HA of 7 Out knocking out all the bets except the Don'ts

[mastercrapsman]

Has anyone seen a math computation of the House Advantage on the 7- out wiping out all the bets except for the Don'ts. It only takes one
wrong roll by the DI to wipe out his many gains.

Perhaps this particular 7-out HA should be included to the HA calculation for each bets.

Not sure if I am making sense on this...

[SHOOTITALL]

MC: I believe the HA is a constant. Numbers to not change it.
I think more of what you are looking for is here in MPs fine articles in #11. (Dodging Bullets on the Dark Side)
This is a link to many of them
http://dicesetter.com/mp/MP_toc.htm
It is always possible that I misunderstood what you are looking for.

[mastercrapsman]

Thanks Shootitall.

What made me think this is that: When I make my Point, all I get is the payment on the Point. When I roll a number not a 7, I get paid on my bet
on that number unless I am on that Dark Side. When I bet on the Dark Side, all I get is the payment on my Dark Side bet, I don't get a piece of the action on the other numbers. Of course I don't share the payment on the winning numbers not a 7 on the Dark Side.

I will check out the link. thanks.

[London Shooter]

Hmmm, I'm not sure what you are getting at MCM. What SIA says is correct. HA is HA, nothing changes it apart from the ability to throw less 7s and more other numbers which a DI tries to do. Other than that, over the long term, the edge you are giving up on each bet will equal the defined HA that the maths tells us. Over the short term, anything can happen, but no individual 7 out can be counted differently to any other.

[mastercrapsman]

How can I explain the fact that a 7 will shoot down all the bets except the Don't. So do you combine all the House Edge on each of your bets against the 7? Let's say, if all your bets are in one basket and that basket was held up by a hook (representing the no.7) if shooting that 7 makes you lose the basket containing all your bets versus shooting each one of your bets where for every round (bullet) you get (paid)on that one bet alone, then you make another shot and shoot another bet...and so on. Don't you have an advantage if you just shoot the 7. Granted that the Odds are already built in on the ways to make that nos vs the 7. Since you don't get paid on anything else but bet on that number vs losing everything else but the Don't.

this is probably as clear as mud or I muddied the water more.

[mastercrapsman]

also may I add that if you try to shoot the 7 and hit it, you only get the payment on your Wrong Bets and nothing else while the House takes all the
Right Side bets when the 7 shows. Don't you think there is a House Advantage or Edge on this fact?

[London Shooter]

I'm stuck in your mud I think

There is no doubt that if you have a "basket of bets" as you put it, such as say $64 across for example, then each of those bets is subject to a house edge and on average you will lose stake x house edge %age on all the action you have in play.

[mastercrapsman]

"The house advantage is basically the fee that the casino charges you to play the game. The way the house advantage is figured for craps is to determine what the odds are of winning and then compare that to what the actual payout odds are.
If you are playing a game that has a house edge of 1% then you would lose on average $1 for every $100 that you wager" - MP/dicesetter

if we add up all the HA of Pass Line Bet with the Point of 8 and Place bet of 6.

HA on PL = 1.42
HA on 8 = 1.52
HA on 6 = 1.52
total = 4.46

So the HA on all my bets is 4.46? meaning I am losing $4.46 per $100 I am waging when I am winning.

These are Odds of Winning vs Actual Payout Odds.

IMHO, Nothing to do with the fact that you lose both your PL and Place Bets when the 7 shows but only gets paid on the 6 or 8 when you roll it, not both.

[rhythm roller]

You are making me think too hard so I may muddy this more. I do see your point about winning all on the don't when a 7 is tossed but I don't think it is reasonable to look at all bets in one basket and here are my thoughts.

Each bet on the craps table is an individual game with the casino. It doesn't matter if right or wrong side. True that you only win one bet at a time on the right side and if you get too many bets on the table then that is over exposure of your bankroll.

You Place bet the 6 and 5. You might win three times on the 6 and none on the 5 before a 7 out, etc. If you lay or Don't come these same numbers you will lose on the 6 and win on the 5. Still each bet is individual. Yes, if you have several bets on the don't when the 7 comes you will win all of them.

Don't forget about the Come Out portion of the Craps game. The math suggests that the right side will win more times and the wrong will lose more times on that portion of the game and the same if using the Come or Don't Come box.

I think it is still dependent on the most basic thing which is how many times you can roll a certain number considering the faces of both dice. The HA doesn't change in my mind and it is up to me how many individual games I want to play against the casino at one time. That depends on my bankroll and my tossing skills and yes, a little luck as well.

I think it comes back to the math being the math and the casino has the upper hand in the long run.

[Dylanfreake]

Let`s put it this way. Say you are making a $10 Pass Line wager taking odds , then placing the 6 and 8 for $12 each, also placing the 5 and 9 for $10 each , and finally placing the 4 and the 10 for $10 each on each shooter.

What you are doing is paying the casino $2.64 ,which is the dollar and cent total that the 13.6% house edge percentage computes to (yes , I ended a sentence with a preposition.).

So at the end of a session , no matter whether you have had a winning or losing session , you have paid the casino $2.64 to play the game for every shooter. Say you played 20 shooters for that session. 20 shooters x the house edge of $2.64 per shooter equals $52.80 that you have paid to play the game regardless of whether you have won or lost money for that session.

At the end of a lifetime of play , if you add that $2.64 per shooter (which would amount to thousands of shooters) you should be even ---unless your shooting ability is such that it could lower the house edge.

Now someone can compute how good your shooting ability must be to over come a 13.6% house edge for the above wagers.

[DanF]

keeping your logic. Playing 3 box bets after CO would be a worst bet then placing only the field??

[London Shooter]

DF house edge isn't cumulative. You need to look at it as an average over all the bets you make. If you bet 5 an 9 for $10 each, the house edge is still 4% of the total bet. It doesn't become 8% because you have two bets in play.

[Dylanfreake]

Are you sure?

[mastercrapsman]

I believe HE is only cumulative on the winning number on that particular roll like Box Numbers which are also in the Field and Hardways e.g. 4 and 10. If we lose our Bet (s), the HE is moot.

I found the computations on the House Edges Per Bet Made, Per Bet Resolved and Per Roll. The HE Per Roll is calculated with this formula: HE Per Bet Resolved divided by the Average Number of rolls. The latter is used to know how good a particular Bet is. Not sure if this is useful.

Here's another thing that's puzzling to me: Why do we lose our turn whenever we roll a 7, even if we are betting on the 7 on the DP? In other words we lose (the dice) by winning.

If we don't lose our Turn on the 7 betting on the DP, the best play in my opinion is to just bet on the DP and nothing else so it will be 7 vs a number which will give the player the advantage on the Odds of making it.

Does this make any sense?

[London Shooter]

DF, yes I am sure. I am looking at this in percentage terms. Your figures for actual (theoretical) cost are correct and they are cumulative for each additional bet you make.

But think about this logically, if you were up against a 13.6% house edge for all the bets you mention you'd never play the game as your bankroll would almost certainly be wiped out within a few hands.... and the more bets you have, the more the house edge percentage increases and eventually you'd get to 100% before even rolling the dice

Maybe we are just looking at the terms differently, but the total given up of £2.64 for the bets placed is not disputed. Now it looks like you have $74 in action in your example - passline and 6 place bets. Let's leave odds bets out of this for now. In this case the house edge to me as a percentage figure is $2.64/$74 = 3.56%

Regardless, it also helps us realise how tough craps is to win at. If a $64 across bet is putting $2.50 in the casino's pockets each time, that's a tough edge to overcome and turn into a winning session.

[Knick111]

Mastercrapsman,the way i see it WHAT you are saying is that the 7 will beat every number on the table, so next time you go to your local casino BUY in for 20 units of 100 dollars chips, and bet one black chip on every shooter on the don't pass line, when you are ahead $200 dollars stop playing and go home ---you will sleep like a baby that night.

Jaime Garcia. not the baseball player.

[mastercrapsman]

Yeah Jaime, that's the best way in overcoming the HA, but not much fun for a DI. Perhaps if a DI lives in Vegas and plays often, that would be
the best MO.

Looking at the HA/HE from a practical standpoint, it is a cheap cost of entertainment if we play it right since except for the Buy Bet
it only mattered when we win and we hardly noticed it.

Keep on influencing!

[Knick111]

Mastercrapsman, are you looking to win at the game of craps OR are you looking to have fun,

if you are looking to have fun FIND yourself some redheads.

Jaime Garcia- not the baseball player.

[mastercrapsman]

Both. If we can win while having fun is much better. then the fun is paid for by the Casinos.
We have to skew that Perfect Distibution of Dice Frequency to Vary our way with what we know in DI.

Hard to tell the Redheads anymore, but red dice are for sure. he he

[220Inside]

If you're winning big, the redheads will follow...