justin5689
Joined: May 5, 2012
• Posts: 3
May 5th, 2012 at 5:17:27 PM permalink
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
Joined: Mar 28, 2010
• Posts: 2001
May 5th, 2012 at 5:32:19 PM permalink
oops did math for exactly 3.

1-((.507)^3+3((.493)(.507^2))+3((.493^2)(.507)))
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 5th, 2012 at 5:48:21 PM permalink
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))

at least in a row then loseProb1 in
20.2429792884.1
30.1197716098.3
40.05903893416.9
50.0291020234.4
60.01434523869.7
70.007071188141.4
80.003485596286.9
90.001718152582.0
100.0008469281,180.7
110.0004174752,395.4
120.0002057864,859.4
130.0001014389,858.3
145.00017E-0519,999.3
152.46473E-0540,572.4
161.21494E-0582,308.7
175.98878E-06166,978.8
182.95205E-06338,748.0
191.45515E-06687,214.1
207.17286E-071,394,143.4
I Heart Vi Hart
Doc
Joined: Feb 27, 2010
• Posts: 6935
May 5th, 2012 at 7:33:56 PM permalink
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?
7craps
Joined: Jan 23, 2010
• Posts: 1975
May 5th, 2012 at 11:12:19 PM permalink
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.

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
winsome johnny (not Win some johnny)
justin5689
Joined: May 5, 2012
• Posts: 3
May 6th, 2012 at 9:40:08 PM permalink
Quote: 7craps

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?