Most don't pass/don't come losses without a 7without a 7
July 2nd, 2016 at 12:02:33 AM
permalink
I've seen 2 or 3 monster rolls in my life, but there were always a few 7s mixed in on the come out rolls. As a fan of the don't pass/don't come with odds it got me thinking -- what is the record for most don't pass/don't come losses with odds without ANY 7 hitting (and a loss on the 11 is irrelevant)?
There are 24 ways to hit the 4,5,6,8,9,10 and 6 ways to roll a 7 so one has a 80% chance of hitting a box number before a 7. But it's possible to have a point and the other 5 numbers rolled on the don't come before a loss is forced, so how to calculate from there?
I guess I'm trying to figure out what are the odds of losing say 40 straight don't pass/don't come decisions with laying odds before a 7 hits are. Can't imagine it realistically goes much higher than 40 place numbers in real life very often.
There are 24 ways to hit the 4,5,6,8,9,10 and 6 ways to roll a 7 so one has a 80% chance of hitting a box number before a 7. But it's possible to have a point and the other 5 numbers rolled on the don't come before a loss is forced, so how to calculate from there?
I guess I'm trying to figure out what are the odds of losing say 40 straight don't pass/don't come decisions with laying odds before a 7 hits are. Can't imagine it realistically goes much higher than 40 place numbers in real life very often.
Last edited by: BigJohn22 on Jul 2, 2016
July 2nd, 2016 at 2:48:43 AM
permalink
Ignoring the come out wins/losses......
What's the probability a 4 becomes the point? 5? 6? 8, 9, or 10? Well, 24 ways for a box number (3 4's, 4 5's, 5 6's, 5 8's, 4 9's, 3 10's). And what's the probability of a 7 out before that point is established?
4: 3/24 to be established, then 1/3 to win (hit) and 2/3 to lose (7-out). Probability to lose on a 4, including probability of setting the 4, is (3/24)*(1/3) = (3/72) = (1/24)
5: 4/24 = 1/6. Then 4/10 or 2/5 to win. So (1/6)*(2/5) = 2/30 = 1/15.
6: 5/24 to establish. 5/11 to win. (5/24)*(5/11) = 25/264.
same for 8/9/10, so multiply all by 2. And u get:
4/10: 2//24 = 1/12
5/9: 2/15
6/8: 50/264
Add 'em all up and you get 0.406060606...
Which makes sense. You have a 33% chance to lose on 4 or 10, but those are rather rare. 40% chance to lose on 5/9. And 45% chance to lose on 6/8, which are more frequent.
If you wanna lose X in a row, then it's gonna be (0.4060606)^X.
40 in a row is going to be (0.406060606)^40 = 2.2061464e-16 = 1 in 4.5327907e+15 ~= 1 in 4,500,000,000,000,000.
I think I did all that math right. I'm sure Must'y will come in with her fancy X-Cell (lulz because [be-cuz] it's spelled Excel, teehee) graphs and stuff.
What's the probability a 4 becomes the point? 5? 6? 8, 9, or 10? Well, 24 ways for a box number (3 4's, 4 5's, 5 6's, 5 8's, 4 9's, 3 10's). And what's the probability of a 7 out before that point is established?
4: 3/24 to be established, then 1/3 to win (hit) and 2/3 to lose (7-out). Probability to lose on a 4, including probability of setting the 4, is (3/24)*(1/3) = (3/72) = (1/24)
5: 4/24 = 1/6. Then 4/10 or 2/5 to win. So (1/6)*(2/5) = 2/30 = 1/15.
6: 5/24 to establish. 5/11 to win. (5/24)*(5/11) = 25/264.
same for 8/9/10, so multiply all by 2. And u get:
4/10: 2//24 = 1/12
5/9: 2/15
6/8: 50/264
Add 'em all up and you get 0.406060606...
Which makes sense. You have a 33% chance to lose on 4 or 10, but those are rather rare. 40% chance to lose on 5/9. And 45% chance to lose on 6/8, which are more frequent.
If you wanna lose X in a row, then it's gonna be (0.4060606)^X.
40 in a row is going to be (0.406060606)^40 = 2.2061464e-16 = 1 in 4.5327907e+15 ~= 1 in 4,500,000,000,000,000.
I think I did all that math right. I'm sure Must'y will come in with her fancy X-Cell (lulz because [be-cuz] it's spelled Excel, teehee) graphs and stuff.
July 2nd, 2016 at 9:44:24 PM
permalink
The odds of hitting a 4,5,6,8,9 or 10 before a 7 are 24/36= 0.8
The odds of hitting 12 of these in a row before a 7 are 0.8^12= 0.54976 or 1 in 20.
Yet the odds of losing 6 don't pass/don't come's in a row before a 7 are 0.4060606^6 = 0.00448 or 1 in 223
Once the 4,5,6,8,9,10 fill up on the don't come, every 4,5,6,8.9.10 from then on is a loss on the come out roll for the one of the previous numbers.
So, shouldn't the 0.54976 number be the same as the 0.00448 number?
Does there need to be some way to factor in the come out roll for the previous bets? Where have I gone wrong?
The odds of hitting 12 of these in a row before a 7 are 0.8^12= 0.54976 or 1 in 20.
Yet the odds of losing 6 don't pass/don't come's in a row before a 7 are 0.4060606^6 = 0.00448 or 1 in 223
Once the 4,5,6,8,9,10 fill up on the don't come, every 4,5,6,8.9.10 from then on is a loss on the come out roll for the one of the previous numbers.
So, shouldn't the 0.54976 number be the same as the 0.00448 number?
Does there need to be some way to factor in the come out roll for the previous bets? Where have I gone wrong?
