What are the odds that....? (sorry, another odds question)

If someone did all 6 place bets....

What are the odds that the shooter will roll either a 4,5,6,8,9,10 5x before rolling a seven, if they keep all the bets on until big red?
What are the odds they will roll a 4-10 3x before rolling a seven, if they remove the bet when they hit?

What am I missing here? 7=16%... 4-10 combined equal 66.66%. That right there is 3 and 3/7 to one. What odds would you be playing against the house if you did place bets (buy 4/10) on all 6 points and
A) Removed them after hitting them?
B) Kept them all up until seven out?

I ran 1,000 up to 10,000 today. Then I ran 10,000 up to 100,000. Then I ran 100,000 up to a million. With play money on my iPhone of course... lol. I'll post a screenshot if anyone doubts me. I have only played using the latter method (b) about 30-40 times in the last day or so, always starting out from 1,000. The record goes approximately something like this: 1,000 to 10,000 quit. 1,000 bust. 1,000 to 5,000 quit. A couple of busts, no more than 4. Then the 1,000 to 1 million run which took a good 4 hours playing solo on an iphone app which obviously entails me quickly rolling much faster than live. I "cashed out" (in my head) at just over a mil. Since then I've had about 10-15 busts mixed in with a few 3-5000 dollar "cash outs" (in my head) and a couple of double ups. I know that this is nothing for a sample size, and might just be sheer luck, but I'm interested in the odds of both of these methods. I googled it and couldn't even find people talking about betting all the place bets and keeping them on until the seven, except for one vague mention where a guy said "bet all the place bets and the casino will love the action". I'm loving the action.

My preference right now is buy-in of 1,000. Then bet 10 on place bets until double up, then double up the bets accordingly. It seems like a run where you can 10x your buy-in should come 3 times consecutively sooner than you buy-in and completely bust 1,000 times, right??? Did I just goober luck out today? Every roll, you have a 16 percent chance of losing 6 units, and a 66.66 percent chance of winning anywhere from 1.15 to 1.95 units. I feel like streaks of no sevens for 5+ rolls come more frequently than streaks of a 7 every 4 rolls consistently. I'm especially interested in the method A as well, where one removes the bets after hitting them, as they only need to hit 3 total, the last of which (and therefore least likely) is still up while 4 total place bets are out there. Even then you've got a 2 to one favorite against the seven, even if the 6/8 are gone and you're still shooting for the 4,5,9,10. And the fact that the payout is screwy is irrelevant because hitting the 3rd place bet (in method a) already insures that this roll was profitable even if the other 3 bust to the seven.

Anyways, I've been up all night playing with stupid money, so excuse the rambling.

Quote: 2Infinity

Did I just goober luck out today?

This sounds about right. No combination of craps bets can overcome the house edge. You've done a lot of the math already: your expectation is negative.

Quote: 2Infinity

If someone did all 6 place bets....

What are the odds that the shooter will roll either a 4,5,6,8,9,10 5x before rolling a seven, if they keep all the bets on until big red?
What are the odds they will roll a 4-10 3x before rolling a seven, if they remove the bet when they hit?

32.77% 5 place bets before a 7 wipe out
(51.20% of winning just 3 place bets before a wipe out)

0.5^3 or 1 in 8 for 3-4,10s before a 7
start here. A table of results and how to figure it on your own
Place Bets

Quote: 2Infinity

What am I missing here? 7=16%... 4-10 combined equal 66.66%. That right there is 3 and 3/7 to one. What odds would you be playing against the house if you did place bets (buy 4/10) on all 6 points and
A) Removed them after hitting them?
B) Kept them all up until seven out?

A lowers your variance
B, without buying the 4 and 10 has the highest house edge of about 3.73%

Place Bets HE

Quote: 2Infinity

What am I missing here? 7=16%... 4-10 combined equal 66.66%. That right there is 3 and 3/7 to one.

6 ways to roll a 7
24 ways to roll a 4,5,6,8,9,10
Place bettors do not care about the 2,3,11 and 12. They do not count

6/30 = 20% a 7 before ANY place number
24/30 = 80% probability of any place number before a 7.
80% is NOT 100%

Quote: 2Infinity

My preference right now is buy-in of 1,000.
Then bet 10 on place bets until double up,

Sounds like your iphone craps app has errors in it.
I trust only the code I can see and verify.

Running my Craps simulator, WinCraps and risk of ruin calculator I get, at best, a 23% chance to double a $1000 bankroll with your $10 bets across.
So in 100 attempts to do this you will average about 23 times being successful.

IF 1 million others attempted this because they believe you and you iphone app results, about 770,000 would be in bustville and you would not be a god to them, more like a devil.

IMO, Step up to some real craps simulators like WinCraps and dump the iphone crap crap app.

"place bettors don't care about the 2,3,11,12. They don't count." So does the crapsgod just remove those when determining probability, for my convenience as a place bettor? No. There is always only a 16% chance of hitting the 7, and a 66.66% chance of hitting any place bet. You can't leave out numbers not pertaining to your bet and expect the probability to change. That's lol. Maybe my iPhone app doesn't have a good random dice generator, but I think you have the wrong idea. I didn't say "I am a god because I have figured out how to beat craps, so follow me." I have merely expressed an interest in learning the probability when betting the way I described, which you didn't provide any help with.

"(51.2% of winning just 3 before a wipeout". Is this considering if you were to remove bets upon hitting them? If so, then that means you have a 51.2% chance of making a profitable shoot. If you are meaning a 51.2% chance with all 6 bets still up then nm... If you mean the latter, then what are the probabilities of the former?

Quote: 2Infinity

"place bettors don't care about the 2,3,11,12. They don't count."

Place bets are a bit different since a 2,3,11, and 12 does not cause a bet to be resolved.
6/36 is the probability of rolling a 7 before ANY other number
but 6/30 is the probability of rolling a 7 before any place number.

24/36 is the probability of rolling any place bet before a 7 or a horn number
24/30 is the only probability a place bettor should be concerned with. A resolved bet is what makes the money move.
That is where the 80% win rate on any roll for the place bets comes from.

My simulations also show about a 22% chance of doubling a bankroll of $1000 with both your methods. Method A is lower by a few points.
I used buy bets for the 4 and 10 with the vig paid on a win only. The only way to Buy.

read another view here
Alan Krigman - You Can Win 80 Percent of the Time... And Still Lose

Quote: 2Infinity

"(51.2% of winning just 3 before a wipeout". Is this considering if you were to remove bets upon hitting them? If so, then that means you have a 51.2% chance of making a profitable shoot. If you are meaning a 51.2% chance with all 6 bets still up then nm... If you mean the latter, then what are the probabilities of the former?

Yes both ways. resolved bets.
You can not go $64 across and hit just 3 numbers and show a profit unless you press and hit a repeat.
You need 4 or 5 wins.

I don't know which 'ap' you are using, but don't be 100% sure it uses a random number generator. I play Pai Gow Tiles with an ap with a co-worker, and we have turned $10,000 into many billions of dollars. The ap has a poor house ways, but I have a hunch that there is something non random about the distribution of hands, somehow it favors the player. As you know, many online casinos 'allegedly' do something like what is happening to you on their 'free' games, then when you think you know how to 'beat' them and use real money.... well.... you know what happens.

No, if you take them down then you have made 370$ (even if it's only the 5,6,8)off $100 bets after hitting 3, and since you bring down the bets when you hit them (that's what lowers the % on the 2nd and 3rd hit, making me wonder if the 51.2 applies to bets that are taken down upon hitting) then you only lose $300 if the seven hits immediately after the 3 place bets. Only when you leave all bets up do you need to win 4 or 5 to make a profit when you're expecting to lose all 6 place bets when the 7 hits. But using my goofy app, there seems to be a lot of hot streaks to be had from leaving them up and just getting paid nearly every roll while waiting for the seven... I guess it's the fact that the 51.2 percent of time you make a "profitable" shot, you are only guaranteed 370, while the 48.8 percent of time when you are likely to lose, you are losing 400+ dollars, depending on how many bets you hit/have remaining that keeps it from being profitable.

Here's another question... What are the odds of playing the Labouchre betting system on the Don't Pass bar and removing it when the point is a 6,8?

Quote: 2Infinity

No, if you take them down then you have made 370$ (even if it's only the 5,6,8)off $100 bets after hitting 3, and since you bring down the bets when you hit them (that's what lowers the % on the 2nd and 3rd hit, making me wonder if the 51.2 applies to bets that are taken down upon hitting) then you only lose $300 if the seven hits immediately after the 3 place bets.

I really messed up when I read your first post.
OK.
My sim results still show about a 22% chance to double a $1000 bankroll with $64 across, Buy the 4 and 10.
19% chance to double $1000 after removing the 3 place bets that hit.

You are right that removing the place bets after a hit lowers the overall probability of winning the very next place bet.
It is worse when the first nunber rolled is a 6 or 8 and less with a 4 or 10.
This can be calculated exactly by listing all the combinations, but for me a simple sim show that the calculations are not all needed.

I get a 44.4% chance of hitting 3 place bet numbers while removing them after a hit.
Instead of removing the bet, just reducing the bet will give slightly better results.
Compare that to the 51.2% for 3 numbers while not removing the bets.

24 ways to win and 6 to lose (4 to 1) 0.80 win probability for one place bet
Place 6 hits, remove place6
Now, 19 ways to win and 6 ways to lose or .76 probability of hitting another place bet since you removed the Place6 already.

Hitting 4 or more has really gone down.
0.4096
to
0.29144

hits	prob	or less	or more
0	0.20012	0.20012	1
1	0.18803	0.38815	0.79988
2	0.1674	0.55555	0.61185
3	0.15301	0.70856	0.44445
4	0.12887	0.83743	0.29144
5	0.1008	0.93823	0.16257

Quote: 2Infinity

Here's another question... What are the odds of playing the Labouchre betting system on the Don't Pass bar and removing it when the point is a 6,8?

I missed your question.
Really do not understand what you are asking.

One should never remove a don't pass or don't come bet once there is a point established for it.
Have no fear, the odds are now with you.

The Cancellation Betting System

Well the labouchre system still only increases your (dis)advantage to slightly under 50%, depending on bankroll or buy-in. I'm wondering if removing the dont pass bet when the 6 or 8 is the point, since you are better off as the 7 against the 4,5,9,10, would increase the labouchre advantage to greater than 50% when playing the don't pass line...

Quote: 2Infinity

Well the labouchre system still only increases your (dis)advantage to slightly under 50%, depending on bankroll or buy-in.

Any house edge game, no matter the betting system used, can never double a starting bankroll with better than a 50% chance. There is none.
The best it can be on a NO house edge game, for these even money type bets, is 50%.
Using the Wizards figures, to double a 10 unit stake is about 47.8% chance with a 1.4% house edge.
0% HE would be 10/20 or 50%.
It does not get any better than that and only slightly better using the free odds bet.

Quote: 2Infinity

I'm wondering if removing the dont pass bet when the 6 or 8 is the point, since you are better off as the 7 against the 4,5,9,10, would increase the labouchre advantage to greater than 50% when playing the don't pass line...

Do not wonder. It will not.

Not one math formula says removing a don't pass or don't come point of 6 or 8 is in any way advantageous to any craps player at any time.

If you want the best chances to double a starting bankroll, try the minimum bet on the line bet and take or lay the max odds.
Bold play says to bet $1000 to win $1000 if you want the highest probability of hitting a win goal.
Playing against a house edge, do not mess around if you want to win.
Just bet and be done with it.

Quote: 2Infinity

Here's another question... What are the odds of playing the Labouchre betting system on the Don't Pass bar and removing it when the point is a 6,8?

Just to pick nits...

It is called the "Don't Pass" and 'bar' is a verb. The layout says "Don't Pass Bar 12" but is should be read "Don't Pass, Bar 12" meaning that they push, don't pay, or 'bar' a crap 12.

This question is related to a question I previously asked on this post so I'll just add it here...

What are the odds of losing an initial buy-in in the following situation:
Betting on the don't pass using the Labouchre system
Lines consist of following:
5 5 5 5 10 10 10
X 10 lines total until stop
Buy-in= 500

Also, what are the odds of losing this 7x in a row?

Thanks and best regards.

Quote: 2Infinity

"place bettors don't care about the 2,3,11,12. They don't count." So does the crapsgod just remove those when determining probability, for my convenience as a place bettor? No. There is always only a 16% chance of hitting the 7, and a 66.66% chance of hitting any place bet. You can't leave out numbers not pertaining to your bet and expect the probability to change. That's lol. Maybe my iPhone app doesn't have a good random dice generator, but I think you have the wrong idea. I didn't say "I am a god because I have figured out how to beat craps, so follow me." I have merely expressed an interest in learning the probability when betting the way I described, which you didn't provide any help with.

7Craps is 100% correct, those numbers are completely irrelevant to the bet. Is any result except for 1-3, 3-1, 2-2 or any Seven now somehow relevant to the resolution of a Hardway bet on Four? They are relevant only to expected loss per roll, but no other reason.

Gotcha. Now how's about my question from October? ;)

Quote: 2Infinity

This question is related to a question I previously asked on this post so I'll just add it here...

What are the odds of losing an initial buy-in in the following situation:
Betting on the don't pass using the Labouchre system
Lines consist of following:
5 5 5 5 10 10 10
Buy-in= 500

Also, what are the odds of losing this 7x in a row?

Thanks and best regards.

The question is irrelevant without a Stop-Win goal. Losing an initial buy-in compared to winning what? Do you just mean winning the $50 that the Labouchere Line you posted would give you, or are you wanting to complete such line multiple times?

Either way, I defer to my good friends 7Craps and Guido111 on this one, it requires a simulation. There are too many possibilities (in terms of Win/Loss combos) for the Labouchere Line. The House Edge would be no different, of course, and you are better to Lay the Odds rather than play DP straight-up, though even that does not do anything to the DP HE taken alone because the Odds are technically a separate bet.

Quote: 2Infinity

Gotcha. Now how's about my question from October? ;)

LOL

I noticed that after I posted!

Sorry... Edited... Duh

Quote: 2Infinity

Sorry... Edited... Duh

2I, this is your lucky day!


https://wizardofvegas.com/forum/gambling/betting-systems/9787-cancellation-blackjack/

Quote: Mission146

I have extensively studied the Labouchere System and have done independent research on the System.

M146 is your man.

The Wizard has posted his Lsystem code and M146 is the expert here.

2+2=4

Good Luck guys!

Quote: 7craps

2I, this is your lucky day!


https://wizardofvegas.com/forum/gambling/betting-systems/9787-cancellation-blackjack/
M146 is your man.

The Wizard has posted his Lsystem code and M146 is the expert here.

2+2=4

Good Luck guys!

Did I say I ran simulations on it? Does, "Extensively studied," also mean that I have proclaimed myself to be an expert? Specifically, I simply meant that I have operated the Labouchere System, read about it in quite a few Gambling-Related texts, did some manual simulations. As stated, I also wrote the vast majority of the Wikipedia article about it which has gone unrefuted.

By the way, I'm still waiting for those simulations in AHigh's video thread to refute the Math that you didn't like about the extra Hardway with otherwise perfect roll distribution, do you have an ETA on that?

Quote: 2Infinity

Sorry... Edited... Duh

Perhaps Guido111 will run simulations.

There's one other thing that needs specified, though, since you are saying you want to win a total of $500 with Labouchere Lines that, when completed, win $50: Does the amount of a winning Labouchere Line (prior to winning ten times) get added to the buy-in of $500, or is the next Line still based on the $500? IOW, if you were to complete nine Labouchere Lines for +$450, would you then have $950 with which to pursue the tenth Line or still just $500?

Thanks for the informative link. I'm hoping I find something answering my specific question. I'm mostly interested in knowing the odds of losing the 10line set described above 7x consecutively.

Good question. I assumed one would add the previous lines won to the bankroll of 500 but, which would be preferable?

As stated, nothing can improve the HE of the bet, so nothing is, "Preferable."

You either put more at risk with a greater liklihood at winning a set amount or you put less at risk with a lesser liklihood of winning the same amount. For example, if you make your buy-in $50, based on the Line provided, you would be done after two consecutive losses because you cannot make the next bet as demanded by the system. In this event, the liklihood of completing the line would be reduced, but you would only be losing $35 everytime you failed.

If your buy-in is a greater amount, you will succeed more often, but lose more when you fail.

Final question:

I'm looking for the odds of losing while using the following betting method:

10x Labouchre lines i.e.
5 5 10 10 10 10
where I buy-in for 10x a single line (500).
If I buy-in for 500 and play until I complete 10 lines by betting on the don't pass, what are the odds of me losing the initial buy-in if I add every previously won line to the 500 buy-in as they are won?
Also, what are the odds of me losing the buy-in 7x consecutively?

Thanks and best regards.

Most importantly, I'm curious as to how probable/improbable it is to bust 7x in a row specifically.
For the more simple math-minded folks:

Labouchre betting method on the don't pass,
buy-in=10x line total
set is complete when 10 lines are finished
Line looks like: 5 5 10 10 10 10
What are the odds of doing this and losing 7 times consecutively?

Thanks and best regards.

2I,
your best bet is still with M146.

I just have no interest in the L method currently.
But I now like Laura Nyro and really did not much in the past, the 60s and 70s.

I still say he is the expert here. A compliment taken as not.

He has the Wizards C++ code, everyone can get it, he just needs to learn how to use it.
I am certain the Wizard would assist him in this if M146 just asked.
They are both nice guys that do not mind helping others

It is not rocket science.


Maybe JB should just write a few apps for running all sorts of simulations.
People would pay $10 here and there for them, even more if they did more.

I appreciate the compliments, 7Craps. If this has not been done by the end of G2E, then I may ask Wizard if he is interested in helping me understand how to run his simulation. He will be quite pre-occupied until G2E is over. I can't promise that he will show me how, programming might as well be rocket science to me because I know less than nothing about it.

i hate to change the subject can you guys or ladies explain off and on in relation to the come bet??? thank you

Quote: leftend22

i hate to change the subject

at least you didn't use a blog post to ask a question!

To start a new thread go to "forums", then click on the proper subject. An option to start a thread will be an icon in the upper left.

Quote: leftend22

can you guys or ladies explain off and on in relation to the come bet??? thank you

The only bets that are normally turned 'off' are those that are still unresolved when it is a new come-out roll, but you don't want to take the bet down. You want the 7 rolled as a right bettor on the come-out, but a 7 would take out this unresolved bet. So it can be called "off" during the come-out. This can be automatic and generally a particular casino will assume certain bets are off without marking them so. Some variation does occur, as a local rule.

Quote: leftend22

explain off and on in relation to the come bet??? thank you

Winning come bets are normally paid after removing them from the point box in the come box.
A new come bet is then placed in the point number box that just rolled.
This is a lot of extra work just to have the same bet with or without odds.

From Precision Dice Dealing by Dale Yeazel
“Off and On”
“Off and on” refers to the procedure that can be used when a player has a winning come bet on a
point number box that is the same amount as his new come bet.
Instead of taking the come bet on the number down,
paying it in the come and putting the new bet on the number,

you can pay the new come bet and say to the player; “Off and on.”
Come bets may be paid “off and on” anytime the new come bet is the same amount as the flat bet on the point number box.
However, if the new come bet or the flat bet on the number are unreadable, it may be easier to just pay the come
bet on the number and then bring the new come bet to the number.

From Taucer/Cutolo Craps Manual p.75
...the bet on the 10 and the come bet are the same amounts (the come 10 has odds or does not)
The roll is 10
The Dealer may now pay the winning bet DIRECTLY and say "OFF and ON", or "OFF and ON, with odds"
without touching the bet on the #10

If the Dealer took the bet off the #10 and paid it and put the new bet on the #10 it would be the same thing.
Note: to "off and on" a bet, the come bet and flat bet in the box must be the same amount.

if my come bet hits on the 10 and i have a 5 flat and 20 odds and the new come bet is also 5 flat and i want the odds on how much do i get paid ????

because the flat be on the #10 is the same as the come bet $5

You would be "Off and On with odds" with a total payoff of $45
placed right next to your $5 come bet.
Take the $45, leave the $5 come bet right there, and you are ready for the next roll.
You have a $5 come bet and a #10 with $5/$20 just like before the last roll

Instead of this.
Your $5/$20 #10 wins, is moved to the come to be paid and
is paid a total of $45 ($5 flat + $40 odds) for a total bet and winnings on the come of $70

Your $5 come bet moves to the #10.

You want the same odds of $20 on it.
The Dealer takes $20 from your $70 pile (leaving you with a $50 pile)
and you now have your $5/$20 on #10
You want the same $5 come bet, So you pick up your $50 and remove $5
and place that on the come line.

The winnings in your hand that is left is $45
You have a $5 come bet
Your #10 is $5/$20odds

off and on is faster and cleaner
and results in exactly what you want
without the Dealer ever touching your #10 bet $5/$20 or the new $5 come bet

thanks johnny that clears up what i needed to know