Craps - Multiple Wagers House Edge
Poll
 | 2 votes (33.33%) | ||
| | 1 vote (16.66%) | ||
| | 2 votes (33.33%) | ||
| | 1 vote (16.66%) |
6 members have voted
Quote: MathExtremistThe Santa Ana Star in New Mexico offers free (no-vig) buy bets on the 4 and 10, and also offers a zero-edge field bet that pays triple on both 2 *and* 12.
http://www.santaanastar.com/gaming/table-games/craps.
Will the Silverton let you go higher than $25 on the 4 or 10 for $1 vig? I've heard some casinos will go as high as $39 because they take the floor of the vig rather than rounding to the nearest dollar. I've also heard that some casinos will treat each bet buy separately for vig calculations vs. others lumping them together. In other words, you'd pay $4 vig on $50 each buy 4 and 10 vs. $5 vig, which is what you'd pay if you put all $100 on one number. But that varies from property to property and is almost always something you have to ask.
Yeah, I know, huh? The free buy bets and free field bets are absolutely something that if I could prove control would be totally exploitable no problem.
To be clear, I haven't proven even to myself that I can control the dice.
Nice observation.Quote: RaleighCrapsIt all comes down to, is there a house edge on a bet that was made, but was cancelled prior to a resolution?
So in scenario #2, I have $50 at risk, with a 50-50 chance of winning $49, and a 50-50 chance of losing $50.
If I have the bet active for 4 rolls with no resolution, and then pull it down, was the bet subject to a HE ?
Let us start with an easy related question. Just simple math. No formulas.
$25 Buy4.
A very popular Craps bet.
You walk up to a table with $25 in hand. You know that there is a craps bet that pays 2 to1.
You toss your quarter on the table and tell the dealer you want to Buy the 4.
He books the bet, places your cheque at the proper location in the 4 point box and puts a blue "BUY" button on it.
You are excited because your $25 will pay $50 and you will have $75.
The very next roll is the 4, you scream and wait your winnings.
The dealer asks if you want the bet down, and you almost pee your pants telling him "yes"
He hands off 2 green, 4 red and 4 white cheques to you.
You pick it up and say thank you and as you walk away you think you should have $75.
$25 + $50 (you really did not understand that vig or juice talk you heard before)
Did the Dealer take out a $1 for a tip?
(I had 2 players in my dealing career ask that)
But, Hey you are happy to have the $74 and you skip off to your next adventure.
House edge, well you were shorted $1 from the $75 total you should have received.
That is the edge -1/75
(the John Scarne method. for simple bets we do not even have to know the winning and losing probabilities)
Now let us re-wind and say after 12 rolls your bet has not been yet decided. You are playing scared.
You ask the dealer for it back. He sets it down and you walk away so very happy that you did not lose it.
House edge? for those 12 rolls?
Now remember HE per bet resolved is calculated from total action or bets that actually became resolved.
Your winnings were 0. You total at risk was $25 so you first think maybe 0/25 = 0%
Wait, the bet was not resolved. winnings was 0 and I had a 0 resolved bet.
0/0?
This method does not work for all situations.
Since not all craps bets resolve on every roll we should know the per bet and per roll house edge values so we can tackle these type of questions.
You asked AHigh to explain how he arrived at his per roll house edge, maybe he will answer for you.
You did ask in a very nice way BTW.
You actually solved the per bet HE, just re-do your calculations with the actual total probabilities and account for every outcome.
Or read ME's post in this thread and he shows how to do this with a little math.
No.Quote: AhighI'm done about the 1.00% edge. If you say it's impossible to have a bet that's not marked on the felt and/or that's not programmed into your simulator, then you are right.
I say the bet is possible whether they mark it or not. It is a bet that you can make. But for it to be that bet, you have to define the bet as being done when either one of the 4 or 10 hits.
We'll have to leave it at that.
You and one other (maybe 2) say the edge when Buying the 4 and 10 for $25 and taking both the bets down on a win is 1%.
5 others say it is 1.33%
See, this is a good thread.
Many do not all agree for different reasons.
This thing about your "one" bet for a buy4 and buy 10.
You really need to let that go.
Every casino I have played Craps in allows a $25 Buy on the 4 and the 10 and at any time they can be removed.
I stopped my simulation at 100,000,000 bets resolved.
In 1,000 bets per player.
The HE at 1.33% should show, per 1000 bets, that the total EV was -$500
Yep it was just that.
(1,000* HE(-1/75 or 1.33%)* 75/2 (the average resolved bet))
If the true house edge was 1% as two have shown by their calculations,
my EV for the 1000 bets resolved should have been -$375
(1,000* HE(-1/100)* 75/2 (the average resolved bet))
Nope.
It was -$500.
Current tally
5 for 1.33%
2 (maybe 3) for 1%
This is really sad that 7 or 8 can not agree on a simple multiple Craps bet house edge.
And many wonder why the world is the way it is.
No politics, no religion here.
Just math.
I am done too
Quote: AhighBut to help people that want to understand imagine if you had no bet on the felt for doing a place bet on the 9.
But rather instead you could place the 5-4, and you could place the 6-3. Each would be a $2.50 bet that would pay $7.00. They would both lose on a seven.
So the 5-4 edge would be $0.50 / $2.50 + $7.50 = 5% house edge.
And the 6-3 edge would also be $0.50 / $2.50 + $7.50 = 5% house edge
by betting them both and agreeing the either one of them hitting would end the event, you could have what we currently know as placing the 9.
And that has an edge of $0.50 / ( $5.00 + $7.50 ) = 4% house edge.
Did nobody read and understand this? If you split the place bet on the five and nine into two separate bets, the edges are 5% instead of 4% per event!
Nobody can understand this? Really? Nobody can connect the dots? If they forced you to make two separate place bets on the five and nine depending on which outcome came out, would the place bet on the 5 and 9 (when you bet both) all of a sudden be 5% then by your logic?
Quote: buzzpaffLet's have a poll that 2 and 2 =5. Make a lot more sense.
You have a high number of negative comments and you post a lot of short comments that don't really add anything of value to the conversation.
Quote: buzzpaff" To be clear, I haven't proven even to myself that I can control the dice. " Someday you will wish you had spent those 3 hours a day more productively.
I dislike many of the things that you say. Sometimes I wish you didn't respond as I can't think of anything of value that you have contributed so far.
Quote: 7crapsThe loss is $50.
There are Two $25 Buy bets there.
As I said, I was considering each bet separately.
Quote: 7crapsMission 146 is adding house edge, that is the wrong method to use as stated by ME.
We add ev and variance not house edge and standard deviation.
Try one $25 Buy4 and one $100 Buy10 and see what you get adding edges.
Adding standard deviation is obviously wrong, but averaging house edge, as Mission146 and I both did, is fine, provided that you weight them by bet size. The fact that adding them up and dividing by two only works if both bets are the same size doesn't need to be explicitly stated unless you suspect your audience doesn't know what a percentage is.
Oh, 1.33% is wrong. There are two correct answers: Ahigh's 1%, which is the expected loss every time you take it down, and 0.33%, the edge per roll, which you'll note is the same as the edge per roll on either buy bet alone. The edge per bet resolved has changed, not because the edge has changed, but because bets are now "resolved" more frequently.
Quote: buzzpaff" New ideas welcomed. "
If you work out the mathematics of dice control, you will find that it only takes one controlled throw every 43 rolls of the dice to eliminate the house edge for the 6 and 8 place bets.
This, I think, was one of the few positive comments.
Quote: AhighI dislike many of the things that you say. Sometimes I wish you didn't respond as I can't think of anything of value that you have contributed so far.
That would be as opposed to the last two, "Valuable," posts you've made? The latter of which being to complain of the lack of value of another's post and for no other reason?
Quote: buzzpaffI am Buzz Paff and I approve this message .
Wow.
Quote: Mission146That would be as opposed to the last two, "Valuable," posts you've made? The latter of which being to complain of the lack of value of another's post and for no other reason?
Just trying to fit in.
Quote: buzzpaffAhigh " Trust and know that I never expect an apology for all this abuse. "
Ahigh " You and guido also don't care to apologize about how you guys have tried so hard to discredit me on every single issue that you could possible hope to achieve to discredit me on. "
I feel this calls for a 3 day suspension for Ahigh contradicting Ahigh !
This makes absolutely no sense to me at all.
Quote: buzzI missed you at the last Mensa meeting. OOPs forgot 3 digit IQ is required.
Cool. That's awesome, bro.
Quote: buzzpaff" So how can I combine those 2 bets and have a HE less than the 1.33%? That makes no sense."
That's because you did not factor in Dice Setting !
Oh. Thanks.
That will be quite enough of this post-after-post-after-post one-liner nonsense. It would be one thing if there was, at a minimum, someone posting in between your one-line posts.
In any event, you may be interested in your video thread. I have developed a metric for you called, "Hardway Roll Replacement Value," and have determined, betting the same amount on all four Hardways, exactly how much more frequently than the expectation you must roll them.
Quote: Mission146AHigh,
That will be quite enough of this post-after-post-after-post one-liner nonsense. It would be one thing if there was, at a minimum, someone posting in between your one-line posts.
In any event, you may be interested in your video thread. I have developed a metric for you called, "Hardway Roll Replacement Value," and have determined, betting the same amount on all four Hardways, exactly how much more frequently than the expectation you must roll them.
Yeah, I saw that and it's awesome. I am going to post later tonight.
I just had an electrician at the house to install custom lighting for the slow motion stuff that I have planned.
The 240 Hz slow motion recording has some very specific lighting requirements to make it work. I'm planning to buy an $8,000 slow motion camera as soon as possible. But it may be a few months before I'm ready to drop the cash.
There has to be two correct answers.Quote: 24Bingo
Oh, 1.33% is wrong. There are two correct answers:
Ahigh's 1%, which is the expected loss every time you take it down,
per bet
and
per roll.
The 1% is not the EV.
What simulation results prove this mathematics statement?
He claims, as you, it is the house edge per bet resolved. Or is this not the case?
I just ran a quick 1 million bet resolved sim.
$25 buy4 and $25 Buy10 at the same time.
When one won, I removed both and waited 9 rolls, no bets at all, and then made the same 2 bets.
When both lost, I waited 18 rolls, made no bets, then I made the same two bets.
What do you think the house edge came to?
-1/100
or
-1/75
Hint: It was not the first one.
Fact remains, everyone will believe what they want to believe and throw away the rest.
Guido, My sim was different from yours except the end results.
And I did not do the 1,000 bets per player.
No need, the math was easy.
Quote: 7crapsThere has to be two correct answers.Quote: 24Bingo
Oh, 1.33% is wrong. There are two correct answers:
Ahigh's 1%, which is the expected loss every time you take it down,
per bet
and
per roll.
The 1% is not the EV.
What simulation results prove this mathematics statement?
He claims, as you, it is the house edge per bet resolved. Or is this not the case?
I just ran a quick 1 million bet resolved sim.
$25 buy4 and $25 Buy10 at the same time.
When one won, I removed both and waited 9 rolls, no bets at all, and then made the same 2 bets.
When both lost, I waited 18 rolls, made no bets, then I made the same two bets.
What do you think the house edge came to?
-1/100
or
-1/75
Hint: It was not the first one.
Fact remains, everyone will believe what they want to believe and throw away the rest.
Guido, My sim was different from yours except the end results.
And I did not do the 1,000 bets per player.
No need, the math was easy.
There is only 1 bet. That's where you made the mistake. I can't believe you are still stuck!
You made the mistake!Quote: AhighThere is only 1 bet. That's where you made the mistake. I can't believe you are still stuck!
You are 100% wrong and you know it.
This is great comedy!
Thanks man, I needed a good laugh after yesterday.
There are TWO bets.
It was the agreed question #2 that you even spelled out in my first post
Quote: Ahigh
#2) But [sic] the 4 AND 10 for $25 each and take $99 and down on the first hit, the edge is 1/100 or 1.00%
There are two separate bets made. Here is a pic of what it looks like.
Both can NOT win on the same roll, absolutely impossible, but both CAN LOSE on the same roll.
So you have to create a bet that does not exist on a Craps table to prove your point of what?
The question still remains
Quote: Ahigh
#2) But [sic] the 4 AND 10 for $25 each and take $99 and down on the first hit, the edge is 1/100 or 1.00%
House edge of this exact style of play, (not some other mystical craps bet that does not exist)
as spelled out in Q#2 is exactly -1/75
This was fun
Enjoy!
I did not believe your sim data anyways.Quote: AhighI'm editing a sim and I will post it soon.
No one at the forum believes in sim data results anyways. Just a big waste of time.
No computer can simulate the game of craps perfectly.
We all just wasted our own time.
The math is easy when one does it correctly.
Done
Yes, a 4 letter word
Quote: guido111I did not believe your sim data anyways.
No one at the forum believes in sim data results anyways. Just a big waste of time.
No computer can simulate the game of craps perfectly.
We all just wasted our own time.
The math is easy when one does it correctly.
Done
Yes, a 4 letter word
Well I can finally agree on one thing: I am wasting my time with you.
That is the first constructive comment you have posted as far as I know.
Quote: 7crapsThere has to be two correct answers.Quote: 24Bingo
Oh, 1.33% is wrong. There are two correct answers:
Ahigh's 1%, which is the expected loss every time you take it down,
per bet
and
per roll.
The 1% is not the EV.
What simulation results prove this mathematics statement?
He claims, as you, it is the house edge per bet resolved. Or is this not the case?
1% is per bet resolved, counting either a win or a loss as one bet resolved (as I said, per roll is 0.33%). Hence my question regarding this simulation:
Quote: 7crapsI just ran a quick 1 million bet resolved sim.
$25 buy4 and $25 Buy10 at the same time.
When one won, I removed both and waited 9 rolls, no bets at all, and then made the same 2 bets.
When both lost, I waited 18 rolls, made no bets, then I made the same two bets.
What do you think the house edge came to?
-1/100
or
-1/75
Hint: It was not the first one.
The problem is in how you counted. I consider both bets to be "resolved" when one wins or loses. If you, like guido, considered the average bet resolved to be $37.50, of course you got the edge you got. However, I would say that since $50 is at risk on any roll, even though half technically pushes when a 4 or 10 is rolled, that is the amount from which the edge should be taken in either case. Strictly speaking, you have the option of picking up your winnings and waiting for the other bet to resolve itself, but that kind of goes entirely against the point of the exercise. What we do agree on is that the EV, from placing the bets through to a 4, 10, or 7, is -50¢, with an SD of $49.50.
There was absolutely nothing that you could say to change his mind and I had to give it up.
This really isn't any different.
These forums where people argue over things like this are an absolute waste of my personal time.
I have no idea why I am trying to help someone who refuses to even consider what I am saying.
Don't waste your time either Bingo.
They just aren't being reasonable.
There was no problem and your method places you in the minority.Quote: 24Bingo
The problem is in how you counted.
I consider both bets to be "resolved" when one wins or loses. .
To quote the Wizard
"The house edge is not the ratio of money lost to total money wagered."
Disagree if you must, you are disagreeing against the best.
(The Wizard, that is, do not even consider anything I say)
The roll is a 4.
My Buy4 won.
The Buy 10 pushed.
Both bets come down
I come back in another hour to make another bet, 2 Buy bets, maybe.
HE = (0.5*49 + 0.5*-50)/(0.5*25 + 0.5*50)
I think we all agree on the first term (EV)
The second one tells us the average bet resolved
"not the ratio of money lost to total money wagered"
Half the time, blah, blah, blah
you believe what you want, you are King in your Kingdom but not in mine.
Kind of cool you actually disagree with the Wizard on this one.
added:
Yes, I did not read all your post. I never really do.
If I disagree with one part, I could care less with what follows.
What?
You want to sue me?
Stand in line man.
One more for the boat, I say.
AND YET YOU PERSIST, EVEN WITH THE CLOCK TICKING AWAY TILL G2E.
WHO HAS A PROBLEM WITH REALITY, I WONDER !
Quote: 7crapsThere was no problem and your method places you in the minority.
To quote the Wizard
"The house edge is not the ratio of money lost to total money wagered."
I know you're not reading this, but...
"The house edge is not the ratio of money lost to total money wagered. In some games the beginning wager is not necessarily the ending wager. For example in blackjack, let it ride, and Caribbean stud poker, the player may increase their bet when the odds favor doing so. In these cases the additional money wagered is not figured into the denominator for the purpose of determining the house edge, thus increasing the measure of risk."
So in other words, the house edge is the ratio of money lost to money initially wagered in games where further bets are possible. If thait has anything to do with this at all, it's a blow against your side, since it's the $50 that was initially wagered.
It's one thing to immediately stop reading someone you've decided to disagree with - you must be brimming with pride that you've never let anyone prove you wrong to your own satisfaction - but not reading one sentence further in the paragraph you yourself are citing as an authority? Color me impressed.
How will you feel once you grasp all the math perfectly, and then go to the casino and still lose your shirt?
Roger Snow is rumored to be in a bidding war for this game .
Ahigh claims this should be treated as one bet, so my casino has just created this new bet called the 104.
It pays true odds (like any Buy bet), but it has a 2% Vig on a $50 bet (my casino is player friendly).
There are 6 ways to win (any 10 or any 4).
The bet loses on a 7 out.
Because there are 6 ways to win, 6 ways to lose, the bet pays 1:1.
At 2% Vig, a $50 bet pays $49 to win
(.5*49-.5*50) / (50) = 1% HE
------------------------------------------------------------------------------
Ahigh's point is, by playing the Buy 4/10 as one play, and taking both bets down when either is resolved, it is truly one bet.
I believe my 104 bet described above will play EXACTLY like the Buy 4/10, when played as one bet. Can anyone show me where there is a deviation from normal casino play on how my bet above would get paid differently?
Does the fact that I needed to change the Vig from 5% to 2% to make the payout equivalent mask the problem? I don't think so, but I am not sure.
Assuming there is no deviation, and the Buy 4/10 bet plays EXACTLY equivalent to the 104 bet described above, does it not stand to reason that the HE on the 2 bets should be the same, since the bet plays the exact same ? Color me confused.
I 'think' this is wrong, but I can't see how it can be, unless it has to do with the Vig..........
Quote: RaleighCrapsOkay, 24Bingo and Ahigh have me back to uncertain, not that it really matters ;-)
Ahigh claims this should be treated as one bet, so my casino has just created this new bet called the 104.
It pays true odds (like any Buy bet), but it has a 2% Vig on a $50 bet (my casino is player friendly).
There are 6 ways to win (any 10 or any 4).
The bet loses on a 7 out.
Because there are 6 ways to win, 6 ways to lose, the bet pays 1:1.
At 2% Vig, a $50 bet pays $49 to win
(.5*49-.5*50) / (50) = 1% HE
------------------------------------------------------------------------------
Ahigh's point is, by playing the Buy 4/10 as one play, and taking both bets down when either is resolved, it is truly one bet.
I believe my 104 bet described above will play EXACTLY like the Buy 4/10, when played as one bet. Can anyone show me where there is a deviation from normal casino play on how my bet above would get paid differently?
Does the fact that I needed to change the Vig from 5% to 2% to make the payout equivalent mask the problem? I don't think so, but I am not sure.
Assuming there is no deviation, and the Buy 4/10 bet plays EXACTLY equivalent to the 104 bet described above, does it not stand to reason that the HE on the 2 bets should be the same, since the bet plays the exact same ? Color me confused.
I 'think' this is wrong, but I can't see how it can be, unless it has to do with the Vig..........
First off, the effective vig on a $25 buy 4 is 4%, not 5%. And if you pay vig only on win, you're only paying that 4% vig on 3 out of every 36 rolls: when the 4 rolls. On your bet, you're paying a 2% vig on 6 out of every 36 rolls, either on the 4 or the 10. 2% * 6/36 = 4% * 3/36. Therefore, loss per hour per dollar is the same, and that's what you really care about (probably).
Percentages rarely tell the whole story. Consider a hypothetical prop bet called Black Aces which, despite the name, has nothing to do with cards. It's only available in increments of $100 (hence "Black") and it's a bet on the 2 ("Aces"). It pays 3475 to 100 for a house edge of 0.694%. And don't you all wish that prop bets had an EV that low? But is $100 on Black Aces a better bet than a $100 passline bet? If you're not paying attention, you might think so: 0.694% is less than 1.414%. But the passline resolves in over 3 rolls while Black Aces resolves every roll, and the per-roll EV on the passline is only 0.42% -- less than the Black Aces bet. So don't just go by the percentages. Look at the loss rates instead. If you only make one Black Aces bet, you're better off than with one $100 pass bet. But if you make them constantly for an hour, you'll lose less on the line.
I think the point of this bet, though, isn't about rolls, but to have an approximately even-money bet with better odds than the passline, to give a better press.
Although actually writing that down has the smartass in me saying, "I know an even better one - look for the space marked 'banker.'"
(That's actually worse, if you actually found a 4% Buy commission, but nevermind.)
Quote: 24BingoYeah - unless Ahigh's coefficients are wrong, he's found a place with very generous Buy bets. With a 5% vig paid on a win, the effective house edge goes up to 1.25% (and the edge on each bet resolved to 1.67%) with a standard deviation of $9.75 per $10 bet.
I think the point of this bet, though, isn't about rolls, but to have an approximately even-money bet with better odds than the passline, to give a better press.
Although actually writing that down has the smartass in me saying, "I know an even better one - look for the space marked 'banker.'"
(That's actually worse, if you actually found a 4% Buy commission, but nevermind.)
I pointed this out in another thread on this topic. The edges range from 1.00% for $50 up to 1.25% for $200 (with a different shapes in between depending on how the house determines the vig). Baccarat goes from 1.00% to 1.24%. But there is a direct analogy to a bet that has a 50/50 chance (or close) of resolving in your favor with such a low edge. The edge varies based on the bet amount instead of player versus banker.
To me, if this fact were more well-known, I would think there would be typical baccarat style betters making these bets and pressing and doing fibbonacci and martingale and betting very large sums.
Also you can be twice the maximum table limit bet since you're constructing one bet out of two bets. So at Caesar's you can win $97,500 half the time from $100,000 on the felt within their limits ($10 to $50,000) on the normal floor without any special reserved table with higher limits.
I appreciate that you seem to be in tune with what I am thinking, even though I know others must have discovered this before me. That's at least one person who sees it the way that I see it anyway, which is something that makes me feel like my efforts were not for naught.
Quote: guido111
1000 bets * $bet * HE = EV
Most use the 1.33% for the HE in the EV formula.
Hey, Guido, where did you go?Quote: 7crapsThat $bet should be the average $bet resolved or you will end up with a value your simulations do not agree with. In your thread example that would be 25/2 + 50/2
No more or no less
Do not listen to me. Unless you want to.
The 75/2 is the average resolved wager to calculate the HE.
You already know that.
The average bet is just $25 per bet.
That NEVER changes.
In 1 million bets resolved, 1/3 are a win and 2/3 are a loss but each bet is just $25.
So something ain't right in your numbers for the EV per 1000 bets.
EV = 1000*$25*(-1/75)=-333.33
I also did your 1000 bet sim as you asked and have the results for you.
I see that WC does add the vig paid on a win to the total handle. This is not correct.
The $5000 example really shows this.
Two bets at $5000 each and after one wins and the bet comes down WC shows the total $ wagers to be $5250.
The average number of rolls per a resolved wager is 2.
That is because of what I said above.
Some of your posts here are just crazy.
Quote: 24BingoI think the point of this bet, though, isn't about rolls, but to have an approximately even-money bet with better odds than the passline, to give a better press.
If that's the goal, just convince your local casino to offer the Odd and Even bets.
Odd: pays 1-1 on 5, 7, 9, 11; pays 4-5 on odd craps (3); loses on even numbers.
Even: pays 1-1 on 4, 6, 8, 10; pays 4-5 on even craps (2, 12); loses on odd numbers.
Both have exactly 50% chance of winning, and both have 1.11% house edge. These are the ideal bets for pressing or progression systems because there's no waiting - these are both one roll bets and they have the lowest edge (per bet) on the table.
The bets are already approved in Nevada and Mississippi, patented, trademarked, etc. As a courtesy to WoV members and friends: PM me for a free trial and custom dice layout (edit: for your casino, not for your living room. Nice try, Buzz. That's what I get for not being excruciatingly specific. :)
Quote: MathExtremistIf that's the goal, just convince your local casino to offer the Odd and Even bets.
Odd: pays 1-1 on 5, 7, 9, 11; pays 4-5 on odd craps (3); loses on even numbers.
Even: pays 1-1 on 4, 6, 8, 10; pays 4-5 on even craps (2, 12); loses on odd numbers.
Both have exactly 50% chance of winning, and both have 1.11% house edge. These are the ideal bets for pressing or progression systems because there's no waiting - these are both one roll bets and they have the lowest edge (per bet) on the table.
The bets are already approved in Nevada and Mississippi, patented, trademarked, etc. As a courtesy to WoV members and friends: PM me for a free trial and custom dice layout (edit: for your casino, not for your living room. Nice try, Buzz. That's what I get for not being excruciatingly specific. :)
I know that many people are never going to normalize all house edges as being "per roll" in craps, but that's the _only_ way you can compare one bet to another bet in terms of average loss per roll and then you can count ties as an outcome.
This game is intrinsically complicated to the novice, and as this thread illustrates, it is also illusive to the self-proclaimed experts.
It's nice at least that we can all agree that the 4 and 10 for $25 vig on the win and at 0.33% HE per roll is the lowest edge (per roll) non-free (not counting odds or lay odds) bet on the table.
Is there anyone who has yet to come to that answer?
Also, Main Street Station offers this bet for $5.00 and pays $9.75 using $0.25 chips, so you can buy the four and ten at $10.00 total giving you more breathing room for a martingale or a fibonacci betting scheme than starting at $25 and still keep your edge down. It's true you have to wait 3 times as long as a single roll bet, but your expected loss would still be lower than $5 on the bet that Math Extremist is describing on a per roll basis because of the multi-roll properties.
Quote: AhighAlso, Main Street Station offers this bet for $5.00 and pays $9.75 using $0.25 chips, so you can buy the four and ten at $10.00 total giving you more breathing room for a martingale or a fibonacci betting scheme than starting at $25 and still keep your edge down. It's true you have to wait 3 times as long as a single roll bet, but your expected loss would still be lower than $5 on the bet that Math Extremist is describing on a per roll basis because of the multi-roll properties.
That raises another interesting question. For betting systems players, what's more important? I'd argue it's edge per bet (not per roll) because betting systems are inherently based on winning and losing bets. If you're going to play a 10-step Martingale anyway, then it's most important to do it with a low edge-per-bet wager. Moreover, unless you like standing around waiting, faster bets are preferable. Buying 4 and 10 leads to an effective 50/50 proposition, but it's really a 16.67/16.67/66.7 proposition where nothing happens 2/3 of the time. For someone who wants to play a few quick rounds of Martingale to pay for dinner, it will take 3x as long with the buy 4/10 combination as it will with a single-roll 50/50 proposition. Of the traditional dice bets, only the field comes close, but that's more of a 45/55 wager than 50/50.
Note: I do not actually advocate "playing a few quick rounds of Martingale to pay for dinner"...
Does that mean you actually advocate...Quote: MathExtremistNote: I do not actually advocate "playing a few quick rounds of Martingale to pay for dinner"...
maybe...
"to pay for dinner then playing a few quick rounds of Martingale"
ME, nice to see your website is back in it's full glory.
I saw a couple at Primm Valley last week playing your craps shotgun method.
was not the "The Four-Barrel Shotgun" but two-barrel variant.
They got shot down, the table was ice cold, but not for me.
Good Luck to you
