Chance Of Winning Craps

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If you have noticed 3 non-field bets pass in a row and then you decide to bet on the field, your chances of winning will go up. Under these circumstances, your chances of losing are reduced to 9.5%. Part of a winning strategy for online craps is knowing what bets to avoid because they offer low odds. Look for the proposition area of the craps. 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.

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))

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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

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?

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

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

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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?

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$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?

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