kauboj
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January 6th, 2011 at 8:15:21 AM permalink
So i ending last year on a loosing streak but stayed ahead of the casino overall. Nice...!

I started this year on a loosing streak as well... not good... lol

But i was thinking of ways to enjoy the game longer and limit my losses... Now i fully understand that there is no gambling system that will ever break the house or even guarantee me to win..

But what i am curious about is what would my expected chances be if i follow either of these betting structures..

1. Bypass the pass/dont pass bet and wait for point to be established. then place minimum bets on the place numbers allowing 1 roll then turning off the place bets and waiting for the (7) or the point to hit. and repeat.

2. Make the pass line bet and wait for the point then place minimum bets on the place numbers minus the point and place 1x odds behind the pass. allowing 1 roll and turning off the place bets and wait for the (7) or the point to hit and repeat.

curiosity kills me.
odiousgambit
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January 6th, 2011 at 9:23:29 AM permalink
I would stay away from the place bets, even on the 6 and 8, and go with the free odds to have the best odds of keeping a bankroll alive. No guarantee that you won't lose money faster than ever though.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
kauboj
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January 6th, 2011 at 9:34:47 AM permalink
by no way are odds free.... many times i lose just as much on the pass line as i win...
kauboj
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January 6th, 2011 at 9:34:52 AM permalink
by no way are odds free.... many times i lose just as much on the pass line as i win...
TIMSPEED
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January 6th, 2011 at 9:49:36 AM permalink
Just play minimum pass line with only single odds. This is the greatest move you can do to reduce HA while keeping the variance low.
I play on a $5 table, and I'll usually buy in for $100 and play for as long as I care to, this way.
Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ
superrick
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January 6th, 2011 at 9:52:28 AM permalink
kauboj

I am not a math guy, but I can tell you trends do happen on craps tables. I know that there is no system that will beat the game of craps every time I go to the tables.

If I didn't have the dice in my hands, I would just place one or two points that everybody has been hitting, for at lease two units. Wait three rolls of the dice and then take them down or turn them off!

If everybody is having roll's that are more than 8 rolls of the dice and you do get one hit on you bets before you took them down, then you can regress your bet down to one unit, and when you get one hit on that bet, press for the moon.

I would never bet more that one or two bets, don't get greedy, thinking you can bet all the points, because the more bets you have, they will kill you when there is a short roll.

Bet the points that you see hitting, it might take you some time standing at the table to see what points are hitting, but it will be worth your time. All the books say bet the 6 and 8, but if no one is rolling them why would you bet on them, trends do happen.

Now I know that all the math guys will tell you that what happened in the past has no effect what is happening now, true in a math problems, but not on the tables.
The next time you go into a casino, just take paper and pen write down how you would be betting and see what the outcome would have been, if you tried to bet this way!

Remember there are days that you will see everybody on the table never get over a few roll's of the dice most will never make it past the 4 rolls of the dice, stay off that table.

This is not a system, just a smart way of playing if you have plenty of time, it will not work all the time, but it will give you time on the tables and you might win, if there is a good trend happening that day!

Living in Vegas you will see some of the retired guys using this type of betting, they are not looking for big wins, they just want to make a few dollars a day, and to socialize with they rest to the guys, it gets them out of the house for the day! These guys can't afford to lose anything as they are all on fixed incomes, for the most part...!


...
Note, all my post start with this is just my opinion...! You do good brada ..! superrick Winning comes from knowledge and skill when your betting and not reading fiction http://procraps4u2.myfanforum.org/index.php ...
FleaStiff
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January 6th, 2011 at 9:55:36 AM permalink
Quote: kauboj

by no way are odds free....

You ain't being charged for them... so they are free.
If you make a Place Bet you will be paid off at place-bet odds. If you want to have your place--bet paid off at true odds, you have to buy that pay out and they put a little "buy" lammer on your bet.
The Free Odds bet requires no such lammer. You can think of it as not being charged for the true odds or as the charge for the true odds as being included in your flat bet or you can think of the charge for your Free Odds bet as being the risk of your going through your bankroll faster and allowing the casino to shear another sheep sooner. Whatever way you think o fit is up to you, but you don't have to shell out for getting the Free Odds.
kauboj
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January 6th, 2011 at 10:05:06 AM permalink
of course there is no guarentee how long the dice will roll but on average you can see a roller roll 5 times before the 7 hits... of course gotta hate the days when point then 7 rolls...

but im thinking yeah you are betting more... but lets say i play the 6 and 8 and it takes 4 rolls too get a 6 or 8 but roll number 3 was a 7... i still lost... however lets say i bet 4,5,6,8,9,10 and roll number 2 is 4 i win and turn off... and lets say roll 2 is an 11 i didnt win but didnt loose and i still turn off... i figure the shortest amount of time my money is in play the lower the chance my money goes to the house
kauboj
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January 6th, 2011 at 10:07:37 AM permalink
of course they arent charging you for the odds payout behind the line.. but the longer the money is in play the better the chance to loose.

to me free odd is where you dont lose your come odds on a new comeout roll...anytime you lose something and dont get it back it isnt free...
FleaStiff
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January 6th, 2011 at 10:10:05 AM permalink
Quote: kauboj

but the longer the money is in play the better the chance to loose.

Don't you mean the better the chance to win?
goatcabin
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January 6th, 2011 at 10:12:05 AM permalink
Quote: kauboj

So i ending last year on a loosing streak but stayed ahead of the casino overall. Nice...!

I started this year on a loosing streak as well... not good... lol

But i was thinking of ways to enjoy the game longer and limit my losses... Now i fully understand that there is no gambling system that will ever break the house or even guarantee me to win..

But what i am curious about is what would my expected chances be if i follow either of these betting structures..

1. Bypass the pass/dont pass bet and wait for point to be established. then place minimum bets on the place numbers allowing 1 roll then turning off the place bets and waiting for the (7) or the point to hit. and repeat.



This will certainly keep you at the table longer, since you are only risking your money for one roll. You are not going to be very welcome at the table, however, taking up a spot and staying out of action much of the time. However, by betting on all the numbers, you are paying a much higher "tax" than on the pass line. The place bets on 4/10 carry a house edge of almost 7%, the 5/9 4%. Suppose the point is 6 and you place all the other point numbers. You have 19 ways to win one bet and 6 to lose them all, combining for about 70% of the outcomes; the other 30%, you will call your bets off. BTW, if you win one of those place bets, do you still turn them all off? Depending on the mix of points, you're going to be paying something over 4%.

q=kauboj]2. Make the pass line bet and wait for the point then place minimum bets on the place numbers minus the point and place 1x odds behind the pass. allowing 1 roll and turning off the place bets and wait for the (7) or the point to hit and repeat.



Same issue here -- the place bets on the outside four numbers have too much house edge for my taste. Certainly, the pass w/single odds makes this strategy more attractive. I assume you're not calling your odds off after one roll.

Superrick tells you to chart the table and bet accordingly, but that's nonsense, of course. He has absolutely no understanding of the principle of independent, random events.
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
kauboj
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January 6th, 2011 at 10:34:23 AM permalink
not if your looking realistically with the fact that you have a 1 in 6 chance to roll a 7... the longer you play... the odds increase in the favor of the house that a 7 will roll... why do you think that odds on the dont pass are layed and paid opposite what you would get on passline odds
kauboj
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January 6th, 2011 at 10:41:25 AM permalink
One roll only win/lose or craps or 11 rolled...

on the second option i would leave the odds on....

i am just figuring that with the average roller rolling 5 times before a 7 chances are pretty good that id win more times than i lose... question will the number of times i win outway the money i loose.

if i bet 1 unit at $32 i would need to win 5 times before i lose once to not lose money and be $3 ahead.

i play at a small casino in delaware and i usually play during the day when the tables are slow... only problem is the casino rarely puts up a $5 minimum. and sometimes on cold tables they will even jack the tables to $25 minimum......

kinda makes me hot... i think if you have a 25 table you should have a 5 table if you have a 15 table you should have 10 table.
goatcabin
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January 6th, 2011 at 10:59:29 AM permalink
Quote: kauboj

not if your looking realistically with the fact that you have a 1 in 6 chance to roll a 7... the longer you play... the odds increase in the favor of the house that a 7 will roll



That, of course, is the Gambler's Fallacy. The probability of a seven rolling never changes, nor does the probability of any of your place numbers. On every roll of the dice, your chances of winning or losing a bet are exactly the same. The only affect of calling your bets off is that you are out of action -- which does reduce your expected loss. Of course, the logical extension of that reasoning is never to play, reducing your expected loss to zero! >:-)

Quote: kauboj

... why do you think that odds on the dont pass are layed and paid opposite what you would get on passline odds



Obviously, because the seven has more ways to roll than any of the point numbers. The payouts on odds bets are symmetrical, and designed to result in no edge, for the casino or the player. This, however, is irrelevant to the issue.
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
kauboj
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January 6th, 2011 at 11:08:26 AM permalink
its not really the fallacy.. im not a mathmetician but calculate it out... the odds of not rolling a seven decrease the more rolls that are made.. obviously the the chance on each roll is independent of itself.. but realistically the odds are your gonna roll a seven the question is when.

take for example the hard eight... bet 1 unit on h8 win parlay win parlay win keep 900 is it possible to roll 3 hard 8 in a row yes.. but the probability of 3 in a row hitting before a soft 8 or a 7 ... greatly decrease as you continue to roll..
goatcabin
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January 6th, 2011 at 11:12:09 AM permalink
Quote: kauboj

One roll only win/lose or craps or 11 rolled...



OK, so you're placing the point, as well. This means you have 24 ways to win one of your place bets, 6 ways to lose them all at once, 6 ways to call your bets off.
6 ways to win $9 on 4/10
6 ways to win $7 on 5/9
10 ways to win $7 on 6/8
6 ways to lose $32

That nets out to a loss of $26, doesn't it, assuming that when you win a bet you still call them all off.

Quote: kauboj

on the second option i would leave the odds on....

i am just figuring that with the average roller rolling 5 times before a 7 chances are pretty good that id win more times than i lose... question will the number of times i win outway the money i loose.

if i bet 1 unit at $32 i would need to win 5 times before i lose once to not lose money and be $3 ahead.



Well, you don't have five times as many ways to win as to lose, do you?
Keep in mind that every one of those bets is independent of all the others.

I'm still not clear on what you do if one of your place bets wins -- do you leave the others up, leave them up and replace the winning bet or call them all off?
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
goatcabin
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January 6th, 2011 at 11:21:52 AM permalink
Quote: kauboj

its not really the fallacy.. im not a mathmetician but calculate it out... the odds of not rolling a seven decrease the more rolls that are made.. obviously the the chance on each roll is independent of itself.. but realistically the odds are your gonna roll a seven the question is when.

take for example the hard eight... bet 1 unit on h8 win parlay win parlay win keep 900 is it possible to roll 3 hard 8 in a row yes.. but the probability of 3 in a row hitting before a soft 8 or a 7 ... greatly decrease as you continue to roll..



No, your reasoning really is the Gambler's Fallacy. You are confused between the odds of a series of events BEFORE the series starts, and the changing odds as the event unfold. When you calculate the odds of a series of events, how do you do it? You use the same probability for each event, right? Doesn't that tell you something?

Please read my blog on this subject:

https://wizardofvegas.com/member/goatcabin/blog/#post16

Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
MathExtremist
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January 6th, 2011 at 11:24:01 AM permalink
Quote: kauboj

take for example the hard eight... bet 1 unit on h8 win parlay win parlay win keep 900 is it possible to roll 3 hard 8 in a row yes.. but the probability of 3 in a row hitting before a soft 8 or a 7 ... greatly decrease as you continue to roll..



No they don't. The probability of 3 hard 8s in a row is constant, regardless of how long you've been rolling, as long as you start counting from the next roll.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
kauboj
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January 6th, 2011 at 11:28:35 AM permalink
each round consists of point plus 1 roll then all bets are turned off..

i dont think anyone is realistically looking at my question... i understand that on every roll i have the same chance to win or lose based on the fact that the dice can land anyway

what i am trying to figure out is- is there a greater possiblity of losing more after i roll of the dice or winning more..

i could do pass line bets all day and roll a point then roll 2 3 or 4 times and hit a 7 i still lose the passline bet... but that first roll even if in 6 points is a place number or a 2/3/11/12 i dont lose. im only going to lose on the 7. and after the first roll my place bets are off.
kauboj
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January 6th, 2011 at 11:29:34 AM permalink
Quote: MathExtremist

No they don't. The probability of 3 hard 8s in a row is constant, regardless of how long you've been rolling, as long as you start counting from the next roll.



ok explain this mathmatically
MathExtremist
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January 6th, 2011 at 11:34:40 AM permalink
Quote: kauboj

ok explain this mathmatically


The chances of a hard 8 are 1 in 36. Each roll is independent, which means the chances of 3 hard 8s in a row are 1 in 36^3, or 1 in 46656. That is true regardless of whatever past rolls you've had. Even if you have just rolled two, five, or ten hard 8s in a row, or none at all, the chances of rolling *another* three hard 8s in a row are still 1 in 46656.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
kauboj
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January 6th, 2011 at 11:40:47 AM permalink
ok goat.. i read your blog but doesnt that stand for probabilities of events being eliminated from occurring.. in craps all 11 numbers have the same probability of rolling each and every time. all i am saying here is its not if you roll a 7 its when and the more you roll the dice the more likely you are gonna roll a 7. if this isnt true.. casino's would be bankrupt and they wouldnt exist. the casino is in business because of the 7.. same reason why some casinos have crapless craps because they know in the long run a 7 is going to roll when alot of money is on the table. on a regular basis.. just no one knows when this is going to happen
kauboj
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January 6th, 2011 at 11:42:13 AM permalink
Quote: MathExtremist

The chances of a hard 8 are 1 in 36. Each roll is independent, which means the chances of 3 hard 8s in a row are 1 in 36^3, or 1 in 46656. That is true regardless of whatever past rolls you've had. Even if you have just rolled two, five, or ten hard 8s in a row, or none at all, the chances of rolling *another* three hard 8s in a row are still 1 in 46656.



but what happens to that 3rd hard eight from the last roll or from the first roll
MathExtremist
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January 6th, 2011 at 12:04:09 PM permalink
Quote: kauboj

but what happens to that 3rd hard eight from the last roll or from the first roll


What 3rd hard 8? The question is not "what are the chances of rolling a single hard 8 after two others have just been rolled". The answer to that is 1/36, but it's the same answer as "what are the chances of rolling a single hard 8 after no others have been rolled, or after three sevens, or after I have a beer". Whether you roll a hard 8 once is independent of any of those other factors, including the beer, and the probability is always 1/36.

Let's say I just rolled 10 hard 8s in a row. At that point in time, what are the chances of rolling 3 hard 8s in a row?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
kauboj
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January 6th, 2011 at 12:08:27 PM permalink
either way the likelyhood is almost nil
SanchoPanza
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January 6th, 2011 at 12:18:28 PM permalink
Quote: kauboj

the longer you play... the odds increase in the favor of the house that a 7 will roll


What do you say the odds are for a 7 to be rolled after, say, 20 rolls without a 7? How about after 50 or 100 rolls without a 7?
goatcabin
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January 6th, 2011 at 1:12:17 PM permalink
Quote: kauboj

ok goat.. i read your blog but doesnt that stand for probabilities of events being eliminated from occurring.. in craps all 11 numbers have the same probability of rolling each and every time.



No. What are eliminated are possible series of outcomes, not the probabilities of individual events themselves. Let's apply it to the seven in craps, then.

p(7) = 1/6
p(not 7) = 5/6

Let's take three rolls of the dice. Here are the possibilities and their associated probabilities:

7 7 7 1/6 * 1/6 * 1/6 = 1/216
7 7 not-7 1/6 * 1/6 * 5/6 = 5/216
7 not-7 7 same 5/216
not-7 7 7 same 5/216
7 not-7 not-7 1/6 * 5/6 * 5/6 = 25/216
not-7 7 not-7 same 25/216
not-7 not-7 7 same 25/216
not-7 not-7 not-7 5/6 * 5/6 * 5/6 = 125/216

Add 'em up, they come to 216/216, as they must, since certainty is define as a probability of 1.0.

So, the probability of three rolls without a seven is 125/216, or .5787.
Now, suppose the first roll is a seven; what is the probability of THESE three rolls all being seven? Obviously, it's zero. For the NEXT three rolls, it's still 125/216, but for THESE three rolls, it's impossible, probability zero. Not only that, but every other sequence starting with a seven is now eliminated, right?

So, we're left with:

not-7 7 7 1 * 1/6 * 1/6 = 1/36
not-7 not-7 7 1 * 5/6 * 1/6 = 5/36
not-7 7 not-7 1 * 1/6 * 5/6 = 5/36
not-7 not-7 not-7 1 * 5/6 * 5/6 = 25/36

The first probability is 1, because we know that the first roll was a not-7, right? Add 'em up, you get 36/36. At the same time, if we start a NEW sequence, looking at the NEXT three throws, the probability of three non-7's is again 125/216.

Suppose the second roll is also a not-7. Now the first and third possibilities above are eliminated, and we are left with:

not-7 not-7 7 1 * 1 * 1/6 = 1/6
not-7 not-7 not-7 1 * 1 * 5/6 = 5/6
Add 'em up. Now, the probability of three not-7's is 5/6, just like it is on every single roll.

Each time the dice are rolled, some of the possible (at the start) sequences are no longer possible; we know that the sum of the probabilities must always equal one, so the probabilities of the eliminated possibilities must be re-distributed.

The probability calculation for a series of events is only valid BEFORE the series starts.

Quote: kauboj

all i am saying here is its not if you roll a 7 its when and the more you roll the dice the more likely you are gonna roll a 7. if this isnt true.. casino's would be bankrupt and they wouldnt exist. the casino is in business because of the 7.. same reason why some casinos have crapless craps because they know in the long run a 7 is going to roll when alot of money is on the table. on a regular basis.. just no one knows when this is going to happen



Apply the same reasoning to the six, then. Doe the six become more likely the more not-6 rolls go by? Why not?
If your reasoning were valid, why not just wait for five rolls without a seven and bet the "Any 7". Since the probability of six rolls without a seven is just .00002, the next roll is almost certainly going to be a 7, right? They pay 4:1 on Any 7, so you have a huge advantage, right? Why don't more people bet that way?

This is the most common misconception about independent, random events. The casinos are in business because the seven is always more probable than any other number, every roll, but it never becomes MORE likely than 1/6.
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
kauboj
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January 6th, 2011 at 2:26:32 PM permalink
so then in probabilities placing bets across all 6 numbers for 1 roll in a series is no different then placing a pass line bet. seems to me that i would

1/6 chance to loose
1/6 chance to stay even
2/3 chance to win

thats 66%, 16.3% better than the pass line percentage of 49.7%
MathExtremist
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January 6th, 2011 at 2:58:00 PM permalink
Quote: kauboj

so then in probabilities placing bets across all 6 numbers for 1 roll in a series is no different then placing a pass line bet. seems to me that i would

1/6 chance to loose
1/6 chance to stay even
2/3 chance to win

thats 66%, 16.3% better than the pass line percentage of 49.7%


Probability is only half of the equation. The other half is the payoff. What do you win when you win vs. what do you lose when you lose? For the pass line, the answer is 1 and 1, respectively. What is it for the place-across?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
TIMSPEED
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January 6th, 2011 at 3:12:53 PM permalink
Quote: MathExtremist

but it's the same answer as "what are the chances of rolling a single hard 8 after no others have been rolled, or after three sevens, or after I have a beer"


Unfotunately, my chances of rolling a hard 8 go down after I've had a beer, hahaha.
Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ
goatcabin
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January 6th, 2011 at 5:16:38 PM permalink
Quote: kauboj

so then in probabilities placing bets across all 6 numbers for 1 roll in a series is no different then placing a pass line bet. seems to me that i would

1/6 chance to loose
1/6 chance to stay even
2/3 chance to win

thats 66%, 16.3% better than the pass line percentage of 49.7%



But when you lose, you lose all six bets. Figure it out:

6 * 9 (4/10)
8 * 7 (5/9)
10 * 7 (6/8)
6 * -32 (7)

Nets out to -12, by my arithmetic. If your luck is average, you lose $12 every 30 resolved bets. Pass line, BTW, is 49.29%, even money.
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
kauboj
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January 6th, 2011 at 5:39:56 PM permalink
assuming i made i unit bet across the pass line and wait for 6 resolved bets:


4 = 5, 5 = 5, 6 = 6, 8 = 6, 9 = 5, 10 = 5 : Total Bet = 32

Now lets assume the first 5 rolls are only 6's and 8's: Total Won = 35

Now lets assume the last roll is a 7: Total Loss = 32

Net Profit: 3
MathExtremist
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January 6th, 2011 at 5:53:19 PM permalink
Quote: kauboj

assuming i made i unit bet across the pass line and wait for 6 resolved bets:


4 = 5, 5 = 5, 6 = 6, 8 = 6, 9 = 5, 10 = 5 : Total Bet = 32

Now lets assume the first 5 rolls are only 6's and 8's: Total Won = 35

Now lets assume the last roll is a 7: Total Loss = 32

Net Profit: 3


And how likely do you think that is to occur? More than 50% of the time? 25%? 10%?

More importantly, how likely is that scenario compared to the first roll being a 7 and you losing all 32 without winning anything?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
guido111
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January 6th, 2011 at 5:54:44 PM permalink
Quote: kauboj

assuming i made i unit bet across the pass line and wait for 6 resolved bets:

4 = 5, 5 = 5, 6 = 6, 8 = 6, 9 = 5, 10 = 5 : Total Bet = 32

Now lets assume the first 5 rolls are only 6's and 8's: Total Won = 35

Now lets assume the last roll is a 7: Total Loss = 32

Net Profit: 3


Your math looks correct.

Problem with your example is this...
you will win 6 place bet wagers or more only 26.21% =(24/30)^6 ( about 1 in 4 )
and show a variable net loss the other 73.78% of the time ( about 3 in 4 )
20% of the time you will win NO place bets before you lose them all.

But if you have a big bankroll to make that 32 across bet, go for it, it's your money.
You will have some big wins but way more losing sessions the longer you play.
guido111
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January 6th, 2011 at 7:06:12 PM permalink
Quote: kauboj

Now lets assume the first 5 rolls are only 6's and 8's: Total Won = 35


A friend asked me how to calculate the probability of rolling 5 place bet numbers in 5 rolls.
I will go 2 steps better.

Place bettors do not like horn numbers (2,3,11 and 12).
Place bettors really hate that 7.
So I broke 5 rolls down into the 3 possibilities.
7, Horn and place bets across.

Below is just a multinomial distribution table.
In the 5 rolls column, example: 113 (1 Seven, 1 horn #, and exactly 3 place bets)
005, all 5 place bets in 5 rolls comes in at 13.169% = (24/36)^5

What I see interesting is 2 out of the top 3 outcomes include only 1 Seven. And the top 3 outcomes appear to also be the median.
The math could get very ugly to split that 7 between a "winner 7" and a "7 out" 7.
Simulation would be faster.
Enjoy!
5 rolls7hornplacerelative freqCumulative Freq
11311316.461%16.461%
01401416.461%32.922%
10410416.461%49.383%
00500513.169%62.551%
0230238.230%70.782%
2032038.230%79.012%
1221226.173%85.185%
2122126.173%91.358%
0320322.058%93.416%
3023022.058%95.473%
2212211.543%97.016%
1311311.029%98.045%
3113111.029%99.074%
0410410.257%99.331%
4014010.257%99.588%
2302300.129%99.717%
3203200.129%99.846%
1401400.064%99.910%
4104100.064%99.974%
0500500.013%99.987%
5005000.013%100.000%
kauboj
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January 6th, 2011 at 7:58:29 PM permalink
I think some of you are taken away from the original question... i am not looking for 5 place numbers to hit in a row before a 7 out occurs....

i go a little more in depth in my explanation. i will refer to a session as defined by a point being established and either 7 out occurs or point made.

Session 1: Roll 7 (Win)
Session 2: Roll 2 (Lose)
Session 3: Roll 5 (Point), Place $32 Across, Roll 4,5,6,8,9,10 (Win), Turn Bets Off, Roll 7 (Lose) - Place Bets Remain
Session 4: Roll 9 (Point), Turn Bets On, Roll 4,5,6,8,9,10 (Win), Turn Bets Off, Roll 6, Roll 5, Roll 9 (Win) - Place Bets Remain
Session 5: Roll 8 (Point), Turn Bets On, Roll 4,5,6,8,9,10 (Win), Turn Bets Off, Roll 6, Roll 6, Roll 7 (Lose) - Place Bets Remain
Session 6: Roll 4 (Point), Turn Bets On, Roll 4,5,6,8,9,10 (Win), Turn Bets Off, Roll 2, Roll 3, Roll 4 (Win) - Place Bets Remain
Session 7: Roll 6 (Point), Turn Bets On, Roll 4,5,6,8,9,10 (Win), Turn Bets Off, Roll 4, Roll 5, Roll 6 (Win) - Place Bets Remain
Session 8: Roll 5 (Point), Turn Bets On, Roll 11 (Push), Turn Bets Off, Roll 6, Roll 5, Roll 5 (Win) - Place Bets Remain
Session 9: Roll 9 (Point), Turn Bets On, Roll 7(Lose) - Place Bets Lost

Now in this scenerio:
Sessions 1 - 2 No Money Won Or Lost
Sessions 3 - 7 Won $35 - $45
Session 8 No Money Won Or Lost
Session 9 Lost $32

Net Gain Of $3 - $13

Now rolls 3 - 7 is where my focus is. With the average person rolling 5 times before a 7 is rolled, could i aspect to be able to play the game for awhile while limiting my losses.
7winner
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January 6th, 2011 at 9:14:49 PM permalink
Quote: kauboj


But i was thinking of ways to enjoy the game longer and limit my losses... Now i fully understand that there is no gambling system that will ever break the house or even guarantee me to win..


"enjoy the game longer and limit my losses". Does that mean have your bankroll last longer?

Quote: kauboj

But what i am curious about is what would my expected chances be if i follow either of these betting structures..

1. Bypass the pass/dont pass bet and wait for point to be established. then place minimum bets on the place numbers allowing 1 roll then turning off the place bets and waiting for the (7) or the point to hit. and repeat.

2. Make the pass line bet and wait for the point then place minimum bets on the place numbers minus the point and place 1x odds behind the pass. allowing 1 roll and turning off the place bets and wait for the (7) or the point to hit and repeat.

curiosity kills me.


A few simple WinCraps sims would answer your 2 scenarios.
You need to let us know how much you plan on risking in a session.

What is your Buy-in? A win goal and stop loss (if other than the total bankroll.).
That would be a good start.

I already have this programmed so I would just need to plug in a few numbers from you.
When you say to have your bets working for one roll, and say a horn number rolls, you still turn your bets off and wait?

Nothing wrong in running a few computer sims to "see" what you can expect from your type of play.

The problem I see with $26 or $27 or $32 place bets across is the high house edge on those outside numbers.
7 winner chicken dinner!
7winner
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January 6th, 2011 at 9:23:17 PM permalink
Quote: kauboj

With the average person rolling 5 times before a 7 is rolled, could i aspect to be able to play the game for awhile while limiting my losses.


51.77% of all 7s that roll do it in 4 rolls.

11.1111% (1 in 9) of all shooters 7 out within 2 rolls.
22.7881% (2 in 9) of all shooters 7 out within 3 rolls.
33.2647% (almost 3 in 9) of all shooters 7 out within 4 rolls.
42.3871% (almost 4 in 9) of all shooters 7 out within 5 rolls.
50.28% (1 in 2) of all shooters 7 out within 6 rolls.

That shows you what you are up against.
7 winner chicken dinner!
FleaStiff
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January 7th, 2011 at 12:09:08 AM permalink
Quote: 7winner

51.77% of all 7s that roll do it in 4 rolls.
11.1111% (1 in 9) of all shooters 7 out within 2 rolls.
22.7881% (2 in 9) of all shooters 7 out within 3 rolls.
33.2647% (almost 3 in 9) of all shooters 7 out within 4 rolls.
42.3871% (almost 4 in 9) of all shooters 7 out within 5 rolls.
50.28% (1 in 2) of all shooters 7 out within 6 rolls.


How many shooters make their point within 2,3,4,5,6 rolls?
Or is this too much like asking Where Are The Customer's Yachts?
kauboj
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January 7th, 2011 at 8:26:19 AM permalink
7winner what would the percentage of shooters that 7 out after the point is made?

also i am turning the bets after after the first roll after the point is established whether that roll is a point, a craps, or a 7 out(obviously)

i would start with the table minimum and build from their so $5 minimum would put me $32 across

Buy in would be $100 (i never go to the table with more than that)

Stop loss would be the entire bank roll.
Doc
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January 7th, 2011 at 8:37:21 AM permalink
Quote: kauboj

7winner what would the percentage of shooters that 7 out after the point is made? ...

Juvenile, sarcastic answer:
When you roll a 7 immediately after the point is made, it is called a "natural", and right bettors win for two rolls in a row.

Not what you meant?
7winner
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January 7th, 2011 at 10:57:01 AM permalink
Quote: FleaStiff

How many shooters make their point within 2,3,4,5,6 rolls?


from the WoO site:
59.39% chance of not hitting a point
40.61% chance of hitting a point (or more)
https://wizardofodds.com/ask-the-wizard/craps-probability/

So we see that only 4 out of 10 shooters even hit 1 point or more.

From:https://wizardofodds.com/craps/appendix4.html#sharpshooter
and my simulation at:
Average Number of Points Hit Per Shooter in Craps, page 1
https://wizardofvegas.com/forum/questions-and-answers/gambling/3146-average-number-of-points-hit-per-shooter-in-craps/

one can see the probability of a shooter hitting a number of different points. It is also a Shuffle Master side bet in Craps called SharpShooter.

Now from:https://wizardofvegas.com/forum/gambling/tables/2616-pass-odds-only-scheme/
Pass-Odds only Scheme
There is a table on page 1 and boymimbo has an excellent formula on page 2 showing how to

calculate the probability of hitting a point in "n" rolls.
6/24*(1-(27/36)^n)+8/24*(1-(26/36)^n)+10/24*(1-(25/36)^n)
1	0.282407407
2 0.484567901
3 0.629418867
4 0.733304517
5 0.80788109
6 0.861468782
7 0.900011824
8 0.92776081
9 0.947758041
10 0.962183009
11 0.972598561
12 0.980126424
13 0.985572479
14 0.989516248
15 0.992374878
16 0.994448924
17 0.995955148
18 0.997050031
19 0.997846645
20 0.998426776
21 0.998849635
22 0.999158133
23 0.999383395
24 0.999548021
25 0.999668434
26 0.999756581
27 0.999821161
28 0.999868512
29 0.999903258
30 0.999928773


I think to answer your question, out of the 40.61% of shooters that hit at least 1 point, 86.1% ( a bit more than 6 out of 7) will hit a point within 6 rolls after establishing it.
or 34.81% of all shooters will hit 1 point within 6 rolls.
please correct me if I am wrong. I may do a simulation since all this math gets my head spinning.
7 winner chicken dinner!
7winner
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January 7th, 2011 at 12:01:37 PM permalink
Quote: kauboj

7winner what would the percentage of shooters that 7 out after the point is made?


Again,
From:https://wizardofodds.com/craps/appendix4.html#sharpshooter
and my simulation at:
Average Number of Points Hit Per Shooter in Craps, page 1
https://wizardofvegas.com/forum/questions-and-answers/gambling/3146-average-number-of-points-hit-per-shooter-in-craps/

one can see the probability of a shooter hitting a number of different points.
So percentage of shooters hitting exactly 1 point and then 7 out would be:
24.1267%

Quote: kauboj

also i am turning the bets after after the first roll after the point is established whether that roll is a point, a craps, or a 7 out(obviously)


A little confusing. I understand you have your place bets working for just 1 roll after point is established (does not matter what that roll is) and then turn them off to wait for a decision and then wait for a new point number to become established to have them turned on again.


Quote: kauboj

i would start with the table minimum and build from their so $5 minimum would put me $32 across

Buy in would be $100 (i never go to the table with more than that)

Stop loss would be the entire bank roll.


Thank you for the info. I will place a few win goals, say $25, $50 and $100
7 winner chicken dinner!
kauboj
kauboj
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January 7th, 2011 at 12:01:55 PM permalink
no not the question!

Shooter on the come out roll rolls a point... what percentage of rollers roll a 7 on the very next roll?
7winner
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January 7th, 2011 at 12:06:02 PM permalink
Quote: kauboj

no not the question!

Shooter on the come out roll rolls a point... what percentage of rollers roll a 7 on the very next roll?


I see now.
Yes,
11.11% (1 in 9) of all shooters will establish a point on their 1st roll and "7out" on the very next roll. The famous "Point7"
https://wizardofodds.com/craps/appendix4.html#pointseven
or
this is from the Wizard of Odds table (about halfway down the page)
https://wizardofodds.com/ask-the-wizard/craps/
about the lady in AC with her 154 roll hand
or
http://catlin.casinocitytimes.com/article/how-long-is-a-roll?-part-2-1232
bottom of page
7 winner chicken dinner!
goatcabin
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January 7th, 2011 at 2:29:19 PM permalink
Quote: kauboj

no not the question!

Shooter on the come out roll rolls a point... what percentage of rollers roll a 7 on the very next roll?



One out of six, of course, if I understand your situation.
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
7winner
7winner
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January 7th, 2011 at 2:49:55 PM permalink
Quote: kauboj


But what i am curious about is what would my expected chances be if i follow either of these betting structures..

1. Bypass the pass/dont pass bet and wait for point to be established. then place minimum bets on the place numbers allowing 1 roll then turning off the place bets and waiting for the (7) or the point to hit. and repeat.

2. Make the pass line bet and wait for the point then place minimum bets on the place numbers minus the point and place 1x odds behind the pass. allowing 1 roll and turning off the place bets and wait for the (7) or the point to hit and repeat.
curiosity kills me.


Results for #1. $32 across place bets, working for only 1 roll after a point is established and win goals of $25, $50 and $100.
Stop loss when busted out or could not make a $32 across wager.
Results for #2 is in the second table below.

10,000 sessions each
I also show how many rolls a session lasts during a win and loss.
$25 win goal$50 win goal$100 win goal
71.87% won57.26% won37.38% won
average win $28.56average win $53.31average win $103.16
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
rolls (win) average 55rolls (win) average 114rolls (win) average 267
Std Dev 56Std Dev 95Std Dev 183
median 32median 82median 214
mode (most common) 16 to 20mode (most common) 35 to 43mode (most common) 114 to 125
low 6low 18low 50
mid 273mid 473mid 724
high 540high 928high 1399
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
28.13% lost42.74% lost62.62% lost
average loss $85.79average loss $85.91average loss $86.10
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
rolls (loss) average 80rolls (loss) average 118rolls (loss) average 187
Std Dev 67Std Dev 101Std Dev 179
median 60median 89median 133
mode (most common) 17 to 22mode (most common) 31 to 39mode (most common) 19 to 31
low 6low 6low 6
mid 296mid 458mid 718
high 586high 909high 1430
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Mean -$3.60Mean -$6.19Mean -$15.35
HA -3.21%HA -3.02%HA -3.98%



Results for #2. $5 pas line bet with 1x odds. ($6 on point 5 or 9) and $26 or $27 across place bets, working for only 1 roll after a point is established and win goals of $25, $50 and $100.Stop loss when busted out or could not make a $5 pass line bet and $26 or $27 across wager.

$25 win goal$50 win goal$100 win goal
69.64% won55.30% won38.06% won
average win $31.76average win $56.57average win $103.16
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
rolls (win) average 33rolls (win) average 64rolls (win) average 144
Std Dev 34Std Dev 53Std Dev 103
median 20median 64median 115
mode (most common) 4 to 6mode (most common) 26 to 29mode (most common) 50 to 60
low 4low 7low 18
mid 157mid 205mid 588
high 309high 410high 1158
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
30.36% lost44.70% lost61.94% lost
average loss $85.79average loss $83.47average loss $86.10
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
rolls (loss) average 49rolls (loss) average 69rolls (loss) average 104
Std Dev 41Std Dev 59Std Dev 97
median 39median 53median 75
mode (most common) 4 to 7mode (most common) 28 to 32mode (most common) 20 to 28
low 4low 4low 4
mid 190mid 258mid 444
high 376high 512high 883
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Mean -$3.17Mean -$6.03Mean -$11.03
HA -2.14%HA -2.30%HA -2.33%
7 winner chicken dinner!
Croupier
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January 7th, 2011 at 9:02:19 PM permalink
Quote: kauboj


i dont think anyone is realistically looking at my question... i understand that on every roll i have the same chance to win or lose based on the fact that the dice can land anyway

what i am trying to figure out is- is there a greater possiblity of losing more after i roll of the dice or winning more..



There is no difference. It doesnt matter how many rolls there have been or still to come. The chance of rolling any number doesnt change.

Before you roll the dice there are still the same number of ways to roll the numbers meaning the chances of any number occuring is the same.

Just like roulette. Just because a certain number has been span once, it still has the same chance of occuring on the next spin.
[This space is intentionally left blank]
kauboj
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January 8th, 2011 at 8:21:01 AM permalink
i think that after all this back and forth 7winner understood the most of what i was asking... and pretty much gave me the answer i was looking for..

i just see that i have 16.7% chance to loose all money bet, a 16.7% chance to not loose anything, and a 66.6% chance to win on any giving roll betting the way i wanna bet
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