Odds of....

Could somebody tell me what the odds are (or against) the following happening:

(1) someone 7'ing out on the next roll of the dice after the point has been established
(2) what the odds against this phenomenon occurring in (1) twice in a row
(3) what the odds against this phenomenon occurring in (1) three times in a row; and finally,
(4) what the odds against this phenomenon occurring in (1) four times in a row


Thank you in advance!

Quote: kcookejr

Could somebody tell me what the odds are (or against) the following happening:

(1) someone 7'ing out on the next roll of the dice after the point has been established
(2) what the odds against this phenomenon occurring in (1) twice in a row
(3) what the odds against this phenomenon occurring in (1) three times in a row; and finally,
(4) what the odds against this phenomenon occurring in (1) four times in a row


Thank you in advance!

Hello kcookejr, and welcome to the forums.

Every roll in craps in independent from every other roll, much like spins in roulette.

The odds of rolling a 7 at any time are P(7) = 6/36, 1/6, or .1667.

What you're asking are combination probabilities. What are the odds X happens given Y has happened already.

1) P(point number) * P(7) = .8333 * .1667 = .1389, or about a 13.89% chance.
2) P(point number) * P(7) * P(point number) * P(7)
3) P(point number) * P(7) * P(point number) * P(7) * P(point number) * P(7)
4) P(point number) * P(7) * P(point number) * P(7) * P(point number) * P(7) * P(point number) * P(7)

So it looks for you to be able to answer your own question all we need to do is figure out what P(point number) is. This is to say, what's the probability that the shooter throws a 4, 5, 6, 8, 9, or 10?

P(point number) = 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 = 30/36, 5/6, or .8333

I've done the first for you, have fun =).

(please don't ask what the probability of rolling 18 yo's in a roll is. k thx)

Quote: Romes

Hello kcookejr, and welcome to the forums.

Every roll in craps in independent from every other roll, much like spins in roulette.

The odds of rolling a 7 at any time are P(7) = 6/36, 1/6, or .1667.

What you're asking are combination probabilities. What are the odds X happens given Y has happened already.

1) P(point number) * P(7) = .8333 * .1667 = .1389, or about a 13.89% chance.
2) P(point number) * P(7) * P(point number) * P(7)
3) P(point number) * P(7) * P(point number) * P(7) * P(point number) * P(7)
4) P(point number) * P(7) * P(point number) * P(7) * P(point number) * P(7) * P(point number) * P(7)

So it looks for you to be able to answer your own question all we need to do is figure out what P(point number) is. This is to say, what's the probability that the shooter throws a 4, 5, 6, 8, 9, or 10?

P(point number) = 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 = 30/36, 5/6, or .8333

I've done the first for you, have fun =).

(please don't ask what the probability of rolling 18 yo's in a roll is. k thx)

I get P(point) = 2/3 = 0.6667, not 0.8333. I think you may have included 7's in your P(point) calculation, Romes.

Also, since he stipulated in the OP, that the point had already been established for question #1, I think you can remove the first P(point) term in each of the answers above. Other than that, you can use Romes' formulas above to get your odds. Unless...

Are you discounting the possibility of any winners (7 or 11) or craps (2,3, or 12) on the come out? I.e., are you asking what are the odds of making a 7-out immediately after the establishment of 2, 3, & 4 points in a row? If that is the case, it is an easier calculation. Just raise the P(7) value (1/6) to the power of the number of times you want to repeat it:

1) -- 1 time in a row = (1/6)^1
2) -- 2 times in a row = (1/6)^2
3) -- 3 times in a row = (1/6)^3
4) -- 4 times in a row = (1/6)^4

Ha, you're quite right! 6/36 should not be in there =P...

P(point number) = 24/36, 4/6, or .6667

Quote: Joeman

...Also, since he stipulated in the OP, that the point had already been established for question #1, I think you can remove the first P(point) term in each of the answers above...

I first did my post with that assumption, but reading the post I feel as though the OP wants to know the original P(point number) then P(7).

I had just P(7) as the first answer then edited it, figuring if that's what he wanted earlier in my post it says "P(7) = .1667 (1/6).

wow I think the odds are the same every time sir, 6/36, after you get a point.
I think it is so because you said 7 out right after the point is made.

if you want to know the odds of getting your point and then 7 out, or getting 2 points and 7 out... That's is a money question and you can re-ask if that is really what you want to know.

Thank you all for your replies. I think what might have been lost in the translation is that lets say on the come out roll a point is established. I see that the odds of a 7 out on the immediately ensuing roll is 1/6. Now comes a second come out. Another point is established. Again, the odds of a 7 out on the very next roll is 1/6. So I guess the answer to my question of this 7-out occurring on the immediate roll after a point is established twice in a row is 1/6 x 1/6 or 1/36. Would that be correct? Then three times in a row is 1/216? And four times in a row 1/1,296?

6, 36, 216, 1296.....yes that is correct, if I'm understanding the situation properly.

ie:
2
6
7
^ counts

8
4
7
^ doesn't count (because a 4 is rolled between the established point number and the 7-out)

5
7
^ counts

3
2
11
7
7
8
7
^ counts

Thank you. I figure that wold be the case. Trying to develop a play with that, but it's very dangerous!

Quote: kcookejr

Thank you. I figure that wold be the case. Trying to develop a play with that, but it's very dangerous!

I'm going to make a wild stab in the dark and say -- whatever it is you're trying to come up with, almost certainly has no chance at being a winning system. Be careful.

Quote: RS

I'm going to make a wild stab in the dark and say -- whatever it is you're trying to come up with, almost certainly has no chance at being a winning system. Be careful.

Stab in the dark... Stab in the ocean... Either way you're hitting pay dirt.

Craps can't be beat with any "system" OP. Beware.

Quote: Romes

Stab in the dark... Stab in the ocean... Either way you're hitting pay dirt.

Craps can't be beat with any "system" OP. Beware.

There are 'systems' that work, this is true. But the whole damn bunch of you may wind up in prison, or worse. Proceed with great caution...;-)