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

Quote:kwalter827Hi! 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 more`

39.06% 2 or more

24.41% 3 or more

15.26% 4 or more

9.54% 5 or more

5.96% 6 or more

3.73% 7 or more

2.33% 8 or more

1.46% 9 or more

0.91% 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:kwalter827Guido111,

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

Getting back to your Q.

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 exactly`

23.44% 1 exactly

14.65% 2 exactly

9.16% 3 exactly

5.72% 4 exactly

3.58% 5 exactly

2.24% 6 exactly

1.40% 7 exactly

0.87% 8 exactly

0.55% 9 exactly

0.34% 10 exactly

I need to ask why do you want to know this information?

Enjoy

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