craps math

Quote: Ayecarumba

Coming out:
You win with a roll of 7 or 11.
-- There are 6 ways to roll a 7.
-- There are 2 ways to roll an 11.
8 ways out of 36 to win (22.22%).

You lose with a roll of 2, 3, or 12
--There is one way to roll a 2.
--There are two ways to roll a 3.
--There is one way to roll a 12.
4 ways out of 36 to lose (11.11%).

Any other number results in the establishment of a "point".
--There are 5 ways to roll a 6.
--There are 5 ways to roll an 8.
10 ways out of 36 (27.78%).

--Other possible numbers that you somehow "know" will not appear...
-- 14 ways out of 36 (38.89%)

Assume a point of 6 (or 8) is established, you win with a repeat 6 (or 8) before a 7.
-- There are 5 ways to roll a 6 (or 8), (13.89%)
-- There are 6 ways to roll a 7 (16.67%)
-- All other numbers are meaningless to this contract bet.


Blissfully ignoring the fact that you have no way of knowing that points of 4,5,9 or 10 will not roll, you will have a 22.22% chance of winning on the come out, and a 38.89% chance of losing on the come out, OR throwing a 6 or 8. It is important to combine these two outcomes because...

If you don't lose coming out, and manage to establish a point (the most likely outcome), you still have a 2.78% chance of sevening out before making a point of 6 or 8.

In other words, the most likely outcome is that you will lose coming out, or establish a point that has a -2.78% chance of repeating. You are still up against it.

If a streak of 6's or 8's appear, and you are betting on it, of course you will make money; but riding a streak, and having an advantage before a point is established, are very, very different things.

Quote: vert1276

I think we are going to talk ourselves in circles here but, thanks for the discussion! And your math is correct. But what I trying to get at is......EV or a given number of points being established. Example: Over 36 come out rolls....

2,3,12....4/36
7,11.......8/36
4,10.......2/36
5,9.........6/36
6,8.......16/36

Assuming you are betting 1 unit on the pass line for 36 come out rolls (1unit *36 come out rolls) your total bet would be 36 units. The EV on those 36 units would be great than 36. Correct?

Quote: vert1276

I think we are going to talk ourselves in circles here but, thanks for the discussion! And your math is correct. But what I trying to get at is......EV or a given number of points being established. Example: Over 36 come out rolls....

2,3,12....4/36
7,11.......8/36
4,10.......2/36
5,9.........6/36
6,8.......16/36

Assuming you are betting 1 unit on the pass line for 36 come out rolls (1unit *36 come out rolls) your total bet would be 36 units. The EV on those 36 units would be great than 36. Correct?

Quote: Ayecarumba

I don't think you can ignore the fact, even with the pumped up occurences of 6's and 8's, that the order matters. In those 36 rolls, the timing of the appearance of the 7 is the key to making a lot of cash, or cursing the "line away".

I will defer to Miplet's expertise and spreadsheet, but do want to remind you that in the real world, you have no way of knowing if a sequence of rolls is rich or poor in 6's and 8's... until it is too late to do anything about it.

Quote: miplet

I put a spreadsheet in Google Docs The come out column is how often you roll the numbers while coming out. The make point and 7 out columns are how often you make the point vs 7 out. It is set up for your average shooter. Any questions, just ask.

Quote: duckston09

One dice has to stop before the other. You can always make a seven with the other dice.

Quote: duckston09

oops, sorry