vegasrvp
Joined: Jun 15, 2010
• Posts: 53
March 6th, 2014 at 4:34:35 PM permalink
Can someone tell me the odds of back to back crap outs after points are established?

Need a little help with this one. Money on the line!!!!!
hwccdealer
Joined: Jun 4, 2013
• Posts: 365
March 6th, 2014 at 4:42:01 PM permalink
Quote: vegasrvp

Can someone tell me the odds of back to back crap outs after points are established?

Need a little help with this one. Money on the line!!!!!

If you mean "point, 7, point, 7," one in 81. A point-7 is one in 9, happening back-to-back would happen once in 81 such scenarios.
geoff
Joined: Feb 19, 2014
• Posts: 368
March 6th, 2014 at 4:42:54 PM permalink
Quote: vegasrvp

Can someone tell me the odds of back to back crap outs after points are established?

Need a little help with this one. Money on the line!!!!!

If you mean like this point->seven->point->seven then the odds are

(2/3)*(1/6)*(2/3)*(1/6)=0.01234567901
wudged
Joined: Aug 7, 2013
• Posts: 997
March 6th, 2014 at 4:51:55 PM permalink
Quote: geoff

If you mean like this point->seven->point->seven then the odds are

(2/3)*(1/6)*(2/3)*(1/6)=0.01234567901

Interesting number. Too bad .01234567890123456789... isn't something so "nice" looking as 1/81.

13717421 / 1111111111
hwccdealer
Joined: Jun 4, 2013
• Posts: 365
March 6th, 2014 at 4:59:48 PM permalink
Quote: wudged

Quote: geoff

If you mean like this point->seven->point->seven then the odds are

(2/3)*(1/6)*(2/3)*(1/6)=0.01234567901

Interesting number. Too bad .01234567890123456789... isn't something so "nice" looking as 1/81.

13717421 / 1111111111

Spoiler is a little off in terms of equaling 81. The actual number would be:

13,717,421 / 1,111,111,101
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
March 6th, 2014 at 5:14:51 PM permalink
Quote: vegasrvp

Can someone tell me the odds of back to back crap outs after points are established?

You mean to "7out"
(a player can not "crap out" while a point is established, that only happens on come out rolls)

The probability of any 7out is
294/495

294/495 * 294/495 would be for the next 2 in a row (35.2764%)

so the odds against that event would be (1/(294/495)^2) - 1 to 1
or
1.834756352 to 1

Is this what you are looking for?
Sally
I Heart Vi Hart
vegasrvp
Joined: Jun 15, 2010
• Posts: 53
March 6th, 2014 at 5:23:00 PM permalink
Clearification:

I'm looking to compare 7 out 7 out versus hitting the point.

vegasrvp
Joined: Jun 15, 2010
• Posts: 53
March 6th, 2014 at 5:24:48 PM permalink
Clearification:

I am not using the proper language I guess.

I'm looking to see the odds on throwing a 7 after point is established two times in a row without hitting the point?

Hope this is more clear.
sodawater
Joined: May 14, 2012
• Posts: 3231
March 6th, 2014 at 5:32:51 PM permalink
Quote: vegasrvp

Clearification:

I am not using the proper language I guess.

I'm looking to see the odds on throwing a 7 after point is established two times in a row without hitting the point?

Hope this is more clear.

So you want to calculate, given that a point is already established, what are the chances are of sevening out, and then throwing a winner 7 immediately on the come out?

example -- given a point of 4, you want to know the chances of at some point sevening out and then immediately throwing a seven on the come out?

It depends on the point:

for 4 and 10 your chances are 2/3 * 1/6 = 1/9

for 5 and 9 your chances are 3/5 * 1/6 = 1/10

for 6 and 8 your chances are 6/11 * 1/6 = 1/11

If you don't know the point, you have to do weighted probabilities:

5/12 of the time, your point will be a 6 or an 8. So 5/12 * 1/11 = 5/132

4/12 of the time, your point will be a 5 or a 9. So 4/12 * 1/10 = 1/30

3/12 of the time, your point will be a 4 or a 10. So 3/12 * 1/9 = 1/36

Add them all together, 1/36 + 1/30 + 5/132 and you get a grand total of 49/495 -- which I believe is the answer you're looking for.
vegasrvp
Joined: Jun 15, 2010
• Posts: 53
March 6th, 2014 at 5:38:37 PM permalink
Here is a specific example:
Roll 6
6 is the point
no matter how many rolls I am looking for 7 before the point.

Once 7 is rolled a new point is established;

Roll 8
Again no matter how many rolls a 7 is rolled before the point.

That is two 7 outs before a point is won.

The question is what are the odds that no winner will be rolled prior to two 7 outs.

If we call a winner point or loser 7out no point a session then I am looking to find out how many times is average between back to back loser 7 rolls?

Clearer yet. I HOPE.