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I was wondering if any of you could assist me here.

How many inside numbers does the average craps player roll before he 7 out's.

Could you please include the figure that includes the come out roll and excludes it.

The inside numbers are of course 5,6,8 and 9.

Thank U

P.S. The average player rolls a little over 8 times, so just wondering what percentage of those numbers are inside numbers more or less.

The probably of rolling the inside numbers is 50% each roll.

(6,8 are 5/36 or .13888889)

(5,9 are 4/36 (1/9) or .1111111)

So by adding them all together you get 18/36 (1/2) or .5

Hope this helped.

P.s. if anyone spots an error in my math please let me know.

Probability of rolling

0: 0.250000

1: 0.231579

2: 0.298086

3: 0.089252

4: 0.131083

Quote:WizardI get 1.619839371.

Probability of rolling

0: 0.250000

1: 0.231579

2: 0.298086

3: 0.089252

4: 0.131083

Is it really easier to roll 4 than 3?

When it comes to math I am a bit thick headed.

If I understand you correctly, the average is 1.619839371 inside numbers after the come out?

Thank U

Quote:RouletteProdigyHow many inside numbers does the average craps player roll before he 7 out's.

Could you please include the figure that includes the come out roll and excludes it.

Question: at what point does the counting begin?

i.e. do you want to know the average number of inside numbers rolled from the time a shooter is passed the dice to when he sevens out?

For example, in this case:

point 6, 3, 12, 6

natural 11

point 5, 4, 7 out

Do you want the three comeouts counted separately, or combined (in which case, the number is 3 if you include the comeout rolls, or 1 if you do not)?

Quote:ThatDonGuyQuestion: at what point does the counting begin?

i.e. do you want to know the average number of inside numbers rolled from the time a shooter is passed the dice to when he sevens out?

For example, in this case:

point 6, 3, 12, 6

natural 11

point 5, 4, 7 out

Do you want the three comeouts counted separately, or combined (in which case, the number is 3 if you include the comeout rolls, or 1 if you do not)?

Hi Don,

I would like to know the average number of inside numbers rolled after the point is established to when the average player sevens out.

Thank U

Quote:AyecarumbaIs it really easier to roll 4 than 3?

3 ways to roll a 4: 1/3 3/1 2/2

2 ways to roll a 3: 1/2 2/1

Quote:RouletteProdigyI would like to know the average number of inside numbers rolled after the point is established to when the average player sevens out.

I got that part. What I don't quite understand is, is it from when the shooter first establishes a point after taking the dice?

That is, what happens if a point is made after, for example, three inside numbers are rolled? Does the count continue from three, or does it reset to zero?

Quote:ThatDonGuyI got that part. What I don't quite understand is, is it from when the shooter first establishes a point after taking the dice?

That is, what happens if a point is made after, for example, three inside numbers are rolled? Does the count continue from three, or does it reset to zero?

This is an important clarification, as I think the figure "8" for "average number of rolls per shooter" includes all the "non-point establishing" rolls. Consider this sequence of rolls, "New Shooter Coming Out!:

7

12

11

2

3

4 - Mark the 4

10

7 - Out! Line away

It's 8 rolls, but only one after the point is established. After a point is established, I thought the average number of rolls was 3 and change?

Quote:iamnomad3 ways to roll a 4: 1/3 3/1 2/2

2 ways to roll a 3: 1/2 2/1

This was in reply to the table the Wizard posted of inside number probabilities. According to his table, rolling 4 inside numbers is more probable than rolling 3 inside numbers.

Since comeout rolls are ignored, the only numbers that matter are 5, 6, 7, 8, and 9.

In any given roll, there is a probability 1/2 of rolling an inside number and 1/6 of rolling a 7, so if we are limited to just those numbers, there is a probability 3/4 of rolling an inside number and 1/4 of a seven.

The average number of inside numbers rolled is:

(0 x 1/4) + (1 x 3/4 x 1/4) + (2 x (3/4)

^{2}x 1/4) + (3 x (3/4)

^{3}x 1/4) +

= 3/4 x 1/4 x (1 + 2 x 3/4 + 3 x (3/4)

^{2}+ ...)

= 3/16 x (1 + 3/4 + (3/4)

^{2}+ ...)

^{2}

= 3/16 x (1 / (1/4))

^{2}

= 3/16 x 16

= 3

Quote:ThatDonGuyAssuming that the question is, "From the moment a shooter is handed the dice until he sevens out, and ignoring all comeout rolls, including those after successful points, what is the average number of times that an inside number will be rolled?"

Since comeout rolls are ignored, the only numbers that matter are 5, 6, 7, 8, and 9.

In any given roll, there is a probability 1/2 of rolling an inside number and 1/6 of rolling a 7, so if we are limited to just those numbers, there is a probability 3/4 of rolling an inside number and 1/4 of a seven.

The average number of inside numbers rolled is:

(0 x 1/4) + (1 x 3/4 x 1/4) + (2 x (3/4)^{2}x 1/4) + (3 x (3/4)^{3}x 1/4) +

= 3/4 x 1/4 x (1 + 2 x 3/4 + 3 x (3/4)^{2}+ ...)

= 3/16 x (1 + 3/4 + (3/4)^{2}+ ...)^{2}

= 3/16 x (1 / (1/4))^{2}

= 3/16 x 16

= 3

Don,

You answered my question. I appreciate it.

If you feel like answering one more question. How many inside numbers occur from roll number 6 to a seven out on average. You would not count roll number 5. Roll number 5 includes the come out roll. ThankU

Quote:AyecarumbaThis was in reply to the table the Wizard posted of inside number probabilities. According to his table, rolling 4 inside numbers is more probable than rolling 3 inside numbers.

AHHHH...gotcha. Wondered why you were asking. I should've figured it out from rest of thread.

0 .250 000

1 .231 579

2 .208 835

3 .178 503

4 .131 083

Average # hits = 1.709 091

The problem solved is how many of the inside numbers will a shooter roll during any series of rolls which ends on either a 7 or all four of the inside numbers have been rolled.

For instance there are six ways to throw a 7 and eighteen to roll 5 6 8 or 9. So the chances of not throwing any are 6/(4+5+6+5+4) = 6/24 = 1/4.

For one roll is SUM( Pr (5 before 7) * Pr (7 before 689) + Pr (6 before 7) * pr (8 before 589) etc.)

It is obvious that the probability of getting 3 is larger than getting 4. Whilst on 3 you stay of 3 if you roll a 7 or get to 4 if you roll the last number. Thus, as 7 is more likely, the chances of staying on 3 is bigger than getting to 4.

Quote:RouletteProdigyIf you feel like answering one more question. How many inside numbers occur from roll number 6 to a seven out on average. You would not count roll number 5. Roll number 5 includes the come out roll.

Assuming once again that all come out rolls are ignored, the answer is still 3. The dice don't know if the first roll that is counted is the second roll or the sixth - or the 13th, for that matter. They just roll.

The question above is how many of (5689) might one roll before throwing a 7. If we ignore comeout rolls, where a 7 is a winner for the shooter rather than a 7-out, then other rolls are merely trying to make the point or 7-out. Thus they can be considered as a series of rolls that ultimately end with a 7-out.

There are 18 ways to roll (568 or 9) and 6 ways to roll (7) - hence if you rolled the dice often enough there would be 3 times as many inside numbers as 7's. Thus on average you would roll 3 inside numbers before the 7.

Another way of looking at it is in (say) a million rolls where you throw (5678 or 9) (i.e. ignore the ones 2 3 11 12) there are (about) 250k 7's and 750k inside numbers. So consider dividing the million rolls at each point where there is a 7; this creates 250k series where the last roll of each is a 7. The average length of a series is 4 (1m/250k) and the last roll of each series is always a 7. Thus the average series has 3 inside numbers (4-1) before the 7.

Quote:charliepatrick...Another way of looking at it is in (say) a million rolls where you throw (5678 or 9) (i.e. ignore the ones 2 3 11 12) there are (about) 250k 7's and 750k inside numbers. So consider dividing the million rolls at each point where there is a 7; this creates 250k series where the last roll of each is a 7. The average length of a series is 4 (1m/250k) and the last roll of each series is always a 7. Thus the average series has 3 inside numbers (4-1) before the 7.

This may be misleading because in reality, there are many other sequences of rolls that contain Horn/outside numbers with 0 or more inside numbers before a 7. Shouldn't this make the actual number of "post point inside numbers before a 7" much lower than 3?

Interesting question.Quote:RouletteProdigyHi All,

I was wondering if any of you could assist me here.

How many inside numbers does the average craps player roll before he 7 out's.

Could you please include the figure that includes the come out roll and excludes it.

I do not know what an average craps player is (next to possibly being a lifetime loser, imo)

so,

I setup Wincraps real fast

counting all rolls

I get an average of 4.26 rolls

(1 million shooters)

part of the distribution

0 0.057336057

1 0.231608232

2 0.147214147

3 0.122207122

4 0.095114095

5 0.074770075

6 0.058461058

7 0.045865046

8 0.036345036

9 0.028325028

10 0.022298022

11 0.017440017

12 0.013759014

13 0.01044601

counting all rolls EXCEPT 5689 on the come out roll.

I get an average of 3 rolls

part of the distribution

0 0.25000825

1 0.187191187

2 0.140664141

3 0.105567106

4 0.07958508

5 0.058822059

6 0.044466044

7 0.033534034

8 0.025041025

9 0.018961019

10 0.014040014

11 0.01043301

12 0.007785008

13 0.005991006

hope this helps out

this should come close to calculated values

of course

Sally

Quote:mustangsally

counting all rolls

I get an average of 4.26 rolls

counting all rolls EXCEPT 5689 on the come out roll.

I get an average of 3 rolls

this should come close to calculated values

The expected number if you don't count comeout rolls is 3, as shown above.

If you do count them:

First, determine the probability of making a point from a comeout roll that establishes a point.

This is 1/4 (the probability that a point will be 4 or 10) x 1/3 (the probability of making that point) + 1/3 (the probability that a point will be 5 or 9) x 2/5 (the probability of making that point) + 5/12 (the probability that a point will be 6 or 8) x 5/11 (the probability of making that point) = 67/165

The probability of sevening out once you have already established a point is 1 - 67/165 = 98/165

The expected number of comeouts where points are established before sevening out is:

1 x 98/165 + 2 x 67/165 x 98/165 + 3 x (67/165)

^{2}x 98/165 + 4 x (67/165)

^{3}x 98/165 + ...

= 98/165 x (1 + 2 x 67/165 + 3 x (67/165)

^{2}+ 4 x (67/165)

^{3}+ ...)

= 98/165 x (1 + 67/165 + (67/165)

^{2}+ ...)

^{2}

= 98/165 x (1 / (1 - 67/165))

^{2}

= 98/165 x (165/98)

^{2}

= 98/165

The probability that a comeout that is a point number is an inside number is 3/4 (there are 6 ways to roll a 4 or 10, 8 to roll a 5 or 9, and 10 to roll a 6 or 8, so the probability is 18/24 = 3/4), so the expected number of inside numbers on comeouts = 98/165 x 3/4 = 495/392 = about 1.26275,

Add this to the expected 3 inside numbers not on comeouts, and the expected number of inside numbers when you include comeouts = about 4.26275.

I like to play Craps and plan to play the don't pass with continuous Don't come bets but I want to lay odds also. If anyone can tell me at what point should I lay the odds. I was thinking if the average player rolls 8.5 should I not lay odds after the 4th or 5th roll? Would that not give me some advantage (or rather help me lose less) After the 5th roll there is a 50% chance of the player seven out on the 6th roll. Not sure if wincraps can provide the optimal time to lay odds on the DP and DC's.

One more thing I will only lay odds on the inside numbers and if I lose my odds twice on DC or DP I will take off my remaining odds and leave my flat bets on

Thanks for the input everyone.

Quote:RouletteProdigyI like to play Craps and plan to play the don't pass with continuous Don't come bets but I want to lay odds also. If anyone can tell me at what point should I lay the odds. I was thinking if the average player rolls 8.5 should I not lay odds after the 4th or 5th roll? Would that not give me some advantage (or rather help me lose less) After the 5th roll there is a 50% chance of the player seven out on the 6th roll.

Excuse me? "After the fifth roll," the chance of the shooter sevening out on the sixth roll is the same as it was for all of the shooter's previous rolls.

Like I said before - the dice don't remember what they did in the past. Especially if, for whatever reason, the shooter switches to new dice!

Quote:ThatDonGuyExcuse me? "After the fifth roll," the chance of the shooter sevening out on the sixth roll is the same as it was for all of the shooter's previous rolls.

Like I said before - the dice don't remember what they did in the past. Especially if, for whatever reason, the shooter switches to new dice!

Don, I get what you are saying each roll is independent. It still does not change that in a series of rolls before a player seven out these figures are real.

Roll Probability of Seven Out

1 0.00000000

2 0.11111111

3 0.22788066

4 0.33264746

5 0.42387109

6 0.50278913

7 0.57095589

8 0.62980865

9 0.68060930

10 0.72445344

Why would I not lose less if I layed odds after a certain number of rolls regardless if each roll has no memory. If I layed odds on roll number 2 the player is less likely to seven out vs. laying odds after the 5th roll the player is more likely to seven out.

One does not mesh with the other. Perhaps this is where most of us non math gamblers go wrong. Just asking.

Thank U

Ah. I'm sure many of us suspected you were working out a strategy based on your question. If you were asking about a bet with a house edge, the only reason you would lose less is by putting less into action overall - if indeed it was less total action. But for an odds bet, there is no edge and it doesn't matter except for how much Variance you run into.Quote:RouletteProdigyWhy would I not lose less if I layed odds after a certain number of rolls regardless if each roll has no memory.

No. If the dice have no memory, then they don't know what roll it is. The chances of a 7-out are the same each time for the next roll. The reason the odds are changing like you show is *not* from the perspective of "the next roll" but from the perspective of such a sequence altogether. You can't make a bet by saying "I want to make a bet that the 7-out [does/doesn't] take 10 rolls"Quote:If I layed odds on roll number 2 the player is less likely to seven out vs. laying odds after the 5th roll the player is more likely to seven out.

They do mesh, you are struggling with what is a counter-intuitive nature of probability to you at this stage. Keep asking questions and keep an open mind and you will find it becomes intuitive.Quote:One does not mesh with the other. Perhaps this is where most of us non math gamblers go wrong. Just asking.

Thank U

One thing to realize is that billions of gamblers before you have studied Craps, examining each angle, every nook and cranny, and bled all over the table with every conceivable concept. If you were on to something here, it would be known already: "just wait till the 7-out is overdue and lay the odds then for better chances" would be the mantra. Of course, if this was true, you would have an edge, wouldn't you? Not just a guy losing less money, but making money. Right from the get-go you would just not be allowed to do it.

Quote:RouletteProdigyDon, I get what you are saying each roll is independent. It still does not change that in a series of rolls before a player seven out these figures are real.

Roll Probability of Seven Out

1 0.00000000

2 0.11111111

3 0.22788066

4 0.33264746

5 0.42387109

6 0.50278913

7 0.57095589

8 0.62980865

9 0.68060930

10 0.72445344

Why would I not lose less if I layed odds after a certain number of rolls regardless if each roll has no memory.

I think the probabilities you list are the probabilities of sevening out in that many rolls or fewer. This is certainly true for the third roll.

Obviously, you can't seven out on the first roll, which is a comeout.

The probability of sevening out on the second roll = (the probability of rolling a point number on the first roll) x (the probability of a seven on the second roll) + (the probability of rolling a natural/craps on the first roll) x (zero, since the second roll is now a comeout) = 2/3 x 1/6 + 1/3 x 0 = 1/9, which is what you show.

The probability of sevening out on the third roll = (the probability that the third roll is not a comeout roll) x 1/6.

The probability of the third roll not being a comeout roll is determined as follows:

First two rolls | Probability | Prob of sevening out |
---|---|---|

Neither is a point | 1/9 | 0 |

First one not a point, second one is | 2/9 | 1/6 |

First one is a point, second one sevens out | 1/9 | 0 |

First one is 4/10, second one makes the point | 1/72 | 0 |

First one is 5/9, second one makes the point | 2/81 | 0 |

First one is 6/8, second one makes the point | 25/648 | 0 |

First one is 4/10, second one is not the point or 7 | 1/8 | 1/6 |

First one is 5/9, second one is not the point or 7 | 13/81 | 1/6 |

First one is 6/8, second one is not the point or 7 | 125/648 | 1/6 |

The probability of the third roll sevening out is (2/9 x 1/6) + (1/8 x 1/6) + (13/81 x 1/6) + (125/648 x 1/6) = 227 / 1944 = 0.11676955

If you add this to the probabilities of sevening out on the first two rolls that you list, the total is 0.22788066.

Note that my number includes the cases where you seven out on the second roll.

Also note that this is the probability of sevening out on the third roll not already knowing what the previous two rolls were (i.e. you make the bet that the third roll will not seven out before the first roll, not after the second roll).

After nine rolls, the probability of sevening out on the 10th roll is 1/6 if a point is established or zero if it is a comeout roll. It doesn't matter what the first eight rolls were.

If you add this to the probabilities of sevening out on the first two rolls that you list, the total is 0.22788066.

Note that my number includes the cases where you seven out on the second roll.

Also note that this is the probability of sevening out on the third roll not already knowing what the previous two rolls were (i.e. you make the bet that the third roll will not seven out before the first roll, not after the second roll).

[/spoiler]

After nine rolls, the probability of sevening out on the 10th roll is 1/6 if a point is established or zero if it is a comeout roll. It doesn't matter what the first eight rolls were.

Quote:RouletteProdigyWizard,

When it comes to math I am a bit thick headed.

If I understand you correctly, the average is 1.619839371 inside numbers after the come out?

That is correct.

Quote:YoElevenI am new to the forum , and I have a question regarding numbers and come bets. Does anyone have the odds or calculations for a come bet to repeat ? In other words , including all numbers, 4,5,6, 8, 9,10 that hit during a roll that are come bets, what are the collective odds of a come bet hitting or repeating before a 7 out? I assumed it would be true odds to repeat , and somehow I came up with 1.1 come bet repeated per each 7 out. Does anyone have any calculations on this?

Could you give me an example (with numbers, preferably) of what you mean by "a come bet repeating"? I am not quite sure what you are asking.

For that matter, could you show how you get "1.1 come bets repeated per each 7 out"?

If the roll was 8, 4, 5, 4, 6,9, 5, 11, 7 out , that would be 2 come bets ( 4, 5) were repeated or hit during that roll

Was just curious what are the odds of come bets repeating and paying?

Quote:YoEleven...Was just curious what are the odds of come bets repeating and paying?

They are the same as the passline. Remember that when your Come bet is initially placed, you also win on a seven or eleven, and lose on a 2,3 or 12. It is just like a Passline bet. The fact that a shooter's roll is already in progress when the Come bet is placed is immaterial.

Quote:YoElevenExample. in a roll : 5 , 6, 11, 8, 6, 3, 9, 7 out. The 6 was a come bet that was repeated. I just assumed that the odds on a number repeating would be true odds, such as 6, 8 is 6 to 5 , 5, 9 is 3 to 2, and 4, 10 is 2 to 1.

They are. The probability of winning a come bet if the bet's point if 4 is 1/3.

However, you have to take into account the fact that you can have multiple numbers in play at once.

For example, in your example, your first three point numbers are 5, 6, and 8. The next roll has a 1/6 chance of a seven, causing all of them to lose, and a 7/18 chance of a 5, 6, or 8, which means one of them wins - so you are 7/3 as likely to have a point number repeated as you are to seven out at this point.

Also remember that if you have multiple come bets in play, on any particular roll, at most one of them will win, whereas on that same roll, all of them (or all but one, if you make a come bet on every roll, as the 7 out is also a 7 natural) can lose. This makes the number of repeats between 7s misleading.

All the 7 out means is that trial stops.

It also explains why the house advantage on the 5 6 8 9 for say $30 each is

-120 * 1/6 + (42 * 8/36 + 35 * 10/36) = -.94444 per roll or a house advantage of -.787% / roll.

On the first roll the odds are .50 inside and 1/6th 7 out.

On the second roll which you have a (5/6) probability to get to, the odds are .50 inside and 1/6 7 out

So the total odds are the sum of .5*(5/6)^n where n starts at 0 and ends at infinity = 3.

Did you know that the sum of any (x/(x-1)) ^n where n starts at 1 and ends at infinity = x.

Quote:boymimboDid you know that the sum of any (x/(x-1)) ^n where n starts at 1 and ends at infinity = x.

(Warning: Vegas isn't the only place where it's hotter than usual, and on top of that, my A/C was out for a day or so, so I'm in pedantic mode today.)

So if x = 1, then 2 + 2

^{2}+ 2

^{3}+ 2

^{4}+ ... = 2?

I always thought 2 + 2

^{2}+ 2

^{3}+ 2

^{4}+ ... = -2...

Let x = 2 + 2

^{2}+ 2

^{3}+ 2

^{4}+ ...

Divide both sides by 2:

x/2 = 1 + 2 + 2

^{2}+ 2

^{3}+ 2

^{4}+ ...

Subtract the second equation from the first:

x/2 = -1, so x = -2

I'm sure you meant to say x / (x + 1). Also, x cannot be negative.

Since x >= 0, x / (x - 1) < 1

(x / (x-1)) + (x / (x-1))

^{2}+ (x / (x-1))

^{3}+ ... = (-1 + 1) + (x / (x-1)) + (x / (x-1))

^{2}+ (x / (x-1))

^{3}+ ...

= 1 / (1 - (x / (x + 1)) - 1

= 1 / ( (x + 1) / (x + 1) - x / (x + 1) ) - 1

= 1 / ( (x + 1 - x) / (x + 1) ) - 1

= 1 / ( 1 / (x + 1) ) - 1

= x + 1 - 1 = x

If I plan on not hitting more than 12 inside numbers per turn, how should I set up a Lay system that pays better than winning 1.62 inside PB numbers per turn?

My big comebacks were not dozens of inside numbers hitting, although if that were to happen, it'd be a big come-back. No, the come-out 7's with the Lay bet brought me back, and losing my Lay bet time after time after time on the come-out became a front & center problem. I'm already losing with a lack of inside numbers per turn, I thought a Lay hedge would help, but it's a bit of fool's gold.