How is this calculated?

https://wizardofodds.com/games/craps/appendix/1/

Pass/Come
The probability of winning on the come out roll is pr(7)+pr(11) = 6/36 + 2/36 = 8/36.

The probability of establishing a point and then winning is pr(4)×pr(4 before 7) + pr(5)×pr(5 before 7) + pr(6)×pr(6 before 7) + pr(8)×pr(8 before 7) + pr(9)×pr(9 before 7) + pr(10)×pr(10 before 7) =

(3/36)×(3/9) + (4/36)×(4/10) + (5/36)×(5/11) + (5/36)×(5/11) + (4/36)×(4/10) + (3/36)×(3/9) =
(2/36) × (9/9 + 16/10 + 25/11) =
(2/36) × (990/990 + 1584/990 + 2250/990) =
(2/36) × (4824/990) = 9648/35640
The overall probability of winning is 8/36 + 9648/35640 = 17568/35640 = 244/495
The probability of losing is obviously 1-(244/495) = 251/495
The player's edge is thus (244/495)×(+1) + (251/495)×(-1) = -7/495 ≈ -1.414%.

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Everything else is clear but I don't understand these calculations:
pr(4 before 7)=3/9
pr(5 before 7)=4/10
pr(6 before 7)=5/11
How is pr(4 before 7) calculated? Why is it 3/9?

Quote: WangSanJose

How is pr(4 before 7) calculated? Why is it 3/9?

There are 3 ways to win and 6 ways to lose, 9 ways total. None of the other rolls matter

With my lack of math skills, I know I will never grasp this despite the assistance of The Wizard, BBB, Mustang Sally
and her companion, vi heart and all the others here who try valiantly to explain things.

I just accept 1.414, try to remember it and realize that is darned low.

So I play craps rather than slots.

Quote: WangSanJose

Everything else is clear but I don't understand these calculations:
pr(4 before 7)=3/9
pr(5 before 7)=4/10
pr(6 before 7)=5/11
How is pr(4 before 7) calculated? Why is it 3/9?

this is how I do it
pr(4 before 7)
as already mentioned we are only concerned about the rolls for a 4 and the 7
3 ways for 4
6 ways for 7
both are mutually exclusive (cannot happen at the same time) and independent events
so we can add them together

9 ways for sample space
(Sample Space: all the possible outcomes of an experiment)
https://www.mathsisfun.com/data/probability.html

the denominator
now we have roll4 / 9
roll4 is quite easy 3/9

the others are done the same way
pr(5 before 7)=4/10: roll5 / 4+6
pr(6 before 7)=5/11: roll6 / 5+6

Instead of Craps consider a similar mathematical problem such as picking balls out of a bag. In this example consider trying to roll a 4 before a 7.

The bag has (say) 3 green balls, 6 red balls, and 27 white balls. The game is to pick a ball until you either pick a green one or a red one. If you pick a white one then put it back in the bag and pick again.
So every time you draw a ball out of the bag the chances of a green is 3/36, red 6/36 and white 27/36.

The chances of the game ending with a green = (pr(0 whites)+pr(1 white)+pr(2 whites)...)*3/36.
The chances of the game ending with a red = (pr(0 whites)+pr(1 white)+pr(2 whites)...)*6/36.
Since you will evetually either pick a red or green, these add up to 1
(pr(0 whites)+pr(1 white)+pr(2 whites)...)*(3/36+6/36)=1.
(pr(0 whites)+pr(1 white)+pr(2 whites)...)=1/(3/36+6/36)=1/(9/36)=36/9.

The chances of a green = 36/9*3/36 = 3/9.
The chances of a red = 36/9*6/36 = 6/9.

(i) Alternative way to look at it....
The game will eventually end (after you've picked a series of whites) so you must have drawn one of the 9 non-white balls. As 3 are green and 6 are red then the chances of winning with green is 3/9.

(ii) Alternative way to look at it....
For the purposes of deciding whether a green or red ball will be drawn first, as any drawing of the white balls could be ignored, mathematically it's the same as a bag with the nine coloured balls 3 green and 6 red.
This time, without any whites, the game is resolved on the first draw, so the chances of a green is 3/9 and red 6/9.