In my observation when a casino offers high odds like 10X or 20X very few people take full odds. A brave player can exploit this by employing the oddsman strategy in which you convince people to let you bet their remaining odds bet. In this way you get the privilege of taking odds without the penalty of having a pass bet. You won't be able to role the dice but you'll play a break even game.
craps
- Thread starter sagefr0g
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just one more question lmao from the guy who don't have a clue.
i got the understanding that taking the odds on a pass line bet is a good thing, well and i guess it's a good thing from the long term strategy sort of idea.
so ok let's talk about luck (good luck) just a bit here and how we know it's a good chance the shooter is gonna 7 out before making the point.
i mean say you have a unit on the pass line and some point is established.
so ok like stated it's a good long term strategy to take the free odds. so but what i'm wondering is from that perspective is it some kind of big sin to for what ever reason one just decides to forego the odd's bet and just ride out the pass line bet. i mean maybe the shooter 7 outs so you don't lose as much in that instance sort of thing.
in other words is it a losing ploy to try and every so often get lucky on guessing the shooter will 7 out to where you wouldn't lose the free odds money you would have normally bet? i guess it might be a fifty fifty sort of thing where your guessing over time right or wrong would even out so in the end long term you end up losing shaving of some of the HA.
edit: ==>> ok and as a follow up quesiton. say one does do something chicken sh!t like that lmao. well then i guess it would be best to pull that kind of a stunt maybe if the point is a 4 or a 10, something like that since the 4 or the 10 might take longer to achieve the point giving the seven a better chance of ruining your bet? where you might reach the point quicker for the 5,6,8,9 sort of thing hoping to beat the seven rearing it's head?
i got the understanding that taking the odds on a pass line bet is a good thing, well and i guess it's a good thing from the long term strategy sort of idea.
so ok let's talk about luck (good luck) just a bit here and how we know it's a good chance the shooter is gonna 7 out before making the point.
i mean say you have a unit on the pass line and some point is established.
so ok like stated it's a good long term strategy to take the free odds. so but what i'm wondering is from that perspective is it some kind of big sin to for what ever reason one just decides to forego the odd's bet and just ride out the pass line bet. i mean maybe the shooter 7 outs so you don't lose as much in that instance sort of thing.
in other words is it a losing ploy to try and every so often get lucky on guessing the shooter will 7 out to where you wouldn't lose the free odds money you would have normally bet? i guess it might be a fifty fifty sort of thing where your guessing over time right or wrong would even out so in the end long term you end up losing shaving of some of the HA.
edit: ==>> ok and as a follow up quesiton. say one does do something chicken sh!t like that lmao. well then i guess it would be best to pull that kind of a stunt maybe if the point is a 4 or a 10, something like that since the 4 or the 10 might take longer to achieve the point giving the seven a better chance of ruining your bet? where you might reach the point quicker for the 5,6,8,9 sort of thing hoping to beat the seven rearing it's head?
I'm pretty sure you're miscalculating. The Come and Pass bets are independent bets, so their house advantages don't depend on when you make the bet.
Well, survival and becoming wealthy are two different goals. If your goal is to survive, you shouldn't place as many odds because your variance goes up significantly. If your goal is to become wealthy, you shouldn't play craps at all.
This topic has been discussed here before, but the bottom line is that Wong himself admits he hasn't made enough dice throws to separate out his winnings (if there were any) from short-term variance.
http://www.blackjackinfo.com/bb/showthread.php?t=4863
Edit: To be clear, I think all people agree dice control would lead to AP. I believe most people agree dice control can be done by someone in the world. I believe few people believe Wong is one of those people. And I believe almost nobody believes he has proven it mathematically. In that regard, I'm pretty sure people agreed Wong's claims were baseless, even if they are true.
Well, survival and becoming wealthy are two different goals. If your goal is to survive, you shouldn't place as many odds because your variance goes up significantly. If your goal is to become wealthy, you shouldn't play craps at all.
This topic has been discussed here before, but the bottom line is that Wong himself admits he hasn't made enough dice throws to separate out his winnings (if there were any) from short-term variance.
http://www.blackjackinfo.com/bb/showthread.php?t=4863
Edit: To be clear, I think all people agree dice control would lead to AP. I believe most people agree dice control can be done by someone in the world. I believe few people believe Wong is one of those people. And I believe almost nobody believes he has proven it mathematically. In that regard, I'm pretty sure people agreed Wong's claims were baseless, even if they are true.
Yea... Since dice control is a matter of skill and not just making decisions, it seems difficult if not impossible I don't see how you can put an EV on it. I remember reading on Wong's website that the skill is in being able to throw non-7s. If such is the case, on average, what does the probability of a 7 (or non-7) have to be in order to break the HA on... lets say the pass line with laying odds at a casino offering 3x odds (.471 HA)?
dream buster, lmao.
yeah it's to survive and hopefully experience some good luck.
but i'm at odds (pun intended) on your statement ,"you shouldn't place as many odds because your variance goes up significantly" relative to the survival issue. lol, i'm not trying to be argumentative but say if one can afford to take as much of the free true odds as offered then that shouldn't become a survival issue should it? or am i missing something?
i mean i was thinking taking as much of the free true odds as possible would enhance your survival chances. is that wrong?
but say if you have a limited trip bankroll then yeah i could see that. maybe that's your point?
yeah it's to survive and hopefully experience some good luck.
but i'm at odds (pun intended) on your statement ,"you shouldn't place as many odds because your variance goes up significantly" relative to the survival issue. lol, i'm not trying to be argumentative but say if one can afford to take as much of the free true odds as offered then that shouldn't become a survival issue should it? or am i missing something?
i mean i was thinking taking as much of the free true odds as possible would enhance your survival chances. is that wrong?
but say if you have a limited trip bankroll then yeah i could see that. maybe that's your point?
I'm pretty sure I'm not I'm going to open your mind to something that most websites or books get into...because they all just copy each other when it comes to basic craps wager knowledge.
Let's start with this: Do you realize that both the Pass Line and Come bets are actually two-phase bets?
PHASE 1: The come out roll. At this point, there are 8 ways to win (7 or 11) vs. 4 ways to lose (2, 3, or 12) this even-money wager. Any other number rolled becomes the point number (no win or loss...Phase 2 begins). Evaluating this 8:4 (a.k.a. 2:1) ratio, you see the player has a 33.3% edge in the come out phase.
PHASE 2: The point phase. Now, the Pass or Come wager's advantage swings to the house. As you can see in my table earlier in this thread, the house's edge is either ~9% (point of 6 or 8), 20% (5 or 9), or ~33% (4 or 10). This is due to the strength of the 7 vs. the point number.
The 1.41% house edge of the Pass or Come bets are a factor of both phases. If you understand the concept of a two-phase wager, look at this common scenario:
You make a $10 Pass Line bet, a 5 is rolled, and it becomes the table's point. Now this wager is looking at a 20% disadvantage, but you take 1X Odds ($10 more) to cut that in half essentially. Now you're interested in getting a Come bet on the board, but what are the ramifications of this move?
* The Come bet will undergo the come out phase. A 2, 3, or 12 will make it a loser, while the 11 will make it a winner.
* The 7, which is the strongest number of the dice, and the key reason why the Come (or Pass) bet has a 33% edge in Phase 1 (or a relatively low overall edge of 1.4%) will now defeat your Pass Line with Odds bet.
* Therefore, the strength of the follow-up Come bet (the 7) is now neutralized...or even worse, a net loss, unless the wager is at least $20.
Now, the question is: How independent are the Pass Line and Come bets in this very common situation? They aren't. A 7 during the Come bet's come out phase is no longer a victory without doing something like a Martingale system for each subsequent volley. Once you make the 7 your enemy during a Pass or Come bet's come out like this, you're not buying yourself anything.
Just droppin some knowledge down for ya.
good luck
I'm going to open your mind to something that most websites or books get into...because they all just copy each other when it comes to basic craps wager knowledge.Let's start with this: Do you realize that both the Pass Line and Come bets are actually two-phase bets?
PHASE 1: The come out roll. At this point, there are 8 ways to win (7 or 11) vs. 4 ways to lose (2, 3, or 12) this even-money wager. Any other number rolled becomes the point number (no win or loss...Phase 2 begins). Evaluating this 8:4 (a.k.a. 2:1) ratio, you see the player has a 33.3% edge in the come out phase.
PHASE 2: The point phase. Now, the Pass or Come wager's advantage swings to the house. As you can see in my table earlier in this thread, the house's edge is either ~9% (point of 6 or 8), 20% (5 or 9), or ~33% (4 or 10). This is due to the strength of the 7 vs. the point number.
The 1.41% house edge of the Pass or Come bets are a factor of both phases. If you understand the concept of a two-phase wager, look at this common scenario:
You make a $10 Pass Line bet, a 5 is rolled, and it becomes the table's point. Now this wager is looking at a 20% disadvantage, but you take 1X Odds ($10 more) to cut that in half essentially. Now you're interested in getting a Come bet on the board, but what are the ramifications of this move?
* The Come bet will undergo the come out phase. A 2, 3, or 12 will make it a loser, while the 11 will make it a winner.
* The 7, which is the strongest number of the dice, and the key reason why the Come (or Pass) bet has a 33% edge in Phase 1 (or a relatively low overall edge of 1.4%) will now defeat your Pass Line with Odds bet.
* Therefore, the strength of the follow-up Come bet (the 7) is now neutralized...or even worse, a net loss, unless the wager is at least $20.
Now, the question is: How independent are the Pass Line and Come bets in this very common situation? They aren't. A 7 during the Come bet's come out phase is no longer a victory without doing something like a Martingale system for each subsequent volley. Once you make the 7 your enemy during a Pass or Come bet's come out like this, you're not buying yourself anything.
Just droppin some knowledge down for ya.
good luck
I measure AP craps throwing skill differently than most everyone else. Most, if not all, of what you see uses SRR...sevens-to-rolls ratio. This is a measure of whether you can reduce the appearance of the 7 in the grand scheme of things.
I use a proportion of on-axis throws to all throws, which can be stated like 1-in-4...which means that an on-axis throw can be expected in about every 4 rolls. Although there are a lot of wagers, I am an advocate of sticking to the "core bets" of things like Pass Line, Odds, and Place bets. You use this proportion to blend the dice' natural probabilities with whatever pre-selected set for the situation. Without elaborating unnecessarily (since most folks won't buy what I'm selling :joker, you can determine what your advantage is with respect to your skill, wagers, and relative bet amounts between the wagers.
Something more to think about.
good luck
I use a proportion of on-axis throws to all throws, which can be stated like 1-in-4...which means that an on-axis throw can be expected in about every 4 rolls. Although there are a lot of wagers, I am an advocate of sticking to the "core bets" of things like Pass Line, Odds, and Place bets. You use this proportion to blend the dice' natural probabilities with whatever pre-selected set for the situation. Without elaborating unnecessarily (since most folks won't buy what I'm selling :joker
, you can determine what your advantage is with respect to your skill, wagers, and relative bet amounts between the wagers.Something more to think about.
good luck
I can't answer your question precisely, but I can share the results from the math I have done on craps.
(1) I'm considering dice control where two axes have been fixed. That is, dice are thrown such that they tumble only on one axis, thus limiting the selection to 4 numbers instead of 6.
(2) If the dice are not thrown correctly, they end up being random; the alternative that wrongly-thrown dice don't come up random is still a form of dice control.
(3) Given this scenario, certain combinations of starting sets will give you an amazing 70% edge over the house.
(4) Since it's highly unlikely that people will make perfect throws every time, more importantly would be the percentage of throws one needs to make in order to overcome the house edge. It's about 3% with no odds, less with odds. So, realistically, you only need to control the dice 3% of the time in order to win long-term.
The problem, of course, is the ability to statistically demonstrate a 5% success rate. If you assume that no control is 100% random and dice control is 5% non-random, 95% random, you still need an absurd number of trials to demonstrate significance. I had calculated a number on the order of 100,000 throws.
You're missing something - variance. Let's say that you were allowed to take unlimited odds on your bet. The best EV play is to put your entire bankroll down on the odds. Unfortunately, if you lose, you lose it all. That is variance. If you put down a lot of odds, you are decreasing the number of "bad events" you need in a streak to bust you.
You are ignoring the possibility that both bets eventually win. For example, if you include the possibility that you:
(1) Place a pass bet.
(2) Roll a point.
(3) Place a come bet.
(4) Roll a second point.
(5) and (6) Make both points
I think you will see that the EV's come out exactly the same.
Think about it this way: let's say you join a table when point has already been established. You are allowed to make a come bet - does that come bet have higher or lower EV than if you waited until the next comeout roll to place a Pass bet? If you think it does, what is the difference between the two scenarios?
(1) I'm considering dice control where two axes have been fixed. That is, dice are thrown such that they tumble only on one axis, thus limiting the selection to 4 numbers instead of 6.
(2) If the dice are not thrown correctly, they end up being random; the alternative that wrongly-thrown dice don't come up random is still a form of dice control.
(3) Given this scenario, certain combinations of starting sets will give you an amazing 70% edge over the house.
(4) Since it's highly unlikely that people will make perfect throws every time, more importantly would be the percentage of throws one needs to make in order to overcome the house edge. It's about 3% with no odds, less with odds. So, realistically, you only need to control the dice 3% of the time in order to win long-term.
The problem, of course, is the ability to statistically demonstrate a 5% success rate. If you assume that no control is 100% random and dice control is 5% non-random, 95% random, you still need an absurd number of trials to demonstrate significance. I had calculated a number on the order of 100,000 throws.
You're missing something - variance. Let's say that you were allowed to take unlimited odds on your bet. The best EV play is to put your entire bankroll down on the odds. Unfortunately, if you lose, you lose it all. That is variance. If you put down a lot of odds, you are decreasing the number of "bad events" you need in a streak to bust you.
You are ignoring the possibility that both bets eventually win. For example, if you include the possibility that you:
(1) Place a pass bet.
(2) Roll a point.
(3) Place a come bet.
(4) Roll a second point.
(5) and (6) Make both points
I think you will see that the EV's come out exactly the same.
Think about it this way: let's say you join a table when point has already been established. You are allowed to make a come bet - does that come bet have higher or lower EV than if you waited until the next comeout roll to place a Pass bet? If you think it does, what is the difference between the two scenarios?
Nope. Maybe you should define EV as opposed to house edge/advantage.
This is a different situation...it's no different than if you just made a Pass Line bet with nothing else out there.
Do you see a difference in overall house edge with a solo Come bet as opposed to making a Come bet when you have all the numbers Placed? Of course. You have altered your payoffs from just winning on 7 or 11 and losing on 2, 3, or 12. Now, you will win with 4, 5, 6, 8, 9, or 10...but will lose a lot more if the 7 rolls.
You can't tell me that the latter situation has the same house edge than just making a solo Come bet.
Let's go back to our original Pass Line followed up with a Come bet. Once the Pass Line has traveled and the Come bet is coming out, changing the 7's effect on your overall action changes the edge on that overall action. House edge is a function of amount at risk and what can be won with respect to probability of winning and losing.
Think of your action holistically. This is a prime reason how craps differs from blackjack. If a 7 will win one of your bets, but causes another to lose, the edge on your play is different overall than if you just looked at each bet separately.
good luck
This is a different situation...it's no different than if you just made a Pass Line bet with nothing else out there.
Do you see a difference in overall house edge with a solo Come bet as opposed to making a Come bet when you have all the numbers Placed? Of course. You have altered your payoffs from just winning on 7 or 11 and losing on 2, 3, or 12. Now, you will win with 4, 5, 6, 8, 9, or 10...but will lose a lot more if the 7 rolls.
You can't tell me that the latter situation has the same house edge than just making a solo Come bet.
Let's go back to our original Pass Line followed up with a Come bet. Once the Pass Line has traveled and the Come bet is coming out, changing the 7's effect on your overall action changes the edge on that overall action. House edge is a function of amount at risk and what can be won with respect to probability of winning and losing.
Think of your action holistically. This is a prime reason how craps differs from blackjack. If a 7 will win one of your bets, but causes another to lose, the edge on your play is different overall than if you just looked at each bet separately.
good luck
Here's a sample table of how you can like your skill (measured in proportion of on-axis throws to all throws), the pre-set, and the wager. This is a very specific example regarding a particular set and a bevy of place bet combinations.
Just an example of my way of looking at craps...coming soon to a bookstore near you
good luck
Just an example of my way of looking at craps...coming soon to a bookstore near you
good luck
Attachments
If you don't think that house edge changes when you have more than one wager out, do the math for the Place 6 & 8 combo, where you bet $6 on each. I'll start you out...
* Chances of winning on a given roll: 10 (5 for the 6, 5 for the 8).
* Chances of losing on a given roll: 6 (for the 7).
* Amount at risk: $12.
* Amount that can be won with a winning wager: $7.
Calculate the house edge of this combination and tell me if it equals 1.52% (the edge on Place 6 OR 8...a solo bet).
good luck
I think we're making progress. I'm just trying to stimulate your mind beyond the run-of-the-mill copy and paste stuff you see on the net or in most books.
* Chances of winning on a given roll: 10 (5 for the 6, 5 for the 8).
* Chances of losing on a given roll: 6 (for the 7).
* Amount at risk: $12.
* Amount that can be won with a winning wager: $7.
Calculate the house edge of this combination and tell me if it equals 1.52% (the edge on Place 6 OR 8...a solo bet).
good luck
I think we're making progress. I'm just trying to stimulate your mind beyond the run-of-the-mill copy and paste stuff you see on the net or in most books.
yes, i understand your point about variance. what's hanging me up is that it's supposed to be a fact that the more free odds you take the more the house edge is lowered (i guess in the long run). like the snippet from the bottom portion of this site:
http://www.nextshooter.com/odds quoted below:
When the Free Odds bet doesn't help
There's a subtlety of the Odds bet that's important to understand. Let's say you made a $10 Pass Line bet, and a point has been made. Now you have the opportunity to make a Free Odds bet. However, if you make the Odds bet it won't increase your chances of winning, and your expected loss will be the same whether you make the Odds bet or not: 1.41% of your $10 Pass Line bet. So if making the Free Odds bet doesn't increase your chances of winning, and doesn't decrease your expected loss, why would you make it?
To answer this question we have to back up a bit. The Free Odds bet is a good deal when you put money on it that you were going to bet anyway. If you wanted to bet about $10 per round, then you'll get a better deal by betting $2 on the Pass Line and $10 on the Odds (on a 5x table), vs. betting the $10 on the Pass Line and taking no odds. With no odds your expected loss is $10 x 1.41% = $0.14 per roll. But by putting $2 on the Pass Line and $10 on the Odds, your expected loss is only $12 x 0.326% = $0.04. Your expected loss is smaller, and you have a greater chance of walking away a winner. So, for whatever amount you want to bet per round, get as much of it on the Free Odds as possible! This is the single most important thing to know about playing craps.
so to me the point of what the snippet is saying is if the original pass line bet added to the amount bet on the free odds is understood to be your unit (ie. the amount your willing to bet on a round) then you should be willing to take those odds and bet that amount despite variance.
no?
lol, i don't doubt i'm still missing something. totally new at all this.... help :cry:
http://www.nextshooter.com/odds quoted below:
When the Free Odds bet doesn't help
There's a subtlety of the Odds bet that's important to understand. Let's say you made a $10 Pass Line bet, and a point has been made. Now you have the opportunity to make a Free Odds bet. However, if you make the Odds bet it won't increase your chances of winning, and your expected loss will be the same whether you make the Odds bet or not: 1.41% of your $10 Pass Line bet. So if making the Free Odds bet doesn't increase your chances of winning, and doesn't decrease your expected loss, why would you make it?
To answer this question we have to back up a bit. The Free Odds bet is a good deal when you put money on it that you were going to bet anyway. If you wanted to bet about $10 per round, then you'll get a better deal by betting $2 on the Pass Line and $10 on the Odds (on a 5x table), vs. betting the $10 on the Pass Line and taking no odds. With no odds your expected loss is $10 x 1.41% = $0.14 per roll. But by putting $2 on the Pass Line and $10 on the Odds, your expected loss is only $12 x 0.326% = $0.04. Your expected loss is smaller, and you have a greater chance of walking away a winner. So, for whatever amount you want to bet per round, get as much of it on the Free Odds as possible! This is the single most important thing to know about playing craps.
so to me the point of what the snippet is saying is if the original pass line bet added to the amount bet on the free odds is understood to be your unit (ie. the amount your willing to bet on a round) then you should be willing to take those odds and bet that amount despite variance.
no?
lol, i don't doubt i'm still missing something. totally new at all this.... help :cry:
Place 6: (5/36*7/6 + 6/36*-1)/(5/36+6/36) = -0.01515
Pass: 8/36*1 + 4/36*-1 + 6/36*(3/36*1 + 6/36*-1)/(3/36 + 6/36) + 8/36*(4/36*1 + 6/36*-1)/(4/36 + 6/36) + 10/36*(5/36*1 + 6/36*-1)/(5/36 + 6/36) = -0.01414
Pass line, placing a 6 or 8 if you make point: (8/36*1 + 4/36*-1 + 6/36*(3/36*(1+(5/36*7/6 + 6/36*-1)) + 5/36*(7/6+(3/36*1 + 6/36*-1)/(3/36 + 6/36)) + 6/36*-1)/(3/36 + 5/36 + 6/36) + 8/36*(4/36*(1+(5/36*7/6 + 6/36*-1)) + 5/36*(7/6 + (4/36*1 + 6/36*-1)/(4/36 + 6/36)) + 6/36*-1)/(4/36 + 5/36 + 6/36) + 10/36*(5/36*(1+(5/36*7/6 + 6/36*-1)) + 5/36*(7/6 + (5/36*1 + 6/36*-1)/(5/36 + 6/36)) + 6/36*-2)/(5/36 + 5/36 + 6/36))/(8/36*1 + 4/36*1 + 6/36*2 + 8/36*2 + 10/36*2) = -0.01455
Not surprisingly, this is exactly the weighted average of the EV's of the bets separately.
Independent Bets: (1*(-0.01414) + (24/36)*(-0.01515)) / (1 + 24/36) = -0.01455
I'm guessing that if you think you're gaining an advantage over the house, you're probably omitting the fact that after you make point you still have a place bet on the table (which has negative EV) and after you win your place bet you still have your pass bet on the table (which has negative EV at that point).
Ignoring the bets left on the table yields: (8/36*1 + 4/36*-1 + 6/36*(3/36*1 + 5/36*7/6 + 6/36*-1)/(3/36 + 5/36 + 6/36) + 8/36*(4/36*1 + 5/36*7/6 + 6/36*-1)/(4/36 + 5/36 + 6/36) + 10/36*(5/36*1 + 5/36*7/6 + 6/36*-2)/(5/36 + 5/36 + 6/36))/(8/36*1 + 4/36*1 + 6/36*2 + 8/36*2 + 10/36*2) = +0.01264
Pass: 8/36*1 + 4/36*-1 + 6/36*(3/36*1 + 6/36*-1)/(3/36 + 6/36) + 8/36*(4/36*1 + 6/36*-1)/(4/36 + 6/36) + 10/36*(5/36*1 + 6/36*-1)/(5/36 + 6/36) = -0.01414
Pass line, placing a 6 or 8 if you make point: (8/36*1 + 4/36*-1 + 6/36*(3/36*(1+(5/36*7/6 + 6/36*-1)) + 5/36*(7/6+(3/36*1 + 6/36*-1)/(3/36 + 6/36)) + 6/36*-1)/(3/36 + 5/36 + 6/36) + 8/36*(4/36*(1+(5/36*7/6 + 6/36*-1)) + 5/36*(7/6 + (4/36*1 + 6/36*-1)/(4/36 + 6/36)) + 6/36*-1)/(4/36 + 5/36 + 6/36) + 10/36*(5/36*(1+(5/36*7/6 + 6/36*-1)) + 5/36*(7/6 + (5/36*1 + 6/36*-1)/(5/36 + 6/36)) + 6/36*-2)/(5/36 + 5/36 + 6/36))/(8/36*1 + 4/36*1 + 6/36*2 + 8/36*2 + 10/36*2) = -0.01455
Not surprisingly, this is exactly the weighted average of the EV's of the bets separately.
Independent Bets: (1*(-0.01414) + (24/36)*(-0.01515)) / (1 + 24/36) = -0.01455
I'm guessing that if you think you're gaining an advantage over the house, you're probably omitting the fact that after you make point you still have a place bet on the table (which has negative EV) and after you win your place bet you still have your pass bet on the table (which has negative EV at that point).
Ignoring the bets left on the table yields: (8/36*1 + 4/36*-1 + 6/36*(3/36*1 + 5/36*7/6 + 6/36*-1)/(3/36 + 5/36 + 6/36) + 8/36*(4/36*1 + 5/36*7/6 + 6/36*-1)/(4/36 + 5/36 + 6/36) + 10/36*(5/36*1 + 5/36*7/6 + 6/36*-2)/(5/36 + 5/36 + 6/36))/(8/36*1 + 4/36*1 + 6/36*2 + 8/36*2 + 10/36*2) = +0.01264
Nice work, but what's this have to do with determining the combined house edge on Placing 6 AND 8? Why is the Pass Line included?
With your example, I am assuming you are evaluating the Pass Line and a Place 6 OR 8 to compliment it. And that's cool, because at least I see you are on board with the concept of house edge on your overall action.
Not sure why you'd guess that I think that combining wagers can gain an edge over the house. The only way to gain an advantage in craps without cheating is to manipulate the roll of the dice...namely by confining thier spin around 1 of the 3 axes.
I think we've just gotten completely off on two different tracks here callipygan...and that's fine, no hard feelings. But I do want you to go back to the original example of Pass Line with a Come Bet follow-up. Use the math to calculate what happens when you do this. It may only seem to be a small difference to have the Come Bet's advantage with the 7 during its come out neutralized, but just do the math. I've done it, I just want you to work through it and see the light.
It's all good...we'll get there!
good luck
With your example, I am assuming you are evaluating the Pass Line and a Place 6 OR 8 to compliment it. And that's cool, because at least I see you are on board with the concept of house edge on your overall action.
Not sure why you'd guess that I think that combining wagers can gain an edge over the house. The only way to gain an advantage in craps without cheating is to manipulate the roll of the dice...namely by confining thier spin around 1 of the 3 axes.
I think we've just gotten completely off on two different tracks here callipygan...and that's fine, no hard feelings. But I do want you to go back to the original example of Pass Line with a Come Bet follow-up. Use the math to calculate what happens when you do this. It may only seem to be a small difference to have the Come Bet's advantage with the 7 during its come out neutralized, but just do the math. I've done it, I just want you to work through it and see the light.
It's all good...we'll get there!
good luck
It's a general example of adding independent bets. That is, if you bet A and bet B, your combined EV is going to be [EV(A) + EV(B)]/[P(A)+P(B)]. There's no difference between betting Place 6 without a pass bet and Place 6 when point has been established - the EV for the combined play of Pass -> Place 6 when point is made is simply [EV(Pass) + P(point)*EV(Place 6)]/[1 + P(point)].
Likewise, EV(Place 6 AND 8) = [EV(Place 6) + EV(Place 8)]/[1 + 1] = EV(Place 6) = EV(Place 8) = -0.01515.
This is a very simple calculation to make. Betting Pass, then betting Come, carries an EV of:
EV(total) = [EV(Pass) + P(point)*EV(Come)]/[1 + P(point)]
Since EV(Pass) = EV(Come),
EV(total) = EV(Pass) = EV(Come) = -0.01414
There's no difference between placing the Come bet after placing a Pass bet, and placing a Come bet without a Pass bet. The two bets are independent.
Likewise, EV(Place 6 AND 8) = [EV(Place 6) + EV(Place 8)]/[1 + 1] = EV(Place 6) = EV(Place 8) = -0.01515.
This is a very simple calculation to make. Betting Pass, then betting Come, carries an EV of:
EV(total) = [EV(Pass) + P(point)*EV(Come)]/[1 + P(point)]
Since EV(Pass) = EV(Come),
EV(total) = EV(Pass) = EV(Come) = -0.01414
There's no difference between placing the Come bet after placing a Pass bet, and placing a Come bet without a Pass bet. The two bets are independent.
I give up. If you're not willing to perform the same mathematical exercize as you did before with the Place 6 & 8, that's fine with me.
But I will...
The details, again:
* Chances of winning on a given roll: 10 (5 for the 6, 5 for the 8).
* Chances of losing on a given roll: 6 (for the 7).
* Amount at risk: $12.
* Amount that can be won with a winning wager: $7.
Using the same formula you used earlier...
Place 6 & 8: (10/36*7/12 + 6/36*-1)/(10/36+6/36) = -1.05%
Place 6 OR 8: (5/36*7/6 + 6/36*-1)/(5/36+6/36) = -1.52%
Total action vs. looking at each bet in a bubble. And this is mild compared to the Pass then Come example. That part of my work took considerable time via pencil and paper, and it was just to illustrate a simple idea to the reader:
Another test:
Run the edge calculation for a scenario where the player is making a $10 Pass Line bet, but is working a $12 Place 6 on the come out roll.
But I will...
The details, again:
* Chances of winning on a given roll: 10 (5 for the 6, 5 for the 8).
* Chances of losing on a given roll: 6 (for the 7).
* Amount at risk: $12.
* Amount that can be won with a winning wager: $7.
Using the same formula you used earlier...
Place 6 & 8: (10/36*7/12 + 6/36*-1)/(10/36+6/36) = -1.05%
Place 6 OR 8: (5/36*7/6 + 6/36*-1)/(5/36+6/36) = -1.52%
Total action vs. looking at each bet in a bubble. And this is mild compared to the Pass then Come example. That part of my work took considerable time via pencil and paper, and it was just to illustrate a simple idea to the reader:
Another test:
Run the edge calculation for a scenario where the player is making a $10 Pass Line bet, but is working a $12 Place 6 on the come out roll.
As you probably know, in craps, the Place bet is a non-contract wager that is essentially a new bet each roll it is working. So nothing is "left on the table" after a win, loss, or push.
good luck at the tables...I don't think we're getting anywhere with this one The bottom line is that craps has a few solid bets, a lot of bad bets, and only one method to beat the house.
good luck at the tables...I don't think we're getting anywhere with this one
The bottom line is that craps has a few solid bets, a lot of bad bets, and only one method to beat the house.Okay, one last try. No math, this time.
Let's say that I walk up to the table and place a bet on the 6 and 8. You contend the house would expect to make 1.05% off of me.
Let's say that I walk up to the table and place a bet on the 6, and you walk up to the table and place a bet on the 8. You would say that the house would expect to make 1.52% off of each of us.
Let's say that I give you a bet, then we walk up to the table together and I place a bet on the 6 and you place a bet on the 8. What does the house expect to make from us?
How would you explain the difference between these three scenarios?
Let's say that I walk up to the table and place a bet on the 6 and 8. You contend the house would expect to make 1.05% off of me.
Let's say that I walk up to the table and place a bet on the 6, and you walk up to the table and place a bet on the 8. You would say that the house would expect to make 1.52% off of each of us.
Let's say that I give you a bet, then we walk up to the table together and I place a bet on the 6 and you place a bet on the 8. What does the house expect to make from us?
How would you explain the difference between these three scenarios?