to make the math work out. If this was an insurance policy and I could cover 98.59% (100%-1.41%)of my house for $5.00 why would I pay $15.00 to cover 99.394% (100%-0.606%)of my house ?Where is my mistake?

To re-address the original question, it makes more sense if you look at the pass line bet or any other bet that is not a free bet as a bet that is required to be made in order to just stand at the table.

The free bets are easiest to understand if you just don't think about lowering the edge as much as possible. Most people who are new to the game are not even used to making $5 bets in the first place, much less taking $100 odds on top of that to make it a "good deal."

The best way to make your experience a lower cost is to not tip anyone, bring your own drink, walk instead of using a taxi, and fly on the cheapest airfare you can find.

All this talk about lowering an edge per roll from 0.42% to nearly zero only matters to people who bet a metric shit-ton of money and have the balls not to tip because they are penny pinching every corner in spite of their five-figure bankroll.

Just bet minimum on the passline and enjoy it as long as you can. Don't let the math guys convince you that you are short-changing yourself by not taking greater risks.

In fact, betting a $100 pass line with no odds on a $100 minimum table is maybe buying a more fun experience than going to Sam's town and being able to collect an extra $4 to $5 dollars if you win. At least until you are gambling all night long.

IE: don't be a degenerate gambler, and it doesn't even matter that much. Bet beyond your normal limits with no odds (at a $25 table or a $100 table) and see how it goes for your own self. Then make up your own mind about what kind of value you get out of the odds compared to no odds and putting it all on the pass line.

Maybe mingle with the really crazy folks with 10x to 1000x your bets on the felt and have fun with it.

Quote:n708s8thThank all of you for your replies. As a small time gambler who is going to be in Vegas for 6 days managing a bank roll is important. This is why I asked the original question. It seems from the answers that I’m better off placing the minimum on the 6&8 and pressing after two hits to make the money last.

If you want to preserve your bankroll, you're probably better off just making a Pass Line bet (no odds) and then making 1 Come bet (also no odds). Plus, you'll have only $10 in action vs. $12 if you just place the 6 and 8.

Quote:n708s8thThank all of you for your replies. As a small time gambler who is going to be in Vegas for 6 days managing a bank roll is important. This is why I asked the original question. It seems from the answers that I’m better off placing the minimum on the 6&8 and pressing after two hits to make the money last.

Minimum on the pass line and nothing else will last longer than anything. If you want it to last longer still, don't bet on every single shooter.

Quote:n708s8thThank all of you for your replies. As a small time gambler who is going to be in Vegas for 6 days managing a bank roll is important. This is why I asked the original question. It seems from the answers that I’m better off placing the minimum on the 6&8 and pressing after two hits to make the money last.

If you really want to stretch your bankroll, Joker's Wild has a $1 craps table and they pay to the quarter. Club Fortune has a $1 table, with no breakage on the dollar.

In Downtown Vegas, there are two continual $3 craps tables: one at the Freemont (a $3 place bet on six or eight wins $3.50); and another at El Cortez (not sure if El Cortez still offers breakage on the dollar).

I think that occasionally there are $3 tables on the Strip (I almost never go there).

I will add another example that underscores this point.Quote:goatcabin2) compare a given (more-than-minimum) flat bet to a smaller flat bet plus some odds multiple, resulting in roughly the same total bet handle for a session

The result is a LOWER expected loss and somewhat increased volatility.

Let's use the "magic number" 1980. We come up with this number because there are 36 different outcomes when a pair of dice is rolled. We want a number that is divisible by 36. We also need a number that is divisible by both 5 and 11. Therefore, 36 * 5 * 11 = 1980. This 1980 number can be used to also arrive at the calculations for house edges if you are inclined to avoid mathematical formulas and this has been used by other writers to describe the math odds of craps. (Though I prefer the formula route.)

In utilizing our 1980 number, we assume that we roll the dice 1980 times and the outcome is exactly as probability predicts: the number 2 appears 1/36 (or 55 times) and the 3 appears 2/36 (or 110 times) and the 4 appears 3/36 (or 165 times) etc., etc.

Please don't be turned away by the amounts in my examples below. I've used these amounts since they lend themselves to easy comparisons with exact dollar amounts. The conclusions are still the same with minimum betting.

Let's further assume Player One bets a Passline bet in the amount of $70 and does not take any odds. Using our 1980 number, Player One bets 1980 times, for bets totaling $138,600. Without taking the space to show the results of each number thrown, you will find that at the end of those 1980 rolls, Player One will have a loss of $1,960 (wins totaling $68,320 and losses totaling $70,280). This loss of $1,960 divided by the total amount wagered of $138,600 results in 0.01414--which is, of course, the house edge for the Passline without odds.

Now let's assume Player Two bets a Passline bet in the amount of $30 and takes 2x odds (or $60) when odds are available. Again using our 1980 number, Player Two bets are in the total sum of $138,600--which is exactly the same as Player One. (Player Two's total wagers on the Passline are in the sum of $59,400 and the total of his odds bets are in the sum of $79,200.) Again without showing the results of each number thrown, at the end of the 1980 rolls, Player Two will have a loss of $840 (wins totaling $76,320 and losses totaling $77,160). This loss of $840 divided by the total amount wagered of $138,600 results in 0.00606--which is the house edge for Passline with 2x odds.

As you can see, with the same amount bet by both Player One and Player Two, the player that did not take odds lost $1,960 whereas the player that did take odds of 2x had a lower loss of $840--which is as predicted by the formula (and by goatcabin) for the calculated ev.

You will also find that there is increased variance or volatility for Player Two as the result of the bets with 2x odds. For example, consider the Point 4. Player One will have bet only his flat bet in the amount of $70. Whereas, Player Two will have bet his flat bet in the amount of $30 plus $60 odds for a total of $90. The most Player One can lose on the Point 4 with a seven-out is the $70, whereas Player Two will lose his $90. On the other hand, when Player One wins on the Point 4, he wins only $70, but Player Two will win $150. You can also consider the totals wins and losses of the two players for the 1980 rolls. Player Two's total wins and total losses exceed that of Player One.

Compared to Player One, this increased variance will be much more to Player Two's advantage when the dice are favorable, but it will be much more to Player's Two detriment when the dice are not favorable (thereby having a direct bearing on the needed size of the bankroll when taking odds).

Though variance or volatility is still relevant when comparing Player One and Player Two--it is not as extreme or dramatic when both players bet the same total amount due to the fact that Player Two decreased his flat bets.

Let's now consider Player Three and assume he bets the same flat bet as Player One--$70. But Player Three also takes 2x odds. For the 1980 rolls, instead of bets totaling only $138,600, his bets total $323,400--due to the fact that his odds bets total $184,800. Over the course of the 1980 rolls there is considerably more variance. Again, consider the Point 4. Player One will have only his $70 flat bet at risk. Player Three will have bet $210--obviously a much larger bet. More importantly, while Player Three is at risk of losing $210, he also has the potential of winning $350.

At the end of the 1980 rolls for Player Three--assuming, of course, the outcome of the 1980 rolls is exactly as probability predicts--he still has a net loss of only $1,960. The same as Player One who did not take odds. That is because for the 1980 rolls of Player Three and the odds bets totaling $184,800--the wins ($109,760) will equal the losses ($109,760) since, obviously, the odds do not have any house edge.

It should now be obvious--

If two players make the same Passline bets and one of them makes additional odds bets they will lose exactly the same amount in the long run. Which is why the expected loss or ev makes no difference with or without odds--again assuming the flat bets are in the same amount.

Whereas, if one of the two players makes Passline bets and takes odds in such a fashion that his total wagers are the same as the other player, then he will lose less in the long run. As stated by goatcabin, here the expected loss or ev is lower.

The more important consideration, again, is the increased variance or volatility.

Even though Player One and Player Three in our example lost the exact same amount, Player Three bet nearly 60% more in total wagers with much larger swings on individual bets. For Player Three, such swings will result in a much greater win if the dice are good with favorable trends, but Player Three is also exposed to a much greater loss in the event of unfavorable trends of the dice--to the point that if the bankroll is inadequate, such swings could wipe out Player Three's bankroll for the session.

Quote:n708s8thThanks ,you do a great job presenting the facts. But to bet the pass line and 2x odds doesn’t the player need to come to the table with a bank roll of $1,000.00 to $2,000.00 for the session to survive the volatility for 6-8 hours

If you do line bet and no come bets, no. Double odds on a $5 table would require maybe $300. to last a couple of hours.

By comparison a $200 bankroll will be very difficult to exhaust in 24 hours of play with no odds on a $5 table if you only bet the pass line on the comeout roll and just waited for that to resolve before betting again.

Maybe??Quote:AhighIf you do line bet and no come bets, no.

Double odds on a $5 table would require maybe $300. to last a couple of hours.

Come on and do not scare away the average Craps player with small bankrolls.

Now $300 would be too much for many Craps players to buy-in for.

Simulation of Craps Pass Line Wagers

Odds Multiplier . . . . = 2

Session Bankroll . . . = 300.00

Max. No. rolls to quit = 180 (90 rolls per hour)

No. Sessions simulated = 1,000,000

Avg. No. games played . = 54.04

Avg. No. games won . . = 26.64

Avg. No. games lost . . = 27.41

Avg. No. dice rolls . . = 182.44

Bankroll was busted . . = 0.298% of the time ( 2979) <<<<<<<< Not a "maybe" bust rate

Bankroll decreased . . = 51.667% of the time

Bankroll increased . . = 47.954% of the time

But for $200 with $5 and 2X odds. More players can have $200 in the rail to start.

(40 betting units. Same exact as betting $100 flat 2x odds with a $4k buy-in)

How about Risk of Ruin around 1 in 20 sessions on average

Simulation of Craps Pass Line Wagers

Odds Multiplier . . . . = 2

Session Bankroll . . . = 200.00

Max. No. rolls to quit = 180

No. Sessions simulated = 1,000,000

Avg. No. games played . = 53.36

Avg. No. games won . . = 26.30

Avg. No. games lost . . = 27.06

Avg. No. dice rolls . . = 180.10

Bankroll was busted . . = 5.327% of the time ( 53275)

Bankroll decreased . . = 51.682% of the time

Bankroll increased . . = 47.942% of the time

$200 and $300 for a $5 player may still be too much bankroll

But for $100 with $5 and 2X odds. More players can have $100 in the rail to start.

(20 betting units. Same exact as betting $100 flat 2x odds with a $2k buy-in)

How about Risk of Ruin around 1 in 3 sessions on average

Simulation of Craps Pass Line Wagers

Odds Multiplier . . . . = 2

Session Bankroll . . . = 100.00

Max. No. rolls to quit = 180

No. Sessions simulated = 1,000,000

Avg. No. games played . = 45.26

Avg. No. games won . . = 22.31

Avg. No. games lost . . = 22.95

Avg. No. dice rolls . . = 152.78

Bankroll was busted . . = 34.798% of the time ( 347985)

Bankroll decreased . . = 54.735% of the time

Bankroll increased . . = 44.947% of the time

Don't sound right.Quote:AhighBy comparison a $200 bankroll will be very difficult to exhaust in 24 hours of play with no odds on a $5 table

if you only bet the pass line on the comeout roll and just waited for that to resolve before betting again.

"very difficult"

Looks to be close to 1 in 5 busting out

Your "very difficult" is not even close to my "very difficult"

This can be calculated using DS Blackjack Attack trip RoR formula but the math is too challenging for many.

Sim results are more impressive says the Wizard and I agree.

($200 making $5 pass line bets only is the same as

a player buying in for $4k making $100 pass line bets only. 40 betting units)

Simulation of Craps Pass Line Wagers

Odds Multiplier . . . . = 0

Session Bankroll . . . = 200.00

Max. No. rolls to quit = 2160 (24 hours @ 90 rolls per)

No. Sessions simulated = 1,000,000

Avg. No. games played . = 600.99

Avg. No. games won . . = 296.25

Avg. No. games lost . . = 304.74

Avg. No. dice rolls . . = 2028.85

Bankroll was busted . . = 19.068% of the time ( 190683)

Bankroll decreased . . = 63.355% of the time

Bankroll increased . . = 35.179% of the time

Good Luck

see ya at the craps table

Quote:Beethoven9thQuote:goatcabinYou miss the point. The odds of winning a passline flat bet include the player-advantage comeout roll; after a point is established, the odds of winning the flat bet and any odds bet are the same. Ahigh is comparing apples with oranges.

Cheers,

Alan Shank

Maybe I'm missing something, but isn't this the whole reason why Ahigh is saying that a PL bet has a higher winning percentage? For example, if you put $5 bets on the line, and I'm allowed to take odds on that bet whenever a point is established, then you will have a higher winning percentage than me at the end of the session because of the advantageous come-out.

Take the same example, but put it in reverse. Let's say that you are always betting $5 on the Don't on every come-out roll. And whenever a point is established, I lay the point (and pay the vig). At the end of the session, I will end up having a higher winning percentage than you do.

The fact is that, except in that very unusual situation where somebody allows a player to make odds bets behind his/her flat bet, you cannot make an odds bet without having a passline bet. So I don't think it is useful to make that first comparison. If two players are playing side-by-side, one making only flat bets and the other taking odds, they will have exactly the same W-L record, but very, very rarely the same net $$ result.

The second comparison is valid, but is not about odds versus just flat. The lay bet is another story entirely.

In any case, the W-L percentage is only half the story, isn't it?

Cheers,

Alan Shank

Quote:AhighMinimum on the pass line and nothing else will last longer than anything. If you want it to last longer still, don't bet on every single shooter.

I agree here. The passline without odds combines low vig with very low volatiliy, because your net outcome is determined only by your W-L "record", and the probability of winning any given bet is close to .5. However, I think most players would find it rather boring after a time; even if you take odds, you can sometimes stand there for quite a while waiting for your bet(s) to be resolved while other players are getting paid on the other point numbers. Pay your money and take you choice.

Cheers,

Alan Shank