Winning at craps is entirely about learning the true probabilities of certain dice rolls occurring and making sure a player has money on the rolls with the best probability. The surest way to win in craps is to maximize bets with good odds and avoid bets with very poor odds. Most ask the probability of winning 3 pass line bets in a row. And for 3 trials it is simply p^3 OP wanted to add the probability of 3.or more. and.followed by a loss. Sally's math shows 3 in a row in 3 trials and the OPs Q arrives at the same value. Gives the best odds of winning among all casino games. As a matter of fact, as we will show in this article, among all the different bets one can place in the game of craps, the Pass Line bet regards one of the best bets due to its high winning probability. In this article, we will concentrate on deriving results for a Pass Line bet.

In the game of craps, two dice are rolled. If the sum is 2, 3, 7, 11, or 12, the game is over and the player either wins (in the case of 7 or 11) or loses (in the case of 2, 3, or 12). In any other case, the sum obtained becomes the player’s “point” and the game changes: Now, the player continues to roll two dice until the sum equals either 7 or the “point”. In the case of 7, the player loses; in the case of the “point”, the player wins; and in any other case, the player simply rolls again and nothing changes.
Find:
(a) P (player wins | the “point” equals 4)
(b) P (player wins | the “point” equals 5)
(c) P (player wins | the “point” equals 6)
Finally, (d) show that the probability that a player wins a game of craps is exactly 244/495.
I'm sure I can figure out (b) and (c) after I figure out how to do part (a). I know that the chance of rolling a 4 would be 1/12, but I'm pretty sure there has to be more to the probability than that. I broke it up into:
P (player wins INTERSECT point equals 4) / P (point equals 4)
but I'm not sure what to do with the top.

justin5689

What is the probability of winning 3 or more pass line bets in a row before losing your pass line bet from either a 2,3, or 12 on the come out roll or with a 7 after the point has been established?
The 3 pass line wins can be any combination of:
Come out win, Come out win, Come out win
Come out win, Come out win, Point win
Come out win, Point win, Come out win
Come out win, Point win, Point win
Point win, Come out win, Come out win
Point win, Come out win, Point win
Point win, Point win, Come out win
Point win, Point win, Point win

rudeboyoi

oops did math for exactly 3.
i believe its about:
1-((.507)^3+3((.493)(.507^2))+3((.493^2)(.507)))

mustangsally

pass line win probability = 244/495
(244/495)^3 = 0.119771609 exactly 3
you want 3 or more

Probability Of Winning Craps

formula for the sum of a geometric series is a/1-r
where a is the first term
r = the ratio

Probability Of Winning At Craps

a = (244/495)^3
r = 244/495
so (244/495)^3 / (1-(244/495) = 0.236202973
But now you want this to end. Multiply the above result by 1-(244/495)
0.119771609

Probability Of Winning Craps On First Roll

((244/495)^3 / (1-(244/495)) * (1-(244/495))
added my table

Probability Of Winning Craps

at least in a row then lose	Prob	1 in
2	0.242979288	4.1
3	0.119771609	8.3
4	0.059038934	16.9
5	0.02910202	34.4
6	0.014345238	69.7
7	0.007071188	141.4
8	0.003485596	286.9
9	0.001718152	582.0
10	0.000846928	1,180.7
11	0.000417475	2,395.4
12	0.000205786	4,859.4
13	0.000101438	9,858.3
14	5.00017E-05	19,999.3
15	2.46473E-05	40,572.4
16	1.21494E-05	82,308.7
17	5.98878E-06	166,978.8
18	2.95205E-06	338,748.0
19	1.45515E-06	687,214.1
20	7.17286E-07	1,394,143.4

Doc

pass line win probability = 244/495
(244/495)^3 = 0.119771609 exactly 3

Mustangsally: Maybe I'm missing something, but I get different results from what you show. Suppose (for simplicity) that it was a coin flip with p=0.5 instead of 244/495. To get at least one win in a row would be P=0.5. To get at least two in a row is P=0.5^2, etc. To get the answer for exactly n in a row, you need to multiply the 'at least' by the probability of losing on the n+1 try.
For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?

7craps

For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?

rudeboyoi and Sally both arrived at the same value.
Sally did it differently by starting with 3 pass line wins in a row in 3 trials.
The OP asked a unique question.
Most ask the probability of winning 3 pass line bets in a row. And for 3 trials it is simply p^3
OP wanted to add the probability of 3 *or more* and *followed by a loss*.
Sally's math shows 3 in a row in 3 trials and the OPs Q arrives at the same value. It should.
IF the OP had asked what is the probability of winning 3 pass line bets in a row then losing, we would have p^3 * q or 0.06073
Let us see if OP is happy and replies.
added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
Craps

5 or more: 67.765
6 or more: 137.475
Multiple streaks of pass line winners.
15 trials about 30 minutes of play at 100 rolls per hour
30 trials about 1 hour of play
Example: 30 pass line trials
about a 90% chance of at least 3 pass line wins in a row at least one time
about a 58% chance of at least 3 pass line wins in a row at least two times
about a 23% chance of at least 3 pass line wins in a row at least three times
here is the losing streaks (miss) for the pass line per N trials

winsome johnny (not Win some johnny)

justin5689

added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
5 or more: 67.765
6 or more: 137.475

You guys are great. Thanks for the detailed responses.
How did you calculate the average number of trials to see a run of 3, 4, 5, 6 or more pass line wins?
How do you define a trial? Would each shooter be a new trial? Or does a new trial begin after any losing pass line bet, in which case a single shooter could have multiple trials that end and start over with a losing pass line bet from throwing 2, 3, or 12 on a come out roll?

Probability Of Winning A Field Bet In Craps

If a bettor were to power press the pass line bet with a $100 wager:

at least in a row then lose	Prob	1 in	Bet	Win	Lose
1	0.492929293	2.0	$100.00	$100.00	$100.00
2	0.242979288	4.1	$200.00	$300.00	$100.00
3	0.119771609	8.3	$400.00	$700.00	$100.00
4	0.059038934	16.9	$800.00	$1,500.00	$100.00
5	0.02910202	34.4	$1,600.00	$3,100.00	$100.00
6	0.014345238	69.7	$3,200.00	$6,300.00	$100.00

Does this mean that on average you would be betting about $1600 to win $700 on a press of 3 pass line wins for a net loss of $900?
$3,300 to win $1,500 on 4 pass line wins losing $1,800?
$6,700 to win $3,100 on 5 p