Math Help on Multiple Place Bets

I have been grappling with some math that I have put to bed only to have it repeatedly rise again without a permanent resolution. I’m coming here where real math posters hang out because this is obviously beyond my pay level. This is actually several questions but kind of tied together with a common thread.

I understand the calculated 6 or 8 is 1.52%, 5 or 9 is 4% & 4 or 10 is 6.67%. However having read the Wizard’s chart of resolved vs. the per roll of 6 or 8 at .46%, 5 or 9 at 1.11% & 4 or 10 at 1.67% has me wondering what if the bet is resolved on the very next toss after the bet is made. Is it now the resolved rather than the first per roll? This has also rekindled my interest in another mind knot.

I’m curious about multiple Place bets. I’ve always been told and read that regardless of how you combine them the HA remains the same. Then Zeke Feinberg came along and published his book that from a logical view seemed to show that the HA could be altered with the 6 & 8 to 1.04%, the 5 & 9 to 2.86% and the 4 & 10 to 5%. Of course I then read why his method was wrong even if the faulty logic appeared sound.

Even though I understood from math players why it was wrong I kept coming back to Zeke’s faulty logic and thinking “but if one truly locked what ever Place bet combination together as a single bet, raised, lowered, turned off or pulled down as a single unit rather than as individual and independent bets wouldn’t Zeke’s math be correct? Of course most players wouldn’t treat such a bet as a single bet and would raise just the winning number, perhaps pressing another number on a gut feeling, maybe pulling down another number and ending any cohesion between them. But that’s a human failure, not a mathematical one. Still it remained as a gnawing thought in the back of my mind.

As I was reviewing some of my old saved Craps readings as I do to refresh the old noggin I came across an article written by another math author now deceased whom I had always respected. The message it contained was that multiple Place bets functioned as Lay bets because that’s what they are. That reopened this whole Zeke thing for me. We think of Craps’ Lay bets as against a box number and having the 7 as our winning number but is that truly the only kind of Lay there is in Craps or the only one recognized by the casinos? What is the definition for a Lay bet? Isn’t it having the odds in your favor to win one event over another at the cost of winning less money than you bet?

I know that just one bet has won and the others are unresolved but, if treated as a unified Lay with that specified win amount vs. the amount invested and risked, that one win is all that's required. Everyone wants to define these as independent bets and while that may be how they were designed it is not the way they must be played.

Well that got me going and using the Wizard’s formula for calculating Lay bets I ran through Zeke’s combined 6 & 8, 5 & 9, 4 & 10, Inside and Across bets and arrived at Zeke’s figures. So is Zeke correct even if he had the wrong reason? Can these multiple Place bets be treated as a unified Lay bet?

Any insight would be greatly appreciated.

Kelph

Quote: Kelph

I have been grappling with some math that I have put to bed only to have it repeatedly rise again without a permanent resolution. I’m coming here where real math posters hang out because this is obviously beyond my pay level. This is actually several questions but kind of tied together with a common thread.

I understand the calculated 6 or 8 is 1.52%, 5 or 9 is 4% & 4 or 10 is 6.67%. However having read the Wizard’s chart of resolved vs. the per roll of 6 or 8 at .46%, 5 or 9 at 1.11% & 4 or 10 at 1.67% has me wondering what if the bet is resolved on the very next toss after the bet is made. Is it now the resolved rather than the first per roll? This has also rekindled my interest in another mind knot.

I’m curious about multiple Place bets. I’ve always been told and read that regardless of how you combine them the HA remains the same. Then Zeke Feinberg came along and published his book that from a logical view seemed to show that the HA could be altered with the 6 & 8 to 1.04%, the 5 & 9 to 2.86% and the 4 & 10 to 5%. Of course I then read why his method was wrong even if the faulty logic appeared sound.

Even though I understood from math players why it was wrong I kept coming back to Zeke’s faulty logic and thinking “but if one truly locked what ever Place bet combination together as a single bet, raised, lowered, turned off or pulled down as a single unit rather than as individual and independent bets wouldn’t Zeke’s math be correct? Of course most players wouldn’t treat such a bet as a single bet and would raise just the winning number, perhaps pressing another number on a gut feeling, maybe pulling down another number and ending any cohesion between them. But that’s a human failure, not a mathematical one. Still it remained as a gnawing thought in the back of my mind.

As I was reviewing some of my old saved Craps readings as I do to refresh the old noggin I came across an article written by another math author now deceased whom I had always respected. The message it contained was that multiple Place bets functioned as Lay bets because that’s what they are. That reopened this whole Zeke thing for me. We think of Craps’ Lay bets as against a box number and having the 7 as our winning number but is that truly the only kind of Lay there is in Craps or the only one recognized by the casinos? What is the definition for a Lay bet? Isn’t it having the odds in your favor to win one event over another at the cost of winning less money than you bet?

I know that just one bet has won and the others are unresolved but, if treated as a unified Lay with that specified win amount vs. the amount invested and risked, that one win is all that's required. Everyone wants to define these as independent bets and while that may be how they were designed it is not the way they must be played.

Well that got me going and using the Wizard’s formula for calculating Lay bets I ran through Zeke’s combined 6 & 8, 5 & 9, 4 & 10, Inside and Across bets and arrived at Zeke’s figures. So is Zeke correct even if he had the wrong reason? Can these multiple Place bets be treated as a unified Lay bet?

Any insight would be greatly appreciated.

Kelph

That might be a record for the time between joining and a first post. (Sept 6, 2010 til today). A belated welcome, Kelph! I hope you get a quick and definitive answer.

Quote: Kelph

I understand the calculated 6 or 8 is 1.52%, 5 or 9 is 4% & 4 or 10 is 6.67%. However having read the Wizard’s chart of resolved vs. the per roll of 6 or 8 at .46%, 5 or 9 at 1.11% & 4 or 10 at 1.67% has me wondering what if the bet is resolved on the very next toss after the bet is made. Is it now the resolved rather than the first per roll? This has also rekindled my interest in another mind knot.

I’m curious about multiple Place bets. I’ve always been told and read that regardless of how you combine them the HA remains the same. Then Zeke Feinberg came along and published his book that from a logical view seemed to show that the HA could be altered with the 6 & 8 to 1.04%, the 5 & 9 to 2.86% and the 4 & 10 to 5%. Of course I then read why his method was wrong even if the faulty logic appeared sound.

Intuitively you know that making a second independent bet doesn't change the edge of a first one. The question is why it seems that way. To clear it up, instead of using percentages, use dollars.

You begin by addressing PLACE bets and then you switch to Lay Bets.

Firstly, I think that figuring your e.v. by "resolved" bets is confusing and misleading.

Secondly, combining (any) bets cannot alter the combined expectation.

Thirdly, your view of Lay bets is a bit off the mark. View them thusly as ...

They are the converse of a Place Bet. They are poor bets. Bet the Don't

(and/or the Don't Come) and Lay the Odds that way to get the TRUE ODDS

where the house has no edge. Yes, I am a confirmed dark side bettor.

Quote: Kelph

... has me wondering what if the bet is resolved on the very next toss after the bet is made. Is it now the resolved rather than the first per roll?

What if the bet is NOT resolved in the next roll?

You can't selectively assume that event. The Wizard's math combines ALL possible events.

And welcome. Don't worry though. If the math doesn't occasionally give you headaches, you're doing it wrong.

First thank you for the responses.

Second let's forget my question regarding per roll vs. resolved.

Third I need your and any other respondent’s opinion of what meets the definition to constitute a Lay bet. I provided mine and it appears you think I'm confusing Place and Lay but I'm not if you read what I wrote. Whether the Lay bet is bad or not is immaterial to my question.

A Lay bet is not necessarily the converse of a Place bet unless you specifically mean, as almost everyone does, the Don’t side Lay bet that reverses the Place odds by giving you the 7 as the winner. Again please spell out what a Lay bet is in general unless you feel it’s only what the casino and Craps books acknowledge.

I know most will disagree but I ask you to just run the number combinations through Wizard’s Lay formulas. I don’t believe the 7 has to be the winning number and that is what makes it a Don’t side play. Whether a Lay wins all or loses all on a specific number isn’t its defining attribute. It’s the ways to win (more), ways to lose (fewer) and total amount invested/risked (more) for a specific win amount (less than amount risked) IMHO.

What I suggested in no way implies any change in expectation frequency. I’m saying that in addition to the Don’t side Lay there is also a Do side Lay by changing the 7 to a different winning number and played as I have suggested.

Now if I’m wrong, which I can accept, I certainly would appreciate a clear explanation as to the why. That’s why I brought the question here.

I don’t mean to be a pain about this but it’s been an itch for quite a while and if I’m going to scratch it I want it to stay scratched.

Kelph

Kelph,

As BBB pointed out, quite the belated first post, but nice to hear from you all the same!

Quote: Kelph

I understand the calculated 6 or 8 is 1.52%, 5 or 9 is 4% & 4 or 10 is 6.67%. However having read the Wizard’s chart of resolved vs. the per roll of 6 or 8 at .46%, 5 or 9 at 1.11% & 4 or 10 at 1.67% has me wondering what if the bet is resolved on the very next toss after the bet is made. Is it now the resolved rather than the first per roll? This has also rekindled my interest in another mind knot.

The first thing that I should point out is that the House Edge Per Roll is simply reflective of the average number of rolls it takes for a bet to resolve. For example, if you take 152/46 you get 3.30 (rounded) which is approximately how many rolls, on average, it will take for that bet to resolve in one way or another. If you have a Place Six bet, then there are 11/36 possible results that resolve the bet either way, which comprise 30.55556% of the results, 1/.3055556 = 3.27 rolls on average, so these things are just rounding errors.

If you look at the Place Six, then you make a $6 bet to win $7:

(7 * 5/36) - (6 * 6/36) = -0.0277777777--- -0.02777777777/6 = -.00462963

However, let's say you make the bet knowing you will leave it out until it resolves, then it becomes an absolute like so:

(7 * 5/11) - (6 * 6/11) = -0.0909090909 --- -0.0909090909/6 = -0.01515151515

The House Edge per roll is a relevant number (for some people) pursuant to the fact that a player can turn such a bet, "Off," or, "Pick the Bet Up," provided the previous roll did not result in a resolution.

It is irrelevant (prior to the roll occurring) whether or not the very first roll after the bet is made resolves the bet. The House Edge Per Roll comprises all possibilities, including the bet not being resolved. Furthermore, with enough trials (i.e. initial Place Bets) in the long run, the results will converge to the Mean number of rolls needed to resolve the bet, which is roughly 3.27 rolls. None of these things change the House Edge per Bet Resolved or the House Edge per Roll, although, if you know you will leave a bet out there UNTIL it is resolved, the House Edge per Roll is not really a practical concern with respect to that bet.

The one thing it does, I suppose, is it helps you compare one bet to another with respect to more than just the pure dollars and cents Expected Value. For instance, with a Triple 2 or 12, the Field Bet has a lower House Edge than the Place 4/10 or Place 5/9, but if you want to get more, "Bang for your buck," the House Edge per Roll is lower on those Place Bets than on the Field Bet. If you plan to leave the bets out there until they are resolved, though, the House Edge on your action is still worse.

Quote:

As I was reviewing some of my old saved Craps readings as I do to refresh the old noggin I came across an article written by another math author now deceased whom I had always respected. The message it contained was that multiple Place bets functioned as Lay bets because that’s what they are. That reopened this whole Zeke thing for me. We think of Craps’ Lay bets as against a box number and having the 7 as our winning number but is that truly the only kind of Lay there is in Craps or the only one recognized by the casinos? What is the definition for a Lay bet? Isn’t it having the odds in your favor to win one event over another at the cost of winning less money than you bet?

I'm confused, a Lay Bet can only mean one thing and that is a bet upon which you bet against a specific number. There is a House Edge working against you on all Lay Bets and it is also incorrect to suggest that you have, "The Odds in your favor to win," because there is a greater PROBABILITY that you will win that lose, but the pays based on those PROBABILITIES mean the Odds are against you, not with you.

Furthermore, in terms of probabilities, that would only be true if you are making a single Lay Bet. If you make any pair of Lay Bets, then at best (4&10) you are as likely to win both bets as you are to lose one of them.

I don't get how multiple Place Bets, "Function," as a Lay Bet, would you be so kind as to elaborate? Is he suggesting that multiple Place Bets working are in some way, "Laying Against a Seven," in the sense that you are more likely to win one bet or another rather than lose them both? That's true enough, but it has no bearing on the House Edge of each individual Place Bet or on the Expected Loss of your combined Place Bets.

Quote:

I know that just one bet has won and the others are unresolved but, if treated as a unified Lay with that specified win amount vs. the amount invested and risked, that one win is all that's required. Everyone wants to define these as independent bets and while that may be how they were designed it is not the way they must be played.

Okay, so you're more likely to win one of your Place Bets than you are for a seven to be rolled and lose all of them...so what? If the Seven rolls, then you still do, in fact, lose all of them and that's how the House Edge presents itself. The House doesn't need a high probability of winning, it just needs to short pay YOU in comparison to the Odds when you do win.

That's an interesting thing about gambling, as it were. Strictly speaking, you actually lose money on your WINNING bets because you are being short changed compared to true Odds.

Finally, you don't have to calculate them as independent bets if you don't want to. Do you want to make it a dependent bet per roll and per bet resolved? Here you go, let's pretend you don't play the Pass Line and the Point is 10, so you are going to Place the 5, 6, 8, 9 and Buy the Four. You're going to bet $18 on the Six and Eight and $20 on everything else.

(21 * 10/36) + (28 * 8/36) + (39 * 3/36) - (96 * 6/36) = -0.69444444444 --- -0.69444444444/96 = -0.00723379629

Okay, so your House Edge per Roll is 0.723379629%

(21 * 10/27) + (28 * 8/27) + (39 * 3/27) - (96 * 6/27) = -0.92592592592 --- -0.92592592592/96 = 0.00964506172

Your House Edge per Bet Resolved is 0.964506172%

Don't get too excited by the, "Lower House Edge," it is only lower because I am artificially treating this as one bet, and this ONE BET resolves on three out of four rolls. The Expected Loss per roll, per Bet Resolved and thus, the House Edge remain the same on the bets taken individually.

((21 * 5/36) - (18 * 6/36)) * 2 = -0.16666666666

((28 * 4/36) - (20 * 6/36)) * 2 = -0.44444444444

((39 * 3/36) - (20 * 6/36)) = -0.08333333333

0.08333333333+0.44444444444+0.16666666666 = 0.69444444443

Thus, you can see that the Expected Loss per roll is the same for these bets individually (or considered together) which means that the overall Expected Loss and House Edges will remain constant.

Quote:

Well that got me going and using the Wizard’s formula for calculating Lay bets I ran through Zeke’s combined 6 & 8, 5 & 9, 4 & 10, Inside and Across bets and arrived at Zeke’s figures. So is Zeke correct even if he had the wrong reason? Can these multiple Place bets be treated as a unified Lay bet?

I suppose that you could look at multiple Place Bets as a Lay Bet against the Seven if you really wanted to. You are more likely to win the wager than lose, (as with a Lay Bet in which you are hoping for a Seven) but due to the fact that you are being short paid the Odds are against you.

I don't know who Zeke is, but if he was trying to say, "This unified Lay Bet has a reduced House Edge," then that statement is misleading, at best. The overall bet has a lower House Edge Per Resolution, but that is because the, 'Unified Bet,' is more likely to resolve on any given roll than a bet looked at individually.

I think that the most important thing to point out is the fact that it does nothing to change your Expected Loss on the bet(s), so that's what I really hope you take away from this post.

Author Zeke Feinberg

nice, Mission

it's another illustration of the confusion that can arise from bet combinations. There can just be things about them, like the change in HE, that confuse and make them seem attractive or even a secret way to beat the house. Even logic sense. Like I mentioned in another thread, when I first saw the doey-don't with odds played on one side only I thought that made a lot of sense.

Add in HE per roll and you evidently can really roll out the confusion it seems. There's that thing of placing the 6 and 8 and picking up the losing bet when the other bet wins ... this can somehow be shown to be a good way to go for the guy who likes to pick up bets and move them around, but is actually meaningless.

Very good synopsis.

Just IGNORE everything but the House Edge.

Quote: DeMango

Author Zeke Feinberg

DeMango,

What is authored by Zeke Feinberg?

Mission 146,

Thank you for the detailed response.

As I begin I ask for your forbearance as I struggle to dog-paddle my way across depths beyond my understanding. Some questions may seem naive and funny to some or outright stupid to others but they are my earnest attempt to understand the real why (if possible) vs. what’s bouncing around in my sieve of a head. So thanks up front.

If I may I would like to ask if you are or consider yourself to be an avid Craps’ player? Your response will have nothing to do with how I accept your answers but much to do with how I ask my questions. My experiences with those with math backgrounds, especially those with deep ones, has taught me they tend to be very precise and detailed leaving little room for misinterpreting meaning. I guess that really shouldn’t be surprising.

After reading your response it was quite evident to me that my questions were no where near as clear or precise and only created a fog for you to contemplate my intent. So you’re trying to answer what you’re guessing I’m asking. When I ask similar questions to fellow players they pretty much understand what I mean because we mean the same things. That erroneous assumption was my fault and I’m sorry.

I would like to take this one step at a time so I get the question correct and allow you to answer it fully before I say I don’t understand. Acceptable?

Needless to say anyone else who cares to join in is welcome to do so. Sometimes I need several knocks on the head before anything actually seeps in through the sieve.

BTW Happy New Year to all.

Kelph

I have a medium-deep mathematics background - -

Importantly - a couple of years of statistics in Graduate School.

I am an ardent (semi-high stakes) Craps Player.

I first stood up at a Craps Table in 1985. Thirty yrs. ago.

Generally I put in about 20 to 30 hours a month shooting dice.

I would love to see a question or two about my favorite game.

OK I want to dial back my original confuddled (yes I made that up) questions and see if I can ask a simple, coherent one with some explanation.

This is a Lay bet but created by the player rather than the casino and would require player discipline to play properly. The event we are betting against is the appearance of the 7 before a win. To keep this simple let’s say that we will Place the 6 & 8 for $6 each. Furthermore these two bets will walk arm in arm meaning that anything done to one must be done equally to the other so they function as one. Of the possible 16 events that will affect this bet 10 will win $7 and 6 will lose $12.

It doesn’t matter if the bet wins $7 on the 6 or 8, both are removed. Skewness of just a Placed 6 or 8 is +0.28 but Placing both makes it -0.81. This puts a number on the trade-off between a good chance at a small profit and a low probability of a big score. The average bankroll fluctuation on just a Placed 6 or 8 is $0.596 per dollar risked but Placing both makes it $0.511. However to be honest my readings have also said the expected loss due to edge per dollar risked remained at $0.0046 regardless of whether the 6 or 8 was Placed by itself or together.

For those not particularly adept at swimming in an ocean of math which includes my self the appearance of having more winning ways than losing to the six ways of a 7 is seductive. Especially since it seems logical and reduces the HA from 1.515% to 1.042% on the 6 and 8 with even larger reductions on other combinations if calculated as a Lay. Why not? Betting against the 7, more ways to win than lose but risking more to win less just seems like a Lay.

Now I tell myself that normally a Lay bet can only lose one number at a time even if there are multiple Lay bets compared to my multiple Place bets where all combined numbers go down. Just as quickly I think that’s because I lost against the event I was laying against so it still makes sense.

See how I keep confusing and twisting my self up over this. Help.

I need something more than saying each is an independent bet. Why doesn’t combining them really work? It’s not as if one can lose more than the other, they’ll lose to the same 7 but that’s calculated in the loss. Saying only one can win is true but they can’t win 10 ways unless combined and as I said in the beginning both are treated as a single bet.

Please give me a why this is wrong so I can put this to bed once and for all.

God I hope I explained this well enough.

Kelph

What you are missing:

It is a common human foible to focus on what occurs in the present - without seeing the big picture.

What mathematicians cite as "The Law of Large Numbers"

We CAN predict the future BECAUSE we can see the past, as so to speak.

Extremely simple is having a computer "roll" out a billion hours of Craps and tabulate the results.

All data can be easily extracted and viewed. It conforms to the "long run"

WHAT HUMANS DO is see a roll at a time and draw false conclusions based upon that.

The best example is (any form of) "HEDGE" betting, e.g. Pass Bet PLUS "any craps"

Each bet stands alone -- only in the instant moment does that APPEAR otherwise.

It is IMPOSSIBLE to combine two bets to improve performance, let alone using a bet with

a disastrously high House Edge.

To see this clearly -- just take a sample of random rolls -- just 36 works best --

because over the long run -- each roll will occur in its proper proportion.

I will use the example above, as it is so simple.

If I bet $5 on the Pass Line AND $1 on "Any Craps" This is what happens:

I will roll a seven (7) or an 11 eleven (11) EIGHT times in 36 and win [$5-1 = 4] X 8 = $32

I will roll CRAPS 4 times and win [$7-$5 = $2] on each of the 4 occurrences = $8

You invested $5 + $1 = $6 36 times for a total of $216

Your return was worse than betting $5 36 times. [That is $180]

That cost we stipulate at 1.4% = $2.52

In this hedging scenario the cost to you will be higher:

Your Pass line bets still cost you $2.52, but now we factor in the "Any Craps" bet of $1 X 36

That $1 bet will lose 32 times and win 4 times per 36 rolls.

Thus it loses $32 and wins 4 X $7 = $28 for an ADDITIONAL loss of $4.

You have "succeeded" in increasing your loss from $2.52 to $6.52

Not even close ! A very costly percentage increase. Suicidal in fact.

Quote: Kelph

This is a Lay bet but created by the player rather than the casino and would require player discipline to play properly. The event we are betting against is the appearance of the 7 before a win.

The fact that you're betting against something doesn't make this a Lay bet. All bets are FOR something and AGAINST something. By your definition, the Passline bet could also be called a Lay bet because it wagers against a 7-out.

Lay bets in craps are normally defined as being FOR the 7 and AGAINST specific point numbers. So if anything, you're using the term Lay opposite to it's normal usage - specifically you're betting AGAINST the 7 instead of FOR the 7. Of course you're free to use the term as you please but it would be less "confuddling" if you stuck to normal terms.

Quote: Kelph

To keep this simple let’s say that we will Place the 6 & 8 for $6 each. Furthermore these two bets will walk arm in arm meaning that anything done to one must be done equally to the other so they function as one. Of the possible 16 events that will affect this bet 10 will win $7 and 6 will lose $12.

You're starting down a slippery slope. The casino does not offer combined Place6/8 bets that walk arm in arm, therefore you've got to be careful how you calculate your expected value and what you compare it to.

Quote: Kelph

For those not particularly adept at swimming in an ocean of math which includes my self the appearance of having more winning ways than losing to the six ways of a 7 is seductive. Especially since it seems logical and reduces the HA from 1.515% to 1.042% on the 6 and 8 with even larger reductions on other combinations if calculated as a Lay. Why not? Betting against the 7, more ways to win than lose but risking more to win less just seems like a Lay.

Yes, it is seductive. This is why people play multiple Place bets - to have more ways of winning. So why stop with just 6 and 8? Why not bet them all?

Unfortunately, this is where you've jumped out of the pan and into the fire. You're comparing two figures that are not comparable. 1.515% and 1.042% are fundamentally different figures.

1.515% represents the average amount the house earns on Place 6 or 8 per winning or losing decision. The entire amount at stake plays a role in each winning and losing outcome.

1.042% on the other hand represents the average amount the house earns from your collective wagers per winning or losing decision. The difference is that the entire amount at stake does NOT play a role in every winning or losing decision. Yes, I know you'd like to think that it does because you've defined it that way, but unfortunately the house has already defined these wagers and will only take or pay them accordingly. For example, in your scenario:

10/16 chance to win 7 = 70/16 = 4.375
6/16 chance to lose 12 = 72/16 = -4.5
Average loss per decision = 4.375 -4.5 = -0.125
HA = 0.125/12 = 1.042% per decision

You've counted the entire $12 wagered as playing a role in each and every winning and losing decision. But did it really? No. It all played a role when a 7 rolled, but only half of it played a role when a winning number rolled. Now don't get me wrong, there's nothing wrong with calculating your loss as you have, but there IS something wrong when you compare it to a number that means something different.

One way to remedy this is to use loss per roll figures.

10/36 chance to win 7 = 70/36 = 1.944
6/36 chance to lose 12 = 72/36 = -2
Average loss per roll = 1.944 - 2 = -0.056
HA = 0.056/12 = 0.463% per roll

You can also arrive at the same figure by converting your 1.042% figure.
On average there are 36/16 = 2.25 rolls per decision.
So 1.042%/2.25 = 0.463% per roll

So you can see that your Place 6/8 combination yields exactly the same house advantage as either the Place 6 or 8 alone.

Now since most craps references list house advantage figures in loss per decision, I can see why you want to compare your combination to the standard 1.52% figure. That's fine, I personally prefer it and think it has advantages but then you need to properly figure the action. The action, as I'm using it, represents the amount of money wagered that's directly responsible for a winning or losing outcome. When evaluating a single wager, the amount of action is the same as the amount wagered because the entire amount plays a role in determining the amount won or lost. But in multiple simultaneous wagers the amount of action is generally not the same as the amount wagered.

10/16 chance of winning a $6 wager. Action = 60/16 = 3.75
6/16 chance of losing all $12 wagered. Action = 72/16 = 4.5
Total action = 8.25
Average loss per decision (from above) = -0.125
HA = 0.125/8.25 = 1.515% loss per dollar of action

So again, you can see that the Place6/8 combination holds the same house advantage as either the Place 6 or 8 alone. This is what we would expect. Craps bets are not synergistic.

Quote: Kelph

I need something more than saying each is an independent bet. Why doesn’t combining them really work? It’s not as if one can lose more than the other, they’ll lose to the same 7 but that’s calculated in the loss. Saying only one can win is true but they can’t win 10 ways unless combined and as I said in the beginning both are treated as a single bet.

If your goal is to measure your average loss on a given strategy (like Place6/8 combo) then what you've done is fine, but if your goal is to compare your average loss to the standard house advantage (and thereby presumably determine which is better) then you've got to get into the nitty gritty of how the house gains the advantage.

The house earns its advantage through winning and losing wagers. It does not earn money from wagers that push. If I bet $30 Place8 for 10 rolls and neither an 8 nor a 7 shows, then I take down the bet, how much money did the house earn? Nothing! The advantage is gained by structuring the bet so that the house wins more and/or loses less than it should according to true odds. It therefore needs wins and/or losses to effect the advantage.

When you win a Place 6 bet, does the dealer first look to see how much you've also wagered on Place 8 to determine your payoff? No, and therefore the money you wagered on Place 8 plays no role in determining the house advantage of Place 6. When you combine bets as you have, you're treating them as though the amount bet on Place 8 DID have a role in the payoff of Place 6.

Hope that helps.

Steen

Quote: Steen

The fact that you're betting against something doesn't make this a Lay bet. All bets are FOR something and AGAINST something. By your definition, the Passline bet could also be called a Lay bet because it wagers against a 7-out.

Lay bets in craps are normally defined as being FOR the 7 and AGAINST specific point numbers. So if anything, you're using the term Lay opposite to it's normal usage - specifically you're betting AGAINST the 7 instead of FOR the 7. Of course you're free to use the term as you please but it would be less "confuddling" if you stuck to normal terms.


You're starting down a slippery slope. The casino does not offer combined Place6/8 bets that walk arm in arm, therefore you've got to be careful how you calculate your expected value and what you compare it to.


Yes, it is seductive. This is why people play multiple Place bets - to have more ways of winning. So why stop with just 6 and 8? Why not bet them all?

Unfortunately, this is where you've jumped out of the pan and into the fire. You're comparing two figures that are not comparable. 1.515% and 1.042% are fundamentally different figures.

1.515% represents the average amount the house earns on Place 6 or 8 per winning or losing decision. The entire amount at stake plays a role in each winning and losing outcome.

1.042% on the other hand represents the average amount the house earns from your collective wagers per winning or losing decision. The difference is that the entire amount at stake does NOT play a role in every winning or losing decision. Yes, I know you'd like to think that it does because you've defined it that way, but unfortunately the house has already defined these wagers and will only take or pay them accordingly. For example, in your scenario:

10/16 chance to win 7 = 70/16 = 4.375
6/16 chance to lose 12 = 72/16 = -4.5
Average loss per decision = 4.375 -4.5 = -0.125
HA = 0.125/12 = 1.042% per decision

You've counted the entire $12 wagered as playing a role in each and every winning and losing decision. But did it really? No. It all played a role when a 7 rolled, but only half of it played a role when a winning number rolled. Now don't get me wrong, there's nothing wrong with calculating your loss as you have, but there IS something wrong when you compare it to a number that means something different.

One way to remedy this is to use loss per roll figures.

10/36 chance to win 7 = 70/36 = 1.944
6/36 chance to lose 12 = 72/36 = -2
Average loss per roll = 1.944 - 2 = -0.056
HA = 0.056/12 = 0.463% per roll

You can also arrive at the same figure by converting your 1.042% figure.
On average there are 36/16 = 2.25 rolls per decision.
So 1.042%/2.25 = 0.463% per roll

So you can see that your Place 6/8 combination yields exactly the same house advantage as either the Place 6 or 8 alone.

Now since most craps references list house advantage figures in loss per decision, I can see why you want to compare your combination to the standard 1.52% figure. That's fine, I personally prefer it and think it has advantages but then you need to properly figure the action. The action, as I'm using it, represents the amount of money wagered that's directly responsible for a winning or losing outcome. When evaluating a single wager, the amount of action is the same as the amount wagered because the entire amount plays a role in determining the amount won or lost. But in multiple simultaneous wagers the amount of action is generally not the same as the amount wagered.

10/16 chance of winning a $6 wager. Action = 60/16 = 3.75
6/16 chance of losing all $12 wagered. Action = 72/16 = 4.5
Total action = 8.25
Average loss per decision (from above) = -0.125
HA = 0.125/8.25 = 1.515% loss per dollar of action

So again, you can see that the Place6/8 combination holds the same house advantage as either the Place 6 or 8 alone. This is what we would expect. Craps bets are not synergistic.


If your goal is to measure your average loss on a given strategy (like Place6/8 combo) then what you've done is fine, but if your goal is to compare your average loss to the standard house advantage (and thereby presumably determine which is better) then you've got to get into the nitty gritty of how the house gains the advantage.

The house earns its advantage through winning and losing wagers. It does not earn money from wagers that push. If I bet $30 Place8 for 10 rolls and neither an 8 nor a 7 shows, then I take down the bet, how much money did the house earn? Nothing! The advantage is gained by structuring the bet so that the house wins more and/or loses less than it should according to true odds. It therefore needs wins and/or losses to effect the advantage.

When you win a Place 6 bet, does the dealer first look to see how much you've also wagered on Place 8 to determine your payoff? No, and therefore the money you wagered on Place 8 plays no role in determining the house advantage of Place 6. When you combine bets as you have, you're treating them as though the amount bet on Place 8 DID have a role in the payoff of Place 6.

Hope that helps.

Steen

Steen,

You've explained it beautifully. Thank you so much for taking the time! I was in the process of asking for this exact sequence of explanation when I saw you'd posted this.

Yes, thanks it was extremely helpful Steen.

Just by coincidence as I was jumping around online late last night looking for interesting Craps' ideas and I just happened to come across an old blog that had my very question tucked in near the end. I came here to share what I found and saw Steen's most excellent reply. So between that and what I found I think I can safely put that itch as definitely scratched.

I truly appreciate everyone's help and good luck to you all. Here's what I found in that 2006 Wordpress,com blog.

An interesting case (which can also be called a system) in which playing two bets simultaneously in a certain manner results in a lower house advantage. A place bet on the six or on the eight is well known for having a vig of 1.52% (calculated as ((5/11)*7+(6/11)*(-6))/6). However, when you play a place bet on the eight simultaneously and meticulously put both bets up at the same time and remove them both when either wins, the house advantage is only 1.04% (calculated as ((10/16)*7+(6/16)*(-12))/12)!

It's fascinating, but to be perfectly candid, the effect on house advantage is somewhat of an illusion. In both cases, a combination of bets allows us to win on average in fewer rolls of the dice, and then take down other bet(s) that will no longer be in jeopardy. In effect, we’re getting the decrease in house advantage by decreasing the time that the bets are at risk. If you examine the returns of the individual bets over time, they still exactly correspond to the known house advantages. In the end, you are ALWAYS left with the same conclusion, that the house advantage of every bet is firm and unbeatable, and no system can ever change that. In fact, all you can do by combining bets (hedging) is make things worse.

Steen, yours came with a lot of meat on the bone and I found the deli version. Between the two I think I'm just fine. Again thanks everyone.

Kelph