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-Whats the Probability of hitting 5 *consecutive* Passlines?

I also made $$$ when I won 9 Don't Passes in a row. (pressing 1/3 after each win)

-Whats the Probability of winning 9 Don't Pass bets in a row?

edit:

after the point is set.

Odds of shooter making 4/10:

- Throws 4/10 on come out (3/36)

- Throws 4/10 again (3/36)

= .0833 * .0833 = .0069

Odds of shooter making 5/9:

- Throws 5/9 on come out (4/36)

- Throws 5/9 again (4/36)

= .1111 * .1111= .0123

Odds of shooter making 6/8:

- Throws 6/8 on come out (5/36)

- Throws 6/8 again (5/36)

= .1389 * .1389= .0193

Odds of shooter not making 4/10:

- Throws 4/10 on come out (3/36)

- Throws 7 (6/36)

= .0833 * .1667 = .0139

Odds of shooter not making 5/9:

- Throws 5/9 on come out (4/36)

- Throws 7 (6/36)

= .1111 * .1667= .0185

Odds of shooter not making 6/8:

- Throws 6/8 on come out (5/36)

- Throws 7 (6/36)

= .1389 * .1667= .0232

hitting Five 5/9's = .0123 ^ 5 = 2.81531E-10?

7 out on Nine 5/9's = .0185 ^ 9 = 2.53832E-16?

wow.. I really defied the odds with winning 9 Don't Passes in a row!

1/6 (7 on the come out)

+ 1/18 (11 on the come out)

+ (2 x 1/12 x 1/3) (4 or 10 on the come out; a 7 is twice as likely as the point in this case, so you win 1/3 of the time)

+ (2 x 1/9 x 2/5) (5 or 9 on the come out; a 7 is 3/2 times as likely as the point, so you win 2/5 of the time)

+ (2 x 5/18 x 5/11) (6 or 8 on the come out; a 7 is 6/5 times as likely as the point, so you win 5/11 of the time)

= 244/495

Therefore, the probability of winning five in a row is (244/495)

^{5}, or about 1 in 34.36.

The probability of winning a Don't Pass Line bet is 1 - 244/495 - 1/36 (since a 12 (or a 2 in some places) on the come out is a push) = 949/1980, so the probability of winning nine in a row is (949/1980)

^{11}, or about 1 in 3261.

Those of you who like to use Martingale or other "bet against a streak" systems, please note: the probability of the 5th consecutive pass, or the 11th consecutive Don't Pass, did not increase (or decrease) because of the previous run of the same result.

I meant after the point is set and you have odds.

getting a 2,3,7,11,12 on the come out roll doesn't count.

Quote:100xOdds...I meant after the point is set and you have odds.

getting a 2,3,7,11,12 on the come out roll doesn't count.

I get that after a random point is established, the probability that it is made is 67/165. So, the probability of making at least five established points in a row is (67/165)^5 = 0.01104.

Quote:100xOddsThatDonGuy,

I meant after the point is set and you have odds.

getting a 2,3,7,11,12 on the come out roll doesn't count.

In that case, the probabilities of the point being 4, 5, 6, 8, 9, or 10 are 1/8, 1/6, 5/24, 5/24, 1/6, and 1/8, respectively, and the probability of making a point is (2 x 1/8 x 1/3) + (2 x 1/6 x 2/5) + (2 x 5/24 x 5/11) = 67/165, so the probability of making five points in a row is (67/165)

^{5}= about 1 in 90.6.

Also, the probability of sevening out is 1 - 67/165 = 98/165, and the probability of sevening out 9 times in a row is (98/165)

^{9}= about 1 in 108.7.

i really doubt the 1st 5 points established wonQuote:100xOddsGot killed playing Don't Pass when the shooter hit 5 points in a row.

-Whats the Probability of hitting 5 *consecutive* Passlines?

I also made $$$ when I won 9 Don't Passes in a row. (pressing 1/3 after each win)

-Whats the Probability of winning 9 Don't Pass bets in a row?

edit:

after the point is set.

and

the first 9 points established lost

in other words

how many points were established during your play?

size matters!

the more trials attempted, the greater the prob of seeing at least 1 such run

this is where a streak calc comes in real handy dandy too (2)

i show this in me Excel

for 5 point winners in a row, it gets easier just doubling the number of attempts (points established)

for 9 points shot right on down in a row

same thing

easier the more attempts

of course when one does NOT know what to expect, it all looks like a rare event just happened

some call that lucky

Sally

Quote:ThatDonGuyIn that case, the probabilities of the point being 4, 5, 6, 8, 9, or 10 are 1/8, 1/6, 5/24, 5/24, 1/6, and 1/8, respectively, and the probability of making a point is (2 x 1/8 x 1/3) + (2 x 1/6 x 2/5) + (2 x 5/24 x 5/11) = 67/165, so the probability of making five points in a row is (67/165)

^{5}= about 1 in 90.6.

Also, the probability of sevening out is 1 - 67/165 = 98/165, and the probability of sevening out 9 times in a row is (98/165)^{9}= about 1 in 108.7.

ahh.. so the probability is almost the same.

so how come Romes's Math is different?