Craps - Average rolls to hit a point or 7 out

Hi Fellow gamers and Craps enthusiasts,

I've read in an article or reference somewhere that through simulation of at least 1million dice rolls, it takes an average of 4 rolls to EITHER
1. A point is made OR
2. 7-Out

Can someone shed the math on how this was computed?

Also, if the average statistic above is 100% true - Once a point has been established, what are the odds of:
A. Hitting ANY inside numbers (5, 6, 8, 9) on first 3 rolls?
B. Any other non-7's (4, 5, 6, 8, 9, 10 + 2, 3, 11, 12) are roll on first 3 rolls?

Appreciate the help math gurus.

I tried betting DP then DC once then throwing a 7-out. So 3 rolls per turn at an empty Bubble Craps machine. I think the math people have said there's about 8.5 rolls average per turn. So that means people throw 19 times before hitting a 7-out or hitting their first point. I'm trying to figure out how many 2, 3, 11, 12's I throw in a row because it seems obvious I can't hit box numbers to save my life sometimes.

Quote: poli2k01

Hi Fellow gamers and Craps enthusiasts,

I've read in an article or reference somewhere that through simulation of at least 1million dice rolls, it takes an average of 4 rolls to EITHER
1. A point is made OR
2. 7-Out

Can someone shed the math on how this was computed?
link to original post

First, note that if the probability of rolling a particular point number or 7 is p, then the expected number of rolls needed is:
1 x p + 2 x (1-p) x p + 3 x (1-p)^2 x p + 4 x (1-p)^3 x p + ...
= p x (1 + 2 x (1-p) + 3 x (1-p)^2 + ...)
= p x (1 + (1-p) + (1-p)^2 + ...)^2
= p x (1 / (1 - (1-p)))^2
= 1 / p
1/8 of the points established are 4s; the probability of rolling a 4 or 7 is 1/4, so the expected number of rolls to make or miss a point of 4 once it is established is 4.
1/6 of the points are 5s; the probability of rolling a 5 or 7 is 5/18, so the expected number of rolls is 18/5.
5/36 of the points are 6s; the probability of rolling a 6 or 7 is 11/36, so the expected number of rolls is 36/11.
Similarly, 1/8 of the points are 10s, 1/6 are 9s, and 5/36 are 8s, and the expected number of rolls for each are 4, 18/5, and 36/11, respectively.
The total is (1/8 x 4) + (1/6 x 18/5) + (5/36 + 36/11) + (5/36 + 36/11) + (1/6 x 18/5) + (1/8 x 4) = 196 / 55 = about 3.563636.

So putting up 1 DC bet on the 5th roll since the point is established and counting the establishment of the point as roll #1. Add another DC bet for every 5th roll after until a new point is established. Roll 7's on the come-outs.

Quote: ThatDonGuy

Quote: poli2k01

Hi Fellow gamers and Craps enthusiasts,

I've read in an article or reference somewhere that through simulation of at least 1million dice rolls, it takes an average of 4 rolls to EITHER
1. A point is made OR
2. 7-Out

Can someone shed the math on how this was computed?
link to original post

First, note that if the probability of rolling a particular point number or 7 is p, then the expected number of rolls needed is:
1 x p + 2 x (1-p) x p + 3 x (1-p)^2 x p + 4 x (1-p)^3 x p + ...
= p x (1 + 2 x (1-p) + 3 x (1-p)^2 + ...)
= p x (1 + (1-p) + (1-p)^2 + ...)^2
= p x (1 / (1 - (1-p)))^2
= 1 / p
1/8 of the points established are 4s; the probability of rolling a 4 or 7 is 1/4, so the expected number of rolls to make or miss a point of 4 once it is established is 4.
1/6 of the points are 5s; the probability of rolling a 5 or 7 is 5/18, so the expected number of rolls is 18/5.
5/36 of the points are 6s; the probability of rolling a 6 or 7 is 11/36, so the expected number of rolls is 36/11.
Similarly, 1/8 of the points are 10s, 1/6 are 9s, and 5/36 are 8s, and the expected number of rolls for each are 4, 18/5, and 36/11, respectively.
The total is (1/8 x 4) + (1/6 x 18/5) + (5/36 + 36/11) + (5/36 + 36/11) + (1/6 x 18/5) + (1/8 x 4) = 196 / 55 = about 3.563636.
link to original post

I believe that’s the average rolls after a point has been established. Add 1.5 rolls starting from comeout

Quote: poli2k01

Hi Fellow gamers and Craps enthusiasts,

I've read in an article or reference somewhere that through simulation of at least 1million dice rolls, it takes an average of 4 rolls to EITHER
1. A point is made OR
2. 7-Out

Can someone shed the math on how this was computed?

Also, if the average statistic above is 100% true - Once a point has been established, what are the odds of:
A. Hitting ANY inside numbers (5, 6, 8, 9) on first 3 rolls?
B. Any other non-7's (4, 5, 6, 8, 9, 10 + 2, 3, 11, 12) are roll on first 3 rolls?

Appreciate the help math gurus.
link to original post

For many craps calculations, such as this one, you can assume all six point numbers are equally weighted at 4/36 and be quite close to the exact answer. Makes it so simple that you can do it in your head: 36/24 rolls to establish a point plus 36/10 rolls to resolve the point is 5.1 rolls (vs 5.06 actual)

Quote: Ace2

Quote: ThatDonGuy

Quote: poli2k01

Hi Fellow gamers and Craps enthusiasts,

I've read in an article or reference somewhere that through simulation of at least 1million dice rolls, it takes an average of 4 rolls to EITHER
1. A point is made OR
2. 7-Out

Can someone shed the math on how this was computed?
link to original post

First, note that if the probability of rolling a particular point number or 7 is p, then the expected number of rolls needed is:
1 x p + 2 x (1-p) x p + 3 x (1-p)^2 x p + 4 x (1-p)^3 x p + ...
= p x (1 + 2 x (1-p) + 3 x (1-p)^2 + ...)
= p x (1 + (1-p) + (1-p)^2 + ...)^2
= p x (1 / (1 - (1-p)))^2
= 1 / p
1/8 of the points established are 4s; the probability of rolling a 4 or 7 is 1/4, so the expected number of rolls to make or miss a point of 4 once it is established is 4.
1/6 of the points are 5s; the probability of rolling a 5 or 7 is 5/18, so the expected number of rolls is 18/5.
5/36 of the points are 6s; the probability of rolling a 6 or 7 is 11/36, so the expected number of rolls is 36/11.
Similarly, 1/8 of the points are 10s, 1/6 are 9s, and 5/36 are 8s, and the expected number of rolls for each are 4, 18/5, and 36/11, respectively.
The total is (1/8 x 4) + (1/6 x 18/5) + (5/36 + 36/11) + (5/36 + 36/11) + (1/6 x 18/5) + (1/8 x 4) = 196 / 55 = about 3.563636.
link to original post

I believe that’s the average rolls after a point has been established. Add 1.5 rolls starting from comeout
link to original post

I thought that was what was was asked, since it specified that a point had to be made or a 7-out rolled.

Maybe a clarification is in order. If the come-out is a 7, or a 3 for that matter, is that included in the number of rolls?
For example, is a sequence of 7, 4, 8, 12, 4 counted as 5 rolls, 4 (starting from when the point was established), or 3 (starting after the point is established)?

Maybe. But for this question I don’t see any mention of starting the calc after a point has already been established

Either way it’s simple. 1.5 rolls to establish point (maybe include) plus 3.56 to resolve point (definitely included)

First off, Thank you for showing the math. My question is in the context of AFTER the point has been established, but thank you for also showing the comeout portion which is 1.5 rolls.

Given that it's 3.5 rolls before a point is made before a 7-out, if I bet ONLY on the 1st out of the 3.5 rolls AFTER COMEOUT, and bring my inside bets DOWN all the time, do I get any statistical advantage?

Appreciate the help.

Quote: poli2k01

First off, Thank you for showing the math. My question is in the context of AFTER the point has been established, but thank you for also showing the comeout portion which is 1.5 rolls.

Given that it's 3.5 rolls before a point is made before a 7-out, if I bet ONLY on the 1st out of the 3.5 rolls AFTER COMEOUT, and bring my inside bets DOWN all the time, do I get any statistical advantage?

Appreciate the help.
link to original post

You have a great statistical advantage if you don't bet at all.

Quote: AlanMendelson

Quote: poli2k01

First off, Thank you for showing the math. My question is in the context of AFTER the point has been established, but thank you for also showing the comeout portion which is 1.5 rolls.

Given that it's 3.5 rolls before a point is made before a 7-out, if I bet ONLY on the 1st out of the 3.5 rolls AFTER COMEOUT, and bring my inside bets DOWN all the time, do I get any statistical advantage?

Appreciate the help.
link to original post

You have a great statistical advantage if you don't bet at all.
link to original post

Actually that is incorrect. You can neither win nor lose. Ergo no advantage can be derived.

Quote: cowboy

Quote: AlanMendelson

Quote: poli2k01

First off, Thank you for showing the math. My question is in the context of AFTER the point has been established, but thank you for also showing the comeout portion which is 1.5 rolls.

Given that it's 3.5 rolls before a point is made before a 7-out, if I bet ONLY on the 1st out of the 3.5 rolls AFTER COMEOUT, and bring my inside bets DOWN all the time, do I get any statistical advantage?

Appreciate the help.
link to original post

You have a great statistical advantage if you don't bet at all.
link to original post

Actually that is incorrect. You can neither win nor lose. Ergo no advantage can be derived.
link to original post

No advantage relative to a negative EV is a positive.

How much nave you won with your system?

The point is there is no mathematical formula to win at craps.

Quote: cowboy

How much nave you won with your system?
link to original post

Millions.