Probability of winning 5 Passline bets? or 9 Don't Pass bets?

Got 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.

The exact odds would depend on the actual points made... but for generics (this assumes no 'winning' or 'losing' on come out 7's, etc... just what are the odds the shooter points each number):

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

for simplicity sake, the avg point is 5/9.

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!

The probability of winning a Pass Line bet is:

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.

ThatDonGuy,

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: 100xOdds

ThatDonGuy,

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.

Quote: 100xOdds

Got 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.

i really doubt the 1st 5 points established won

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: ThatDonGuy

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.

ahh.. so the probability is almost the same.

so how come Romes's Math is different?