craps math

So what I'm trying to figure out is......what would the house advantage be(or player advantage) if in craps the only 2 point numbers were 6 and 8. So if on the come out, on a pass line bet everything would be the same(7, 11 win and 2,3, and 12 lose) except if you rolled a 4,5,9 or 10 it would just be a push on your pass line bet and no point was established unless it was a 6 or 8.

I'm just trying to figure out if you have an above average amount of points going to the 6 and 8 on the come out if the player at that point would have and advantage over the house? I can just look at the Wizards table for a "put bet" on the 6 or 8(9.09% house edge) but of course thats not taking into account the 2:1 advantage you would have on the come out roll. If there would be a player advantage, if the only point numbers where 6 and 8(which I think there would be) It would also be nice to know where the tipping point is as to how many 6 or 8 points have to be established vs other points to where the game turns from house advantage to player advantage. I hope this makes since, if not just ignore the post LOL!

Any help would be cool, thanks
Jason

If points of 4,5,9, and 10 pushed, the player would have a 8.59% advantage. (17/198 exactly) As for your other question, I have no clue.

thats for the response :)

Quote: vert1276

...It would also be nice to know where the tipping point is as to how many 6 or 8 points have to be established vs other points to where the game turns from house advantage to player advantage...

In a standard craps game (4,5,9,10 can be points) there is never a time when strings of 6/8's as points make the game a player advantage. There will always be more ways to seven-out than make either the 6 or 8 as the point. Playing the don't, you will always have to get past the seven (and push on the 12) coming out. The edges are built in.

Quote: Ayecarumba

In a standard craps game (4,5,9,10 can be points) there is never a time when strings of 6/8's as points make the game a player advantage. There will always be more ways to seven-out than make either the 6 or 8 as the point. Playing the don't, you will always have to get past the seven (and push on the 12) coming out. The edges are built in.

Not that I have the math to prove it but.......If the only points rolled were 6 and 8 there would be a player advantage. since you are a 2:1 on the come out and only a 5:6 on the 6 and 8.

I guess I should have been more clear on my second question..... What im trying to get at is. By probability of 24 points established the break down would look like this:

Point # of times established
------ -------------------------
4 3 times
5 4 times
6 5 times
8 5 times
9 4 times
10 3 times

BUT WHAT IF IT LOOKED LIKE THIS?

4 1 time
5 2 times
6 9 times
8 9 times
9 2 times
10 1 time

Now assuming the come out came out in line with probabilities 2 wins for every one loss. And you were making $5 pass line bets your EV on the $5 pass line bet would have to be greater then $5 dollars correct? Or am I missing something?

Quote: vert1276

Not that I have the math to prove it but.......If the only points rolled were 6 and 8 there would be a player advantage. since you are a 2:1 on the come out and only a 5:6 on the 6 and 8.

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.

For true 6 x 6 die:

2,3,12: lose 1 unit x 4/36
7,11: win 1 unit x 8/36
6,8: lose 1 unit (5/11 - 6/11) x 10/36
5,9: lose 1 unit (4/10 - 6/10) x 8/36
4,10: lose 1 unit (3/9 - 6/9) x 6/36

Add it up: -1 x 4/36 + 8/36 - 1/11 x 10/36 - 2/10 x 8/36 - 3/9 x 6/36 = -.0141414. So far so, good?


There is only 22 possibilities for win/lose, per resolved bet when the 4,5,9,10 magically disappear.

2,3,12: lose 1 unit x 4 /22
7,11: win 1 unit x 8 / 22
6,8: lose 1/11 unit (5/11 - 6/11) x 10 / 22

Add it up: 4 / 22 - (10/11) / 22 = (34/11) / 22 = 34/242 or 14.04 percent player advantage.

If the casino wanted to make the game in their advantage, they would make the 11 a loss instead of a win.

2,3,11,12 lose 1 unit x 6/22
7,11: win 1 unit x 6/22
Add it up: -10/242 = 4.13 percent house advantage.

The game is so well balanced on the pass line that as soon as you get rid of 1 point (like a 4 or 10), the game turns into the player advantage.

Example: Craps with no 10s (10 is a push)

2,3,12: lose 1 unit x 4/33
7,11: win 1 unit x 8/33
6,8: lose 1 unit (5/11 - 6/11) x 10/33
5,9: lose 1 unit (4/10 - 6/10) x 8/33
4,10: lose 1 unit (3/9 - 6/9) x 3/33

Add it up: -1 x 4/33 + 8/33 - 1/11 x 10/33 - 2/10 x 8/33 - 3/9 x 3/33 = 1.49 percent player advantage.

My math is a bit different here.

First off, we only care about rolls of 2, 3, 6, 7, 8, 11 and 12. That's 22 possible rolls. I think this is the big mistake others are making -- the denominator is not 36, as there are 14 rolls we don't care about.

We win on come-out of 7 or 11, or on a 6 and then 6 rolling before 7, or on an 8 and then 8 rolling before 7. The last two events have the same probability.

Win = (6/22) + (2/22) + 2(5/22)(5/11) = 8/22 + 25/121 = (88+50)/242 = 138/242.

We lose on everything else, which should be 104/202, but as a check we can compute:

Lose = 1/22 + 2/22 + 1/22 + 2(5/22)(6/11) = 4/22 + 30/121 = (44+60)/242 = 104/242

Ignoring the possibility of taking odds on points of 6 or 8, the wins return +1 and the losses return -1. So the EV is simply the difference between these two fractions, or 34/242 = 17/121 = .14049. This game has a 14.05% player advantage.

Quote: 7outlineaway

My math is a bit different here.

First off, we only care about rolls of 2, 3, 6, 7, 8, 11 and 12. That's 22 possible rolls. I think this is the big mistake others are making -- the denominator is not 36, as there are 14 rolls we don't care about.

We win on come-out of 7 or 11, or on a 6 and then 6 rolling before 7, or on an 8 and then 8 rolling before 7. The last two events have the same probability.

Win = (6/22) + (2/22) + 2(5/22)(5/11) = 8/22 + 25/121 = (88+50)/242 = 138/242.

We lose on everything else, which should be 104/202, but as a check we can compute:

Lose = 1/22 + 2/22 + 1/22 + 2(5/22)(6/11) = 4/22 + 30/121 = (44+60)/242 = 104/242

Ignoring the possibility of taking odds on points of 6 or 8, the wins return +1 and the losses return -1. So the EV is simply the difference between these two fractions, or 34/242 = 17/121 = .14049. This game has a 14.05% player advantage.

Yep that's what I get. I was just counting pushes in the denominator like the wizard does of the don't pass. (17/121)*(22/36)=17/198.