Sheldon ross two sevens before 6 evens

[mattclgn]

There was a thread on this problem but It was closed and I am still having problems even having read it.

Homework Statement

In successive rolls of a pair of fair dice, what is the probability of getting 2
sevens before 6 even numbers?

Homework Equations

Let A, B and C be the events that you get a seven, and even number, or neither on a given roll,
respectively. Thus
P(A) = 6/36
P(B) = 18/36
P(C) = 1 − P(A) − P(B) = 1/3

The Attempt at a Solution

This had a solution http://www.math.duke.edu/~wka/math135/until.pdf
and there was one at this forum https://www.physicsforums.com/showthread.php?t=656455
but I could not fully understand the answers (doing an independent study to I do not have a teacher I can just ask)
UC davis had an answer that made a bit more sense https://www.math.ucdavis.edu/~gravner/MAT135A/hw4.pdf but still fell short of a coherent explanation.

I'll address my problem with that first.
I get why we subtract the first part from one. probability 6 evens 0 sevens plus probability 1 seven exactly multiplied by 6 evens ALL subtracted from one. But shouldn't that be .5^6 for both of them?

The duke one was even more troublesome
for this part = 1 P(A) + 0 P(B) + P(E0)P(C)
C should be 1/3* probability I roll j number of evens on the first roll given l neither evens nor 7 on the first roll?
Why do I treat it as though C is unknown and as though (E0lC)(C) is the same as E0

I was at a complete loss for the second
= 0 P(A) + P(Ej−1)P(B) + P(Ej )P(C)

why is P(Ej |even on the first roll)P(even on the first roll) equal to P(Ej−1)P(B)...so I factor out the first even?

what are the steps that take (1 − P(C))P(Ej ) = P(Ej−1)P(B)
to the next part? Why is it all to the j power? I think i might be missing some rules of Geometric series. Or something.

 

[Ray Vickson]

There was a thread on this problem but It was closed and I am still having problems even having read it.

Homework Statement

In successive rolls of a pair of fair dice, what is the probability of getting 2
sevens before 6 even numbers?

Homework Equations

Let A, B and C be the events that you get a seven, and even number, or neither on a given roll,
respectively. Thus
P(A) = 6/36
P(B) = 18/36
P(C) = 1 − P(A) − P(B) = 1/3

The Attempt at a Solution

This had a solution http://www.math.duke.edu/~wka/math135/until.pdf
and there was one at this forum https://www.physicsforums.com/showthread.php?t=656455
but I could not fully understand the answers (doing an independent study to I do not have a teacher I can just ask)
UC davis had an answer that made a bit more sense https://www.math.ucdavis.edu/~gravner/MAT135A/hw4.pdf but still fell short of a coherent explanation.

I'll address my problem with that first.
I get why we subtract the first part from one. probability 6 evens 0 sevens plus probability 1 seven exactly multiplied by 6 evens ALL subtracted from one. But shouldn't that be .5^6 for both of them?

The duke one was even more troublesome
for this part = 1 P(A) + 0 P(B) + P(E0)P(C)
C should be 1/3* probability I roll j number of evens on the first roll given l neither evens nor 7 on the first roll?
Why do I treat it as though C is unknown and as though (E0lC)(C) is the same as E0

I was at a complete loss for the second
= 0 P(A) + P(Ej−1)P(B) + P(Ej )P(C)

why is P(Ej |even on the first roll)P(even on the first roll) equal to P(Ej−1)P(B)...so I factor out the first even?

what are the steps that take (1 − P(C))P(Ej ) = P(Ej−1)P(B)
to the next part? Why is it all to the j power? I think i might be missing some rules of Geometric series. Or something.

The answer in your Duke link is straightforward to get, but not using the argument in the link itself. It is much easier to proceed directly.

At some stage during the tossing procedure, we are in state (i,j), meaning that we already have accumulated i 7s and j evens. Another (relevant) toss takes us to either (i+1,j) (with probability q = 1/4) or to state (i,j+1) (with probability p = 3/4). We start at (0,0) and want to know the probability of reaching one of the states (2,j), 0 <= j <= 5. It is quite easy to count paths in this case: there are (j+1) distinct paths from (0,0) to (2,j), and each such path has probability q^2 * p^j. Therefore, the event A = {get 2 7s before 6 Es} has probability
P(A) = <br /> \sum_{j=0}^5 (j+1) q^2 p^j = (1/4)^2 \sum_{j=0}^5 (j+1)(3/4)^j