Craps probability question 6/8s in 7s
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
