The most basic of all craps bet is on the Pass Line, which is also known as “right way”. Mar 06, 2012 Just came back from a trip where I played next to the Greatest Craps Shooter Ever! I have been playing Craps for over a decade and this one takes the cake. After having dinner with my wife.I had $100 on me, gave my wife $20 to play the slots, had $80 left.decided to play a few minutes of craps.the only spot available was on a $15 Craps.

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37 members have voted

Kind of a trick question, because if he sevens on roll #2, he never has an opportunity to seven out on the third or later roll.

thank you for joining the discussion here
now,
not at all
kind of a trick question
really a simple easy question
a craps shooter picks his 2 dice and faces his first come out roll
what roll (from 2 to 2,000,000,000,000) is the most likely roll to 7 out on?
craps players say the skill of the shooter determines the answer, but that is not true, imo, as each roll is an independent event
some also say to 7out on roll 2 is a rare event
I have shown the probabilities to 7out on the 2nd and the 3rd roll so far in other posts
summary
the 2nd roll is easy as the only sequences that qualify are these
4,7
5,7
6,7
8,7
9,7
10,7
because there are 36*36 = 1296 ways the dice can land in exactly 2 rolls
we can easily calculate the number of ways for each sequence
4,7 = 3*6 = 18
5,7 = 4*6 = 24
6,7 = 5*6 = 30
(18+24+30)*2 = 144 ways
so 144/1296
but all this math was not really needed as the case for 2 rolls is simply 2/3 * 1/6 = 2/18 = 1/9 = 11.11%

So if the chance is 1/6 on the second roll,

yes

its only 5/6 x 1/6 on the third, because you don't even get there 1/6 of the time.

you over-counting
the first roll could be a 7 followed by 4,7 for example

The very possibility of it happening on roll two reduces this conditional probability for subsequent rolls.

well, roll 3 has a higher probability than roll 2 as I showed in another post
so your conclusion is not correct
every possible sequence is right here to 7out on that 3rd roll
2,4,7
2,5,7
2,6,7
2,8,7
2,9,7
2,10,7
3,4,7
3,5,7
3,6,7
3,8,7
3,9,7
3,10,7
4,2,7
4,3,7
4,5,7
4,6,7
4,8,7
4,9,7
4,10,7
4,11,7
4,12,7
5,2,7
5,3,7
5,4,7
5,6,7
5,8,7
5,9,7
5,10,7
5,11,7

Best Craps Roll Ever Wins

5,12,7
6,2,7
6,3,7
6,4,7
6,5,7
6,8,7
6,9,7
6,10,7
6,11,7
6,12,7
7,4,7
7,5,7
7,6,7
7,8,7
7,9,7
7,10,7
8,2,7
8,3,7
8,4,7
8,5,7
8,6,7
8,9,7
8,10,7
8,11,7
8,12,7
9,2,7
9,3,7
9,4,7
9,5,7
9,6,7
9,8,7
9,10,7
9,11,7
9,12,7
10,2,7
10,3,7
10,4,7
10,5,7
10,6,7
10,8,7
10,9,7
10,11,7
10,12,7
11,4,7
11,5,7
11,6,7
11,8,7
11,9,7
11,10,7
12,4,7
12,5,7
12,6,7
12,8,7
12,9,7
12,10,7
if one does the math the same way as I did for the 2nd roll
and divide the total by 36*36*36
that should be
5448 / 46656 = 11.67695473251%
higher than roll 2 at 11.11%
both of my results match these here
http://wizardofodds.com/games/craps/number-of-rolls/
'The second column is the probability of a seven-out on exactly that roll'
and the 2nd article here
http://wizardofvegas.com/member/7craps/blog/
now on to roll 4 but with a different method
using the chance of the 4th roll NOT being a 7out

Best Craps Roll Ever To Play

(and will show roll 2 and 3 too this way)
My Excel
len-shooters-hand-recursive.xlsb

Best Craps Roll Ever Caught

link in my blog

Roll	State 1 (cor)	State 2 (4or10)	State 3 (5or9)	State 4 (6or8)	survive past roll x
0	1	0	0	0	1
1	12/36	6/36	8/36	10/36	1
2	0.188271605	0.180555556	0.234567901	0.285493827	0.888888889

Roll 0 = the shooter has selected the 2 dice for the come out roll but has not yet thrown them.
The 1 = that probability, the 3 point states are = 0
after roll#1
we now see the 4 probabilities for the 4 states the hand can be in
we must be in one of the 4 states with the stated probabilities.
(we are not considering a no-roll)
There are also transition probabilities to get to each state from roll to roll
It should be clear there are only two ways to get to a point state.
Either we came from a COR state at rollN-1 (the point was just established)
or we have remained in the point state (the roll was not a 7out or a point winner)
for row = roll 1
C3 = (6/36)*B2 + (27/36)*C2
(6/36)*B2 = the chance of rolling a 4 or 10 * the probability of being on the cor on the last roll
(27/36)*C2 = the chance of rolling any number except a 4,10,7 * the probability of being in that point state on the very last roll
D3 = (8/36)*B2 + (26/36)*D2
E3 = (10/36)*B2 + (25/36)*E2
for B3 we now have 4 ways to get to the cor
= (12/36)*B2 + (3/36)*C2 + (4/36)*D2 + (5/36)*E2
= we stay +we came from + we came from + we came from
F2 = the sum of B2 to E2
these cells get filled down to how far you want to go
I went to row 202
ok
this is done for me for now

Best Craps Roll Ever Game

I get bored quickly doing simple math
this was fun to a point

Roll	7out on
0	0
1	0
2	0.111111111
3	0.116769547
4	0.104766804
5	0.091223629
6	0.078918038
7	0.068166764
8	0.058852758
9	0.05080065
10	0.043844137
11	0.037836136

imo,
it has been reported in the media that this problem is very difficult to calculate.
http://blogs.wsj.com/numbers/crunching-the-numbers-on-a-craps-record-703/
Some (including the Wizard of Odds who was in the news) used a simulation and some a Markov chain
Sally just used simple multiply and addition with the help of a spreadsheet
the world still has a long way to go to get to easy street for everyone

Sally

I refused to answer because, stick change, waitress at the table and somebody spoke to the shooter are not on the choices.

Best Craps Roll Ever Best

^ LMAO ^
Yeah .. the real answer is NEVER .. there is no roll you're ever more likely to roll a 7 .. each and every roll you only have a 1 in 6 chance of rolling 7 .. and a 5 in 6 chance of not rolling a 7! ;)
As for the answer of 'how many rolls' until you roll a seven. The 'average' would 6 .. since you always have a 1 in 6 chance. But obviously some of those rolls are forgotten because not too may people count 2,3,11,12 when there is a point and you're not playing Come bets or props. Then you also need to factor in '7 winners' on come out rolls after the previous roll was won (this is where you ask the Wizard to do his thing .. lol). And bla bla bla ... at the end of the day each roll you roll has a 1 in 6 chance of getting 7! Unless as Ontario Dealer forgot to mention .. it's a full moon on a Friday between 10 and 11 (am or pm doesn't really matter at this point)! ;)

that should be 5448 / 46656 = 11.67695473251%
higher than roll 2 at 11.11%
Sally

Okay, that is 0.56 %.
So if I am out on the Strip panhandling so I can come in and get another twenty dollar buy-in, that 0.56 percent means twelve minutes less panhandling time.I am both of those people. When I was in Biloxi last month, I decided to prove it. I played with a full table but passed on the dice each time. Did ok. Went and did other things. Several hours later, passed the same table, full crew, no players. So I talked to them some about playing the don'ts, because they were available and I'm trying to understand a few things about the game. They were politely incredulous about my (admittedly sparse) throwing claims as a PSO loser. So I threw 12 PSO's in a row. No kidding. Interspersed were 2 come-out 7's and 3 craps, but every time I set a point my next throw was a 7. Even threw one from behind my back trying to change it up.
Ask Al, the PB from Harrah's, if you don't believe me. But I made sure I was the only one at the table so I wouldn't cost anybody else any money. I tipped them well for the tutorial, but at the end, they said, 'um, yeah, you're a darksider'. I should be proud, I guess.
BTW, somebody was looking for crapless craps. Harrah's Gulf Coast has it, and it's usually open.