March 12th, 2013 at 9:36:44 PM
permalink
Can you please educate me on what the exact odds are in the following scenario(s).
1. What is the odds of (3) consecutive Sevens appearing before either a 6 or an 8 appears.
2. What is the odds of (4) consecutive Sevens appearing before either a 6 or an 8 appears.
Thank you for your help.
Cainan
1. What is the odds of (3) consecutive Sevens appearing before either a 6 or an 8 appears.
2. What is the odds of (4) consecutive Sevens appearing before either a 6 or an 8 appears.
Thank you for your help.
Cainan
March 12th, 2013 at 10:28:05 PM
permalink
Nice questions.Quote: Cainan18Can you please educate me on what the exact odds are in the following scenario(s).
1. What is the odds of (3) consecutive Sevens appearing before either a 6 or an 8 appears.
2. What is the odds of (4) consecutive Sevens appearing before either a 6 or an 8 appears.
I get
1 in 18.96 trials for 3 in a row
1 in 50.57 trials for 4 in a row
Each trial is NOT one roll long
A bit more math to the questions
We can ignore any roll other than a 6,7 or 8 for now.
There are 6 ways to throw a 7
5 ways each for the 6&8
7 before a 6 or 8 = 6/16 = .375
3 times in a row
=.375^3 (.375*.375*.375) = 0.052734375
or 1 in 18.96 trials
But each trial is not 1 roll long (of 6,7 and 8s only)
That is the sum of a geometric series
1+(6/16)+(6/16)^2 = 1.515625 = length of each trial
Multiply the two together
18.96 * 1.515625 = 28.74074074 the average number of rolls of 6,7,8s
For the average number of total rolls (your total wait time)
28.74074074 * (36/16) = 64.667 rolls on average
For 4 in a row just extend the math
(6/16)^4 = 0.019775391 or 1 in 50.56790123 (1/0.019775391)
The length of each trial
1+(6/16)+(6/16)^2+(6/16)^3 = 1.568359375
product = 79.30864198 (50.56790123*1.568359375)
This is just the average number of 6,7 and 8s
now Times 2.25
(2.25 is the average number of rolls to see either a 6,7 or 8)
178.44 = average number of total dice rolls
A quick look at the Zumma Craps Tester Book (35097 actual dice roll file) shows
at least 3 in a row 7s before a 6 or 8 happened 536 times, avg=65.48
at least 4 in a row 7s before a 6 or 8 happened 196 times, avg=179.07
There are 6 ways to throw a 7
5 ways each for the 6&8
7 before a 6 or 8 = 6/16 = .375
3 times in a row
=.375^3 (.375*.375*.375) = 0.052734375
or 1 in 18.96 trials
But each trial is not 1 roll long (of 6,7 and 8s only)
That is the sum of a geometric series
1+(6/16)+(6/16)^2 = 1.515625 = length of each trial
Multiply the two together
18.96 * 1.515625 = 28.74074074 the average number of rolls of 6,7,8s
For the average number of total rolls (your total wait time)
28.74074074 * (36/16) = 64.667 rolls on average
For 4 in a row just extend the math
(6/16)^4 = 0.019775391 or 1 in 50.56790123 (1/0.019775391)
The length of each trial
1+(6/16)+(6/16)^2+(6/16)^3 = 1.568359375
product = 79.30864198 (50.56790123*1.568359375)
This is just the average number of 6,7 and 8s
now Times 2.25
(2.25 is the average number of rolls to see either a 6,7 or 8)
178.44 = average number of total dice rolls
A quick look at the Zumma Craps Tester Book (35097 actual dice roll file) shows
at least 3 in a row 7s before a 6 or 8 happened 536 times, avg=65.48
at least 4 in a row 7s before a 6 or 8 happened 196 times, avg=179.07
I gots a feeling I know why you asked the questions.
Good Luck
winsome johnny (not Win some johnny)