Pass Line Probability Question

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

oops did math for exactly 3.

i believe its about:

1-((.507)^3+3((.493)(.507^2))+3((.493^2)(.507)))

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

you want 3 or more
formula for the sum of a geometric series is a/1-r
where a is the first term
r = the ratio
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

((244/495)^3 / (1-(244/495)) * (1-(244/495))

added my table

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

Quote: mustangsally

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?

Quote: Doc

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
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

Quote: 7craps

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?

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 pass line wins losing $3,600?
$13,700 to win $6,300 on 6 pass line wins losing $7,400?

What's the best way to interpret this expected value for power pressing a pass line bet?

I would like to know the %probability of pass line wins of a series of 1 2,3,4,5 or 6 wins. Probability of 1 win is shown above as 49%. However, I would like to see the result of calculations where we ignore the "natural" wins on the comeout, The reason for my question is that box number place bets are normally OFF on a comeout and thus not vulnerable to a 7. So how long do I keep my place bets working after successive points are set? What is the risk of leaving them ON after 2 points have been made, or after 3 points have been made? I think for my purposes we should ignore the Craps numbers on the comeout roll as well. So essentially the question is: assume a point has been set. What is probability that shooter will make 1 or 2 or 3 or 4 or 5 or 6 points before finally SO ?

Quote: barigh

I would like to know the %probability of pass line wins of a series of 1 2,3,4,5 or 6 wins. Probability of 1 win is shown above as 49%. However, I would like to see the result of calculations where we ignore the "natural" wins on the comeout, The reason for my question is that box number place bets are normally OFF on a comeout and thus not vulnerable to a 7. So how long do I keep my place bets working after successive points are set? What is the risk of leaving them ON after 2 points have been made, or after 3 points have been made? I think for my purposes we should ignore the Craps numbers on the comeout roll as well. So essentially the question is: assume a point has been set. What is probability that shooter will make 1 or 2 or 3 or 4 or 5 or 6 points before finally SO ?
link to original post

If you assume a point has been set, the probability of making the point (without specifying what it is) is 1/4 x 1/3 + 1/3 x 2/5 + 5/12 x 5/11 = 67/165

The probability of making exactly N points and then sevening out is (67/165)^N x 98/165.
0 points made: 59.3939%
1 point made: 24.1175%
2 points made: 9.7932%
3 points made: 3.9766%
4 points made: 1.6148%
5 points made: 0.6557%
6 points made: 0.2662%
7 points made: 0.1081%

If you are looking for the probability of reaching a certain number of consecutive points without sevening out:
1 point: 40.6061%
2 points: 16.4885%
3 points: 6.6953%
4 points: 2.7187%
5 points: 1.104%
6 points: 0.4483%
7 points: 0.182%

I like MustangSally's Chart better, just skip the odds bets and do a progression on strictly the PL bets. I might add a progression on 1 or 3 come bets too. If I can hit a PL winner twice per shooter, I'll be on my way and ahead with the progression. But I had the misfortune to have a 50 to 70 roll turn the other night and only made a few PL hits while everybody else hit at least 4 dozen box numbers, so I've got nothing but FOMO. This was the type of night where I needed nearly 20 rolls to get a decision on a point of 6 or 8, while the sister number got clobbered. Then throw in the random craps 12 on the come-out rolls and it ends the PL progression I was working on, over and over again. That's why I like Place Bets, they don't lose per shooter and they can be progressed. But it's always a toss-up whether the PL hits better than a PB 6 or 8; and it's tempting to play all 3.