What is the ratio of Blue 7 vs Red 7 in E-Craps

If the shooter rolls a 7 on the come out roll, Aruze will registered the roll as a Blue 7.

If the shooter rolls a 7 when the point is established, Aruze will registered the roll as a Red 7.

I did a thought experiment and if you take the limit as N gets to infinity, then I get that a Blue 7 occurs 1/3 of the time and a Red 7 occurs 2/3 of the time. My logic is on the come out roll, 1/3 of the series will end and that is when a Blue 7 is present; 2/3 of the series will be a point and that is when the Red 7 shows up.

Does anyone know the exact ratio and help explain why that is the case. Thanks.

Quote: CAD2

Does anyone know the exact ratio

well,
(165/557) for Winner 7
(392/557) for 7 out

ratio follows as
392/165=2.37576 to 1

Quote: CAD2

and help explain why that is the case. Thanks.

Winner 7s only happen on the come out roll and about 29.62% (165/557) of craps rolls are come out rolls

has to do with
rolls per decision=3.375758 (557/165)

Donald Catlin did lots of math to show this
lots of credit to him (RIP)
http://catlin.casinocitytimes.com/article/how-long-is-a-craps-roll-1240

hope this helps

Sally

Quote: mustangsally

well,
(165/557) for Winner 7
(392/557) for 7 out

ratio follows as
392/165=2.37576 to 1
Winner 7s only happen on the come out roll and about 29.62% (165/557) of craps rolls are come out rolls

has to do with
rolls per decision=3.375758 (557/165)

....

hope this helps

Sally

Thank you. That helps a lot.

My original approach was:

6/12 * 1/3 = 1/6 or 0.1667 for Blue 7 &
0.596 * 2/3 = 0.3975 for Red 7

The logic being 1/3rd of rolls ends the series on the come out roll & the shooter 7's out 59.6% for the remaining 2/3rd of the series that has a point.

That would have gotten me to 2:38 to 1 but that was just a swag. The exact answer is 2.376 per your calculations.

There is guy on another website who claims he can control dice on Aruze E-Craps and I wanted to test his claims.

He was 4 for 4 for Red 7's; with a 70.377% success rate for Red 7's, he had a P(X=>x) of 24.53%. However, he then went 4 for 4 again bringing his total to 8 for 8, thus resulting in a P(X=>x) of 6.02%, which is getting close to a level that would be evidence of non-random rolls.