Will casinos let you wager equal amount on both pass and dont pass?
June 16th, 2015 at 12:57:13 PM
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Will land base casinos let you wager a equal amount on both pass and dont pass, and then take full odds "only" on the pass line wager, once point is established?
It seems to me that this is "almost" a sure-fire way to win in the longrun, assuming I am willing to accept the occasional pass line lost when a "12" appears on the come out roll.
What am I missing here?
SK
It seems to me that this is "almost" a sure-fire way to win in the longrun, assuming I am willing to accept the occasional pass line lost when a "12" appears on the come out roll.
What am I missing here?
SK
June 16th, 2015 at 1:02:25 PM
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well, off the bat, the don't pass is a much better bet once the point is established. Why not put your odds bet there?
But, also, the expectation is to break even except on a come-out 12. So how are you winning?
But, also, the expectation is to break even except on a come-out 12. So how are you winning?
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
June 16th, 2015 at 1:07:23 PM
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From a strictly "odds" stance, yes the odds on the "dont" are slightly better. However I personally do not like placing wagers that pay less than the amount laid on the table. For instance, I hate laying $20 odds on the 4, to win only $10. I guess it is a personal hangup I have.
I believe I will be winning as I will receive better than even money when the point hits on my pass line with odds, correct?
back to my original question- will the casino even let me make a pass and dont pass wager at the same time?
I believe I will be winning as I will receive better than even money when the point hits on my pass line with odds, correct?
back to my original question- will the casino even let me make a pass and dont pass wager at the same time?
June 16th, 2015 at 1:14:54 PM
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Yes, the bet is usually allowed.
No, you won't be winning. The odds payments are proportional to the likelihood of success. In other words, making a point of 4 pays 2:1 on the odds, but has a 1:2 chance of success. The odds break even over the long haul, and the system bleeds money to the come-out 12.
No, you won't be winning. The odds payments are proportional to the likelihood of success. In other words, making a point of 4 pays 2:1 on the odds, but has a 1:2 chance of success. The odds break even over the long haul, and the system bleeds money to the come-out 12.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
June 16th, 2015 at 1:44:01 PM
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Quote: rdw4potusYes, the bet is usually allowed.
No, you won't be winning. The odds payments are proportional to the likelihood of success. In other words, making a point of 4 pays 2:1 on the odds, but has a 1:2 chance of success. The odds break even over the long haul, and the system bleeds money to the come-out 12.
What he said. Can you do it? Yes. Is it worth it? My two cents: No. I have played this way, way back when I was younger, trying to keep my bankroll intact. The twelve digs your grave.
Lookup "The Patrick System" along with the "Doey-Don't" sans the Five-Count. The Wizards other page, about half way down gives some explanations. The Wiz explains.
Never count your winnings at hour 23 of a 24-hour drive.
June 16th, 2015 at 1:48:39 PM
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Quote: sikleinWhat am I missing here?
Plenty. Mathematically, you are doubling the cost of playing the game and simultaneously eliminating volatility on the line bet itself. Both of which reduce (or eliminate without odds bets) the opportunity to win.
If you split it out into two problems, you will see that, on average, having a unit on pass and don't pass loses a unit for every 36 comeout rolls in general.
The percentage of rolls, on average, that are comeout rolls is approximately 29%.
https://wizardofvegas.com/forum/gambling/craps/10716-what-percentage-of-rolls-are-come-out-rolls/
So ( 1/36 ) * ( 165/557 ) = ( 1 / 121.5272727272 )
So one in every 121 and a half rolls, you will lose the pass line without winning the don't pass on a 12 outcome.
That's without any chance to win at all. So that's the first half of the problem.
The second half of the problem is the free bets. Yes, you can bet max on the free bets and if you are lucky enough you will win more due to volatility than you will certainly lose (on average) with no opportunity to win on the pass line.
If betting systems based on laying numbers worked (which they don't -- even with no house edge) you could beat the game.
There is an expectation not to lose money in the long run on the free bets. But unfortunately, there is no expectation to win in the long run to make up for the certainly of losing in the long run betting both sides of the line.
In general, keeping the two problems separate, you are equally likely to lose from playing free bets as you are to win. So exactly half the time (in the long run), your free bets will be winning (all else remaining the same). But that doesn't mean that you cannot go on a multi-decade losing streak with free bets! You can!!!
The corollary is that no matter how long the system "works" for you, it's not "working" really. You are just getting lucky.
aahigh.com
June 16th, 2015 at 1:52:17 PM
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This comes up regularly. At one time, years ago, I wondered about it myself.
The one thing that will disabuse you of this kind of thinking the best is to realize that each bet you make, no matter what it is , has a tax on it. So when considering bet combinations of any kind that are known to each be up against a house edge, no other bet that also is up against an HE will be able to neutralize it. You are just compounding what is known as the Expected Value, a negative amount.
What is often said is that 'EV is cumulative'. In this case, you are having to bet twice as much, and each bet has an undefeatable EV.
PS: a lot of nearly simultaneous posting here.
The one thing that will disabuse you of this kind of thinking the best is to realize that each bet you make, no matter what it is , has a tax on it. So when considering bet combinations of any kind that are known to each be up against a house edge, no other bet that also is up against an HE will be able to neutralize it. You are just compounding what is known as the Expected Value, a negative amount.
What is often said is that 'EV is cumulative'. In this case, you are having to bet twice as much, and each bet has an undefeatable EV.
PS: a lot of nearly simultaneous posting here.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
June 16th, 2015 at 1:54:33 PM
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One other part to this problem is that MANY places here in Vegas will rate your "action" at ZERO. I have heard it said, "you're not giving the casino the opportunity to beat you."
So you will also suffer from not being rated, which can reduce the house edge by as much as a third when comps are figured into your "winnings."
I have heard that MGM properties are some of the few that will rate $5 do and $5 don't as $10 action. But I know of at least one MGM property and one employee there who will refuse to rate you with that play.
Dealers want you to win and tip and leave or lose and leave ASAP. There are plenty of dealers who don't want to give up a spot for someone who is just taking up space with no chance to win at all (or who isn't tipping enough for them to want you to be there).
So you will also suffer from not being rated, which can reduce the house edge by as much as a third when comps are figured into your "winnings."
I have heard that MGM properties are some of the few that will rate $5 do and $5 don't as $10 action. But I know of at least one MGM property and one employee there who will refuse to rate you with that play.
Dealers want you to win and tip and leave or lose and leave ASAP. There are plenty of dealers who don't want to give up a spot for someone who is just taking up space with no chance to win at all (or who isn't tipping enough for them to want you to be there).
aahigh.com
June 16th, 2015 at 2:01:16 PM
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If you are to do this, do it with a confederate and make it look as if the two of you are not connected.
DUHHIIIIIIIII HEARD THAT!
June 16th, 2015 at 2:28:41 PM
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I'm pretty much repeating what the others have said, but what you are missing is, you are, in effect, paying the casino a commission of about 3% of your pass line bet for the opportunity to make an odds bet.
You are better off just making the pass line bet, then taking odds when a point is rolled. The fact that you don't have a winning don't pass bet offsetting your losing pass bet when the comeout is 2 or 3 or a point is established but then the shooter sevens out is more than offset by the fact that you don't have a losing don't pass bet offsetting your winning pass bet when the comeout is 7 or 11 or a point is made.
You are better off just making the pass line bet, then taking odds when a point is rolled. The fact that you don't have a winning don't pass bet offsetting your losing pass bet when the comeout is 2 or 3 or a point is established but then the shooter sevens out is more than offset by the fact that you don't have a losing don't pass bet offsetting your winning pass bet when the comeout is 7 or 11 or a point is made.
June 16th, 2015 at 5:21:51 PM
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Quote: rdw4potuswell, off the bat, the don't pass is a much better bet once the point is established. Why not put your odds bet there?
After the point is established, the expectation for any pass line odds bets and any don't pass odds bets are exactly the same: 0. Yes, the original don't pass bet is now a favorite once the point is established, but that's got nothing to do with any odds bets.
OP's strategy is not good, of course. You should never bet an equal amount on the pass and don't pass. That's making two bets at negative expectation when you only need one to take full odds. The casinos will let you do this, of course.
Some uneducated casino staff will penalize you for comps if you do this, which is moronic. Someone who bets $100 each on the pass and don't pass should be classified as a $200 average bet. Some idiotic floormen will instead consider the average bet to be $100 or even $0 because they don't know how to do basic math.
June 16th, 2015 at 7:59:30 PM
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Quote: bushmanWhat he said. Can you do it? Yes. Is it worth it? My two cents: No. I have played this way, way back when I was younger, trying to keep my bankroll intact. The twelve digs your grave.
Lookup "The Patrick System" along with the "Doey-Don't" sans the Five-Count. The Wizards other page, about half way down gives some explanations. The Wiz explains.
I've never seen anyone play this way long enough to see the 12 bleed away their bankroll, but I have seen players do it and pick up their "don't" after the point is established, take odds, and then have a 7 follow right up behind it. Very rarely have I seen someone let both Pass and DP ride out the roll. I've definitely never witnessed anyone taking odds on the Pass Line and Laying odds on the DP at the same time.
June 17th, 2015 at 6:35:03 AM
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The "doey-don't" system Frank talked about many years ago and then abandoned was
supposed to be a way to lose less money on the majority of rolls which lasted less than
six rolls.
As Ahigh has correctly indicated, it does not change the house advantage, and you are
exposing yourself two bets instead of one.
However what is not calculated is the effect of mitigating the terrible loss you can take
the first 5 rolls.
I used the "doey-don't" system for 2007 and 2008 and I can say that during those two years
I stayed as close to even for the entire period than at any time in the 40 years I played
craps. Now it is not exciting and your going to take a lot of grief from other players
and the crew. I can still hear all the " you cant win money that way, your playing scared kinda
talk"
I would wait until roll 4 and place a come and don't come, and same on roll 5. If roll six
was a box number I would take single odds on both and then just play.
It surely wont help you make more money, but is sure as hell helps you lose less. I cant count the
times my friends were out of money and I was still playing. On very hot starting rolls I would make
a little less.
One time at Riverside in Laughlin I was using that and one of the crew was giving me a hard time, "you
can't make money that way" I said I have enough money to live on the rest of my life, I would like to play craps
without losing that" The next day he said he had thought about that and it was a better approach than most
people bring to the table. Later that morning he explained to a lady that asked how I was playing that I could
play the game without more money than I wanted.
I have not used it since the end of 2008, and even though I have won much more at times, I have not finished
any year as close to even.
dicesetter
supposed to be a way to lose less money on the majority of rolls which lasted less than
six rolls.
As Ahigh has correctly indicated, it does not change the house advantage, and you are
exposing yourself two bets instead of one.
However what is not calculated is the effect of mitigating the terrible loss you can take
the first 5 rolls.
I used the "doey-don't" system for 2007 and 2008 and I can say that during those two years
I stayed as close to even for the entire period than at any time in the 40 years I played
craps. Now it is not exciting and your going to take a lot of grief from other players
and the crew. I can still hear all the " you cant win money that way, your playing scared kinda
talk"
I would wait until roll 4 and place a come and don't come, and same on roll 5. If roll six
was a box number I would take single odds on both and then just play.
It surely wont help you make more money, but is sure as hell helps you lose less. I cant count the
times my friends were out of money and I was still playing. On very hot starting rolls I would make
a little less.
One time at Riverside in Laughlin I was using that and one of the crew was giving me a hard time, "you
can't make money that way" I said I have enough money to live on the rest of my life, I would like to play craps
without losing that" The next day he said he had thought about that and it was a better approach than most
people bring to the table. Later that morning he explained to a lady that asked how I was playing that I could
play the game without more money than I wanted.
I have not used it since the end of 2008, and even though I have won much more at times, I have not finished
any year as close to even.
dicesetter
June 17th, 2015 at 6:58:23 AM
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Quote: sodawaterAfter the point is established, the expectation for any pass line odds bets and any don't pass odds bets are exactly the same: 0. Yes, the original don't pass bet is now a favorite once the point is established, but that's got nothing to do with any odds bets.
OP's strategy is not good, of course. You should never bet an equal amount on the pass and don't pass. That's making two bets at negative expectation when you only need one to take full odds. The casinos will let you do this, of course.
Some uneducated casino staff will penalize you for comps if you do this, which is moronic. Someone who bets $100 each on the pass and don't pass should be classified as a $200 average bet. Some idiotic floormen will instead consider the average bet to be $100 or even $0 because they don't know how to do basic math.
I know some floormen do this because they are idiots (ie: what you said) and truly believe the bets cancel out and it's effectively a break-even system.
I know other floormen who completely understand the player is giving up whatever X% of his action to the HE. However, he rates such players as having a $0 average bet (or won't even rate them at all), because the people who're doing this tend to be fleas that think they've discovered some winning system so they stand at the table for 14 hours with $10 on both pass line, don't-pass, and continious come / DC. They order their diet coke, water, and coffee. The player stiffs the waitress and stiffs the dealers over the course of his 14-hour marathon session.
June 17th, 2015 at 7:12:57 AM
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Quote: sikleinWill land base casinos let you wager a equal amount on both pass and dont pass, and then take full odds "only" on the pass line wager, once point is established?
If you want a 0% house edge, play the bill breaker machine.
Did you come to gamble?
May the cards fall in your favor.
June 17th, 2015 at 7:28:58 AM
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Quote: DieterIf you want a 0% house edge, play the bill breaker machine.
Now where
have I heard
that before?
June 17th, 2015 at 7:33:51 AM
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Quote: ThatDonGuy
And it's STILL funny! :)
If the House lost every hand, they wouldn't deal the game.
June 17th, 2015 at 7:38:42 AM
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But don't use the loose change machine in supermarkets!Quote: DieterIf you want a 0% house edge, play the bill breaker machine.
Just in case you don't have them in the US, we have machines that take all your small change, charge a few percent and give out notes. Personally why people don't just put the change in the self-service terminals when buying their weekly shopping beats me - I used to do this for Saturday's paper or like today as I had built up too much small change #pennyviolation .
June 17th, 2015 at 8:00:40 AM
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Quote: charliepatrickBut don't use the loose change machine in supermarkets!
Just in case you don't have them in the US, we have machines that take all your small change, charge a few percent and give out notes. Personally why people don't just put the change in the self-service terminals when buying their weekly shopping beats me - I used to do this for Saturday's paper or like today as I had built up too much small change #pennyviolation .
Yeah, in the US now, nearly all banks charge a % for cashing/depositing your coins, and the commercial machines charge 8% I think. A few exceptions are some of our Savings and Loans, Credit Unions, and a couple of banks (Regions is one) which will accept coin deposits without a fee.
Bizarre, IMO. For all legal tender. Yeah...
If the House lost every hand, they wouldn't deal the game.
June 17th, 2015 at 8:08:46 AM
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Quote: charliepatrickBut don't use the loose change machine in supermarkets!
Just in case you don't have them in the US, we have machines that take all your small change, charge a few percent and give out notes. Personally why people don't just put the change in the self-service terminals when buying their weekly shopping beats me - I used to do this for Saturday's paper or like today as I had built up too much small change #pennyviolation .
One day a few years ago I took four quarters to the cage at Greektown here in Detroit to get a dollar bill. The lady there told me I had to use the CoinStar machine. I laughed and told her I'm not stupid enough to take the four quarters to it, then get a receipt for only 94¢, then come back to cage and redeem the receipt. She gave me the dollar bill.
DUHHIIIIIIIII HEARD THAT!
June 17th, 2015 at 10:41:41 AM
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Quote: beachbumbabsBizarre, IMO. For all legal tender. Yeah...
According to the Treasury, while all coins are legal tender, there is no Federal law requiring that anybody has to accept coins - or currency, for that matter (you try paying for something on a United flight with cash) - as payment. Ever see a "We do not accept $100 bills" sign? The business in question has every right to do that. I would not be a bit surprised to see a sign at a county tax office saying, "We do not accept more than 100 pennies in any transaction." (Apparently, paying an unpopular property tax in pennies as a sign of protest is common.)
As for converting coins, I have never had a problem having a casino's cage convert my coins to bills at 100% value - even when there was a Coinstar machine right next to the cage. In fact, while banks pretty much demand that coins be in wrappers, the casinos make me empty the coins from my wrappers into a tray first.
June 17th, 2015 at 12:09:08 PM
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Even with tolls:Quote: ThatDonGuyI would not be a bit surprised to see a sign at a county tax office saying, "We do not accept more than 100 pennies in any transaction." (Apparently, paying an unpopular property tax in pennies as a sign of protest is common.
"If you’ve ever tried to pay your toll on the Turnpike or the Parkway with pennies, you might have a problem. Some toll takers refuse to accept them, and occasionally they’ll become downright nasty. One member of the New Jersey 101.5 News team- when she attempted to pay a portion of the toll with pennies – had them thrown back at her!
Fortunately the coins landed harmlessly on the ground.
New Jersey Turnpike Authority spokesman Tom Feeney says “we ask people not to use them as a courtesy to the drivers in line behind them, but the official policy is they are legal tender and they are accepted at all the cash lanes…we’ve had cases in the past where we have collectors who didn’t want pennies – refused to take them – and were disciplined for it – there’s a bulletin instructing them that pennies are accepted…but sometimes if you’re stuck, and that’s all you have in the car, there’s no reason we shouldn’t be accepting that.” 101.5
April 20th, 2016 at 10:33:12 AM
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I found this post and I'd like to comment on it, even though it is very old.
I actually play this strategy on cold tables until the table starts to heat up.
I start with $10 on the Pass Line and $10 on the Don't Pass Line. I throw a $1 on the 12. The goal is to cover the DPL and get it on so I can take advantage of odds.
After the point is set, I do the following strategy:
Always accept the money for the first place bet win.
On the next place bet win, place another number (my order is 6,8,5,9,10,4) and pocket the extra chips after placing.
On the next place bet win, take the money.
On the next place bet win, place another number.
Continue until all numbers are covered (except for the point) alternating, taking money and placing money.
Once all the numbers are covered, alternate pressing the number hit and taking money.
After the first money is received right after the first press, increase the DP Odds to $50 risk.
Continue alternating press and taking money until the 7 hits or the point hits.
Example Walk through and you can test this all you want:
--------
Bank Roll of $200.
$10 on Pass and DP. $1 on 12. A 10 is rolled as the set point. Puck is now on.
Place $30 on the DP Odds.
Place $6 on the 6.
Place $6 on the 8.
Current Bank Roll = $200 - $63 or $137.
Roll of 5. Nothing.
Roll of 6. Take the first money given. Plus $7.
Current Bank Roll at $144.
Roll of 4. Nothing.
Roll of 9. Nothing.
Roll of 11. Nothing.
Roll of 6. Place the 5 or 9. I placed the 5 for $5. Received $2.
Current Bank Roll at $146.
Roll of 7.
Pass Line loses and DPL wins. No loss or gain. +$20.
Current Bank Roll at $166.
Receive the initial $30 on the DP Odds +$15.
Current Bank Roll is $211.
----------
Practice it. It is highly unlikely that you won't receive at a minimum the first place bet win before the point is hit or the 7 is hit. At the worst case you lose $30 or $24. If a 7 is rolled right after the point is set and you place your 6,8 or 6,5 or 8,5, you will always end up +$3 or +$4.
I use this strategy on a full table and usually hit one hot roller. If two set points are hit by the same roller, I move to max odds on the DP Line.
When I become the roller, I use the same strategy only I max the DP Odds right from the start and start with 4 units each on the first 2 place numbers. I mainly do this because I trust my rolling technique more than the other person. I don't consider myself some elite control shooter because frankly, none exist. I do trust my probability and math skills though and I've tested this using 10 different testing scenarios which house 1,000,000 atmospheric noise generated random numbers each.
Personally, I believe this to be one of the safest, long term strategies out there.
I actually play this strategy on cold tables until the table starts to heat up.
I start with $10 on the Pass Line and $10 on the Don't Pass Line. I throw a $1 on the 12. The goal is to cover the DPL and get it on so I can take advantage of odds.
After the point is set, I do the following strategy:
Always accept the money for the first place bet win.
On the next place bet win, place another number (my order is 6,8,5,9,10,4) and pocket the extra chips after placing.
On the next place bet win, take the money.
On the next place bet win, place another number.
Continue until all numbers are covered (except for the point) alternating, taking money and placing money.
Once all the numbers are covered, alternate pressing the number hit and taking money.
After the first money is received right after the first press, increase the DP Odds to $50 risk.
Continue alternating press and taking money until the 7 hits or the point hits.
Example Walk through and you can test this all you want:
--------
Bank Roll of $200.
$10 on Pass and DP. $1 on 12. A 10 is rolled as the set point. Puck is now on.
Place $30 on the DP Odds.
Place $6 on the 6.
Place $6 on the 8.
Current Bank Roll = $200 - $63 or $137.
Roll of 5. Nothing.
Roll of 6. Take the first money given. Plus $7.
Current Bank Roll at $144.
Roll of 4. Nothing.
Roll of 9. Nothing.
Roll of 11. Nothing.
Roll of 6. Place the 5 or 9. I placed the 5 for $5. Received $2.
Current Bank Roll at $146.
Roll of 7.
Pass Line loses and DPL wins. No loss or gain. +$20.
Current Bank Roll at $166.
Receive the initial $30 on the DP Odds +$15.
Current Bank Roll is $211.
----------
Practice it. It is highly unlikely that you won't receive at a minimum the first place bet win before the point is hit or the 7 is hit. At the worst case you lose $30 or $24. If a 7 is rolled right after the point is set and you place your 6,8 or 6,5 or 8,5, you will always end up +$3 or +$4.
I use this strategy on a full table and usually hit one hot roller. If two set points are hit by the same roller, I move to max odds on the DP Line.
When I become the roller, I use the same strategy only I max the DP Odds right from the start and start with 4 units each on the first 2 place numbers. I mainly do this because I trust my rolling technique more than the other person. I don't consider myself some elite control shooter because frankly, none exist. I do trust my probability and math skills though and I've tested this using 10 different testing scenarios which house 1,000,000 atmospheric noise generated random numbers each.
Personally, I believe this to be one of the safest, long term strategies out there.
“It’s a dog eat dog world out there and I’m wearing milkbone underwear .” – Norm Peterson
April 20th, 2016 at 1:23:36 PM
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Hi Joel, and welcome to the forums.
Even though it's an old post, you do bring good conversation to it. However, from a strictly "Expected Value" (EV) standpoint, I do believe your strategy is not one of the safest out there.
The safest is of course not to play, as you won't expect to lose anything... But here we are on a gambling site, so let's rule this one out =).
Next it would be to make the best bet, for the lowest amount possible. The best bet on the entire craps table is the 1.36% house edge on the Don't Pass. Betting the table min on this (let's pretend that's $10 for math purposes) would lead to the lowest expectation of loss.
Example: As stated above, $10 min. If you get 30 "games" of craps (point being established and then either won or lost) per hour then your Expected Loss (EL) is easy to calculate.
EL = AvgBet * NumGames * HouseEdge = (10)*(30)*(-.0136) = -$4.08.
This means if you play JUST the Don't Pass for $10 for 1 hour (30 games) then you, in the long run, can expect to lose approximately $4 per hour.
Now how does this compare to your system? Well, I don't need to (and don't want to) do all the math to show it's worse. EVERY bet you make has a built in house edge and thus will add to your overall expected loss.
Example:
Covering the DP and Pass lines together... EL = AvgBet * NumGames * HouseEdge = (10)*(30)*(-.0278) = -$8.34
When you cover the DP and Pass you'll never win on these initial bets but only lose when the 12 is rolled, which is 1/36 or 2.78% of the time. Thus, you've effectively DOUBLED the houses edge from just playing the DP alone!
Now, for EVERY SINGLE bet you add to this (other than odds, which have no long run value) you'll expect to lose more. If you place the 6 or 8 for $6, each one of those place bets carries a -1.515% house edge... and thus your EL for placing the 6 = $6 * (-.01515) = -$0.09. This might not be a lot, but as you can see it ADDS to your overall expected loss. Every other bet will continue to do the same... So the more bets you make, the more you can expect to lose.
The "safest" system, while still playing the game, is to make 1 bet... The table minimum on the lowest house edge table bet.
Even though it's an old post, you do bring good conversation to it. However, from a strictly "Expected Value" (EV) standpoint, I do believe your strategy is not one of the safest out there.
The safest is of course not to play, as you won't expect to lose anything... But here we are on a gambling site, so let's rule this one out =).
Next it would be to make the best bet, for the lowest amount possible. The best bet on the entire craps table is the 1.36% house edge on the Don't Pass. Betting the table min on this (let's pretend that's $10 for math purposes) would lead to the lowest expectation of loss.
Example: As stated above, $10 min. If you get 30 "games" of craps (point being established and then either won or lost) per hour then your Expected Loss (EL) is easy to calculate.
EL = AvgBet * NumGames * HouseEdge = (10)*(30)*(-.0136) = -$4.08.
This means if you play JUST the Don't Pass for $10 for 1 hour (30 games) then you, in the long run, can expect to lose approximately $4 per hour.
Now how does this compare to your system? Well, I don't need to (and don't want to) do all the math to show it's worse. EVERY bet you make has a built in house edge and thus will add to your overall expected loss.
Example:
Covering the DP and Pass lines together... EL = AvgBet * NumGames * HouseEdge = (10)*(30)*(-.0278) = -$8.34
When you cover the DP and Pass you'll never win on these initial bets but only lose when the 12 is rolled, which is 1/36 or 2.78% of the time. Thus, you've effectively DOUBLED the houses edge from just playing the DP alone!
Now, for EVERY SINGLE bet you add to this (other than odds, which have no long run value) you'll expect to lose more. If you place the 6 or 8 for $6, each one of those place bets carries a -1.515% house edge... and thus your EL for placing the 6 = $6 * (-.01515) = -$0.09. This might not be a lot, but as you can see it ADDS to your overall expected loss. Every other bet will continue to do the same... So the more bets you make, the more you can expect to lose.
The "safest" system, while still playing the game, is to make 1 bet... The table minimum on the lowest house edge table bet.
Playing it correctly means you've already won.
April 20th, 2016 at 5:21:51 PM
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Hey Romes...
You can only lose half your bet on 12. You did not calculate that in. Is that even possible? I don't know, but, it looks like a mix of dice odds that cant change and a hedge bet that can modify the EV which is not the copy of the HE here. Its two bets with one resolution. It sure seems possible. It seems that one resolution is greater than the sum of its bets. The other half is completely protected, or hedged. So even though the crap-wagons (12) has a HE of 2.78%, I feel its more important to pay attention to the real goal and that's your win rate in $$. That's my goal anyway. I would not step foot in a casino if I didn't win.
How do you account for that in your analysis?
You can only lose half your bet on 12. You did not calculate that in. Is that even possible? I don't know, but, it looks like a mix of dice odds that cant change and a hedge bet that can modify the EV which is not the copy of the HE here. Its two bets with one resolution. It sure seems possible. It seems that one resolution is greater than the sum of its bets. The other half is completely protected, or hedged. So even though the crap-wagons (12) has a HE of 2.78%, I feel its more important to pay attention to the real goal and that's your win rate in $$. That's my goal anyway. I would not step foot in a casino if I didn't win.
How do you account for that in your analysis?
April 20th, 2016 at 5:35:38 PM
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Quote: JoelDezeBank Roll of $200
...
Current Bank Roll is $211
If you started with $2000, and increased each bet 10-times, you would win $110, instead of $11 and you would be better off by $99. It's like you're just giving $99 of the money you should be earning back to the casino every time you do this.
April 22nd, 2016 at 9:06:03 AM
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Hi Romes,
I appreciate the kind reply.
This strategy is just a minor example of a much larger strategy I use. As you probably know, it's very difficult to try to explain the nuances of every strategy.
In the example above, I'm playing the PL + DPL at $10 each and covering the 12 bet with a $1. Total risk is $1 dollar not $10. The main goal is just to ensure I'm on the DPL with no fuss. If the 12 hits, I've covered the PL since the DPL pushes.
That's the first phase.
At any given time I could choose to play either PL Odds and remove the DPL wager entirely (but since the DPL is up already it makes no sense to do this). I can also choose not to wager on the place bets, wager $52 or $54 across if a roller gets hot, or cover specific number sets. I can also pull the DPL odds entirely and accept the $1 dollar loss.
In the end I use many different examples but I was basically stating that a person can in fact wager on both sides and cover the 12 at very little risk.
I play right way and dark side depending on the table and how much time I have. In the last 4 weeks I've made more than $8,000 in Atlantic City, Foxwoods and Twin Rivers. I thought this would just be a hobby for me but I have a good knack for timing 7s and going off on place bets when playing dark side.
I've built my own craps application which utilizes atmospheric noise when creating random number sets. I'm working through 10 sets of 1 million example rolls to create new combination strategies depending on risk appetite.
When it comes down to it, everything is about money management and risk appetite. I've gone down -$500 only to go back up +$900 in a session. Probability measurements don't inherently cover table momentum swings.
I own and operate Bustthebooks . com which I use for football probability. Craps is just a fun hobby that will keep me entertained until seasonal play kicks in. I toss out $25k-$50k per week once week 9 hits and the data is solid. There's numbers inside of numbers in every game. I like digging. :)
I appreciate the kind reply.
This strategy is just a minor example of a much larger strategy I use. As you probably know, it's very difficult to try to explain the nuances of every strategy.
In the example above, I'm playing the PL + DPL at $10 each and covering the 12 bet with a $1. Total risk is $1 dollar not $10. The main goal is just to ensure I'm on the DPL with no fuss. If the 12 hits, I've covered the PL since the DPL pushes.
That's the first phase.
At any given time I could choose to play either PL Odds and remove the DPL wager entirely (but since the DPL is up already it makes no sense to do this). I can also choose not to wager on the place bets, wager $52 or $54 across if a roller gets hot, or cover specific number sets. I can also pull the DPL odds entirely and accept the $1 dollar loss.
In the end I use many different examples but I was basically stating that a person can in fact wager on both sides and cover the 12 at very little risk.
I play right way and dark side depending on the table and how much time I have. In the last 4 weeks I've made more than $8,000 in Atlantic City, Foxwoods and Twin Rivers. I thought this would just be a hobby for me but I have a good knack for timing 7s and going off on place bets when playing dark side.
I've built my own craps application which utilizes atmospheric noise when creating random number sets. I'm working through 10 sets of 1 million example rolls to create new combination strategies depending on risk appetite.
When it comes down to it, everything is about money management and risk appetite. I've gone down -$500 only to go back up +$900 in a session. Probability measurements don't inherently cover table momentum swings.
I own and operate Bustthebooks . com which I use for football probability. Craps is just a fun hobby that will keep me entertained until seasonal play kicks in. I toss out $25k-$50k per week once week 9 hits and the data is solid. There's numbers inside of numbers in every game. I like digging. :)
“It’s a dog eat dog world out there and I’m wearing milkbone underwear .” – Norm Peterson
April 22nd, 2016 at 10:06:51 AM
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Quote: JoelDezeProbability measurements don't inherently cover table momentum swings.
What does that mean? Are you claiming you can identify times when the dice will behave differently based on "momentum"? I'm not sure what point you are making.
All the craps ideas I have read about making offsetting bets seem misguided to me. Every time you make a bet other than odds at the craps table, you are paying some small amount of expectation. Why would you want to pay the tax, then make another bet which offsets it with its own tax, then make a third bet with even more tax? You could just not bet at all for free. Seems odd.
I guess if you're just desperate to play and the table is a high minimum for you and you're willing to pay a lot of EV to make it more affordable, it makes sense, but that's about all I can think of.
April 22nd, 2016 at 10:18:31 AM
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Agreed there seems to be a lot of voodoo in this thread. Each and every single craps roll is an independent trial from every other roll with the same probabilities and odds. The dice don't get hot or cold. There is no momentum with the dice.
You might think the only bet you're exposing yourself to is the come out 12 cover, but each and every bet has it's own individual house edge against it and the sum of those bets against their edges is your overall expected value.
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet)
So while you think "only my protect 12 bet is at risk" you're actually taking negative expectation from each individual bet... and trying to cover it by hedging other bets.
You might think the only bet you're exposing yourself to is the come out 12 cover, but each and every bet has it's own individual house edge against it and the sum of those bets against their edges is your overall expected value.
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet)
So while you think "only my protect 12 bet is at risk" you're actually taking negative expectation from each individual bet... and trying to cover it by hedging other bets.
Playing it correctly means you've already won.
April 22nd, 2016 at 12:36:02 PM
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Quote: RomesAgreed there seems to be a lot of voodoo in this thread. Each and every single craps roll is an independent trial from every other roll with the same probabilities and odds. The dice don't get hot or cold. There is no momentum with the dice.
You might think the only bet you're exposing yourself to is the come out 12 cover, but each and every bet has it's own individual house edge against it and the sum of those bets against their edges is your overall expected value.
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet)
So while you think "only my protect 12 bet is at risk" you're actually taking negative expectation from each individual bet... and trying to cover it by hedging other bets.
I don't see that math at all. How is this visualized?
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet) would be 2.77 + 2.77 HE when all that can hurt you is a 12? That's doesn't add up.
Lose $1 35/36 you'll be $35 down.
Win 1/36 at $30.
Down -$5 in 36 rolls That's an eventual/evolved 1/36 or 2.77 HE same as 12. Not 5.44% HE as noted above.
Seems pretty basic what wrong with your formula or my story problem?
Because he is down $10 that's why
Last edited by: rushdl on Apr 22, 2016
April 22nd, 2016 at 12:52:21 PM
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Quote: rushdlI don't see that math at all. How is this visualized?
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet) would be 2.77 + 2.77 HE when all that can hurt you is a 12? That's doesn't add up.
Lose $1 35/36 you'll be $35 down.
Win 1/36 at $30.
Down -$5 in 36 rolls That's an eventual/evolved 1/36 or 2.77 HE same as 12. Not 5.44% HE as noted above.
Seems pretty basic what wrong with your formula or my story problem?
You're missing the outcome of the pass/DP bets. when the 12 comes, you lose on the P/DP AND you don't get fair odds on the payout for the 12.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
April 22nd, 2016 at 12:58:17 PM
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Quote: rushdlI don't see that math at all. How is this visualized?
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet) would be 2.77 + 2.77 HE when all that can hurt you is a 12? That's doesn't add up.
Lose $1 35/36 you'll be $35 down.
Win 1/36 at $30.
Down -$5 in 36 rolls That's an eventual/evolved 1/36 or 2.77 HE same as 12. Not 5.44% HE as noted above.
Did you remember to take into account the fact that the pass and don't pass bets don't cancel each other out on the 12?
If you bet $5 on each, anything other than a 12 results in a net -1, and a 12 results in +30 for the 12 bet, -5 for the pass, and a push for the don't pass, or a net +25. so the total in 36 rolls is -10. Of course, since you are betting a total of $11 on the comeout, technically that's a HE of 2.525%. If you make the pass and DP bets $10 each, a 12 is +30 - 10 = +20, over 36 rolls x $21 bet per roll = 2.6455%.
Then again, I am not always certain my definition of HE is the same as everybody else's.
April 22nd, 2016 at 1:05:44 PM
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Quote: rdw4potusYou're missing the outcome of the pass/DP bets. when the 12 comes, you lose on the P/DP AND you don't get fair odds on the payout for the 12.
You just said you lose P/DP... and you do not,
you only lose pass.
April 22nd, 2016 at 1:11:06 PM
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Your math is wrong... as you're not accounting for all scenarios.
EV(pass) = EV(come out 7, 11) + EV (come out 2,3,12) + EV(other number)
EV(don't pass) = EV(come out 7,11) + EV(come out 2,3,12) + EV(other number)
EV(12 bet) = EV(come out 12) + EV(other number)
...For example, if you want to know what your EV is assuming a come out 7, then:
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet) = (+1 unit) + (-1 unit) + (- 12 bet)
EV(pass) = EV(come out 7, 11) + EV (come out 2,3,12) + EV(other number)
EV(don't pass) = EV(come out 7,11) + EV(come out 2,3,12) + EV(other number)
EV(12 bet) = EV(come out 12) + EV(other number)
...For example, if you want to know what your EV is assuming a come out 7, then:
EV(come out) = EV(pass) + EV(don't pass) + EV(12 bet) = (+1 unit) + (-1 unit) + (- 12 bet)
Playing it correctly means you've already won.
April 22nd, 2016 at 1:12:55 PM
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Quote: rushdlYou just said you lose P/DP... and you do not,
you only lose pass.
right, but you push the dont pass. So you net lose money on those bets on a 12. And you're short-paid on the payout on the 12.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
April 22nd, 2016 at 1:14:53 PM
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Quote: ThatDonGuyDid you remember to take into account the fact that the pass and don't pass bets don't cancel each other out on the 12?
If you bet $5 on each, anything other than a 12 results in a net -1, and a 12 results in +30 for the 12 bet, -5 for the pass, and a push for the don't pass, or a net +25. so the total in 36 rolls is -10. Of course, since you are betting a total of $11 on the comeout, technically that's a HE of 2.525%. If you make the pass and DP bets $10 each, a 12 is +30 - 10 = +20, over 36 rolls x $21 bet per roll = 2.6455%.
Then again, I am not always certain my definition of HE is the same as everybody else's.
I missed the extra nickel loss on the DP that's it, so the posters scenario is just like Romes said said, 5.54% HE on his come-out bets with hedge. I don't like that hedge personally and am trying to ask a question that if you do NOT use the 12 hedge. Sorry if I got mixed up.
If you do not use the 12 hedge for a dollar, I say, I am trying to say, that the HE is 2.77 but the EV is 1.39.
Can anybody back that up?
April 25th, 2016 at 7:09:23 AM
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Quote: rushdlSeems pretty basic what wrong with your formula or my story problem?
As others have said, when the comeout roll is a 12, you don't profit $30, you only profit $25 because you win the 12 (+$30) but lose the pass (-$5) and PUSH the don't (=$0). Thus the EV combined is -$10/36 which is about -28 cents. I'm rounding here to avoid too many decimals.
This breaks down as follows:
-1.4% on the $5 pass for -7 cents
-1.4% on the $5 don't for -7 cents
-14% on the $1 12 for -14 cents
As far as house edge goes, there are actually multiple ways to think about this. The way I would calculate it is similar to ThatDonGuy above. For $5 each on p/dp and $1 on the 12, your expected value is -10/36 dollars. Based on your 11 dollars of action, the house edge is:
(-10/36)/11 = -10/396 which is about -2.525%
If you instead went $10 on p/dp, your expected value is -15/36 dollars, and the house edge would be:
(-15/36)/21 = -15/756 which is about -1.984%
Notice the more you risk on p/dp, the lower the house edge will get, approaching 1.4%, because the bet on 12 becomes ever less relevant in the edge calculation. However, your expected value in terms of actual dollars lost will be worse.
Last edited by: MrGoldenSun on Apr 25, 2016
April 25th, 2016 at 9:54:19 AM
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Quote: rushdlIf you do not use the 12 hedge for a dollar, I say, I am trying to say, that the HE is 2.77 but the EV is 1.39.
Can anybody back that up?
Romes' statement was that EV is additive. EV is in dollars. House edge is a percent and is NOT additive--the house edge will be a weighted average of the individual bets. Since pass and don't pass are very similar at about -1.4%, then making a pair of -1.4% bets keeps the overall edge at about -1.4%.
Assuming the don't pass is a push on a 12, then if you bet $5 on pass and $5 on don't, you have the following:
Comeout roll is a non-12: chance of this is 35/36, result is $0
Comeout roll is a 12: chance of this is 1/36, result is -$5
EV is (35/36)*(0) + (1/36)*(-5) = -5/36 which is approximately -0.139, meaning you will lose an expected 13.9 cents. About half of this is from the $5 pass bet and the other half is from the $5 don't pass.
House edge is EV / Money Wagered. At least that's the way it's typically calculated, but note that because some bets take multiple rolls to resolve, there are some alternative house edge methods for craps specifically that some prefer. But taking the method I think makes sense, you have wagered a total of 10 dollars with an expected result of -0.139 dollars, meaning the house edge is:
-0.139/10 = -0.0139 which is -1.39%
April 25th, 2016 at 10:46:16 AM
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Quote: MrGoldenSunRomes' statement was that EV is additive. EV is in dollars. House edge is a percent and is NOT additive--the house edge will be a weighted average of the individual bets. Since pass and don't pass are very similar at about -1.4%, then making a pair of -1.4% bets keeps the overall edge at about -1.4%.
Assuming the don't pass is a push on a 12, then if you bet $5 on pass and $5 on don't, you have the following:
Comeout roll is a non-12: chance of this is 35/36, result is $0
Comeout roll is a 12: chance of this is 1/36, result is -$5
EV is (35/36)*(0) + (1/36)*(-5) = -5/36 which is approximately -0.139, meaning you will lose an expected 13.9 cents. About half of this is from the $5 pass bet and the other half is from the $5 don't pass.
House edge is EV / Money Wagered. At least that's the way it's typically calculated, but note that because some bets take multiple rolls to resolve, there are some alternative house edge methods for craps specifically that some prefer. But taking the method I think makes sense, you have wagered a total of 10 dollars with an expected result of -0.139 dollars, meaning the house edge is:
-0.139/10 = -0.0139 which is -1.39%
Totally backed up thanks bro.
