kwalter827
Joined: Apr 12, 2011
• Posts: 8
April 12th, 2011 at 11:44:13 AM permalink
Hi! My first post, hope you math wizards can answer this one...

In a series of say 300 rolls in craps. How many times will a 6 or 8 appear between the 7s? This is just looking at the series of numbers, no "points" or anything else. I've done some thinking I think it will be around 80% of the time a 6 or 8 will get in between the 7s.

Any help is appreciated! Thanks to the Wizard for creating this site.

Keith
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6512
April 12th, 2011 at 2:08:17 PM permalink
Are you specifically looking for a sequence... (not 6,8 nor 7), 7, (6or8), 7, (not 6,8 nor 7)... Or does it matter how many non-6 or 8 numbers come up between the first 6 or 8, and the seven?
Simplicity is the ultimate sophistication - Leonardo da Vinci
SOOPOO
Joined: Aug 8, 2010
• Posts: 5879
April 12th, 2011 at 3:20:43 PM permalink
If I read your question correctly, after a 7 has already appeared, the odds of hitting a 6 or 8 before the next 7 is 10/16, which is 62.5%.
guido111
Joined: Sep 16, 2010
• Posts: 707
April 12th, 2011 at 3:39:22 PM permalink
Quote: kwalter827

Hi! My first post, hope you math wizards can answer this one...

In a series of say 300 rolls in craps. How many times will a 6 or 8 appear between the 7s? This is just looking at the series of numbers, no "points" or anything else. I've done some thinking I think it will be around 80% of the time a 6 or 8 will get in between the 7s.

Any help is appreciated! Thanks to the Wizard for creating this site.

Keith

Below are the chances up to 10 6s or 8s before a 7.
`62.50%	1 or more39.06%	2  or more24.41%	3 or more15.26%	4 or more9.54%	5 or more5.96%	6 or more3.73%	7 or more2.33%	8 or more1.46%	9 or more0.91%	10 or more`

I think you are asking in your question: in 300 dice rolls what is the probability of getting at least 1 6 or 8 in between 2 7s?
at least 2 6s or 8s in between 2 7s? at least 3 or more? In other words, you want to know the distribution of 6s and 8s between 2 7s. There also can be NO 6s and 8s between 2 7s.
That is a slightly different question from how many 6s or 8s can one get before a 7 rolls 1 time.
I do have results somewhere, maybe even actual calculations. I have to look for them.

In 300 dice rolls, on average, you would see ~133 6,7 and 8s.
expected number of 7s would be 50.
edit:
I have to run
Quote: kwalter827

Guido111,

Thanks for your help! You are on the right track! I am interested in the EVENT of 300 rolls, and looking at the distribition of 7s, 6/8s. I figured from previous posts that there would be around 6 double 7s (2-7s together) and 1 triple 7(3 7s in a row) so obviously there would be no 6s or 8s between them, but what about the other 43 7s?

Given 300 rolls and the expected 50 7s, I guess another way of asking is out of the 49 spaces between 7s, how many would have at least one 6 or one 8? It would also be cool to know how many of those spaces between 7s would contain at least n number of 6s or 8s...but that is another question.

Thanks for looking into this!

Keith

I had the math in a table that I can not find and I be too lazy to re-do the math. Basically your average of 49 spaces will still be filled with the probability of a 6 or 8 before one 7 just as you have sais.
Since you only do not count the 6 and 8s that roll before the first 7 but count every gap between the next 49.

A few sim results then you can take it from there.
Average # of 7s in 300 rolls 49.6725 that have 6 and 8s before. standard deviation of 6.391. That is close to the expected number of 49.

exactly 0 6/8s 37.458% right close to the 37.5%
exactly 1 6/8s 23.828% right close to the 23.44%
exactly 2 6/8s 14.696% right close to the 14.65%
here is the distribution from the probabilities at the beginning of this post.
`37.50%	0 exactly23.44%	1 exactly14.65%	2 exactly9.16%	3 exactly5.72%	4 exactly3.58%	5 exactly2.24%	6 exactly1.40%	7 exactly0.87%	8 exactly0.55%	9 exactly0.34%	10 exactly`

I need to ask why do you want to know this information?
Enjoy
kwalter827
Joined: Apr 12, 2011
• Posts: 8
April 13th, 2011 at 5:34:06 AM permalink
Guido111,

Thanks for your help! You are on the right track! I am interested in the EVENT of 300 rolls, and looking at the distribition of 7s, 6/8s. I figured from previous posts that there would be around 6 double 7s (2-7s together) and 1 triple 7(3 7s in a row) so obviously there would be no 6s or 8s between them, but what about the other 43 7s?

Given 300 rolls and the expected 50 7s, I guess another way of asking is out of the 49 spaces between 7s, how many would have at least one 6 or one 8? It would also be cool to know how many of those spaces between 7s would contain at least n number of 6s or 8s...but that is another question.

Thanks for looking into this!

Keith
kwalter827
Joined: Apr 12, 2011