All 6 point numbers before a 7?
Do you mean before ANY 7, or just before a 7-out?
Is it OK to hit numbers repeatedly? Is it OK to hit 2, 3, 11, 12?
Quote: WizardHe probably means winning the Fire Bet, which has been discussed before. I indicate the answer in my blackjack appendix 4.
craps Wiz
( ( 3/36 ) * ( 4/36 ) * ( 5/36 ) * ( 5/36 ) * ( 4/36 ) * ( 3/36 ) ) / ( 6/36 ) = 0.0000099229
My brute force method gets 0.062168. Is there a mathematical way that's easier.
Any one point number before a 7 is 24/30 = .8
Any two different point numbers before a 7 is .614017
Any three different point numbers before a 7 is .443960
Any four different point numbers before a 7 is .292429
Any five different point numbers before a 7 is .163184
4 - 1/3
5 - 2/5
6 - 5/11
8 - 5/11
9 - 2/5
10 - 1/3
The odds of hitting them all before a 7 would be the product of those probabilities = 0.3673%
If you think about a pyramid (D4), what is the probability of hitting a 1, 2, and 3 before a 4? The odds of hitting x before 4 is:
1: 1/2
2: 1/2
3: 1/2
Is the odds therefore of hitting 1, 2 and 3 before a 4: 1/8.
No. The experiment starts when a 1, 2, 3 is rolled.
What are the odds, that a different one of the 3 numbers is thrown before a 4. Well it's 2/3rds as there are only three other possibilities as the original number thrown is discarded.
So the odds of 1-2, 1-3, or 2-3 are 1/3 x 2/3 + 1/3 x 2/3 + 1/3 x 2/3 = 6/9
What are the odds then of getting the 3rd digit before the 4? Well it's 1 half as there is one way of getting the 3rd number and 1 way of getting the 4.
6/9 x 1/2 = 6/18 = 1/3.
So, the odds of hitting the 1,2, and 3 before the 4 is 1/3, not 1/8th.
24/30 (6 stand offs,6 losses, 24 wins)
x 21/27 (9 stand offs,6 losses, 21 wins)
x 17/23 (13 stand offs,6 losses, 17 wins)
x 12/18 (18 stand offs, 6 losses, 12 wins)
x 7/13 (23 stand offs, 6 losses, 7 wins)
x 3/9 (27 stand offs, 6 losses, 3 wins) = 1/18.171519 or 17.17 to 1
But if the order is 6,8,5,9,4,10 the answer becomes:
24/30
x 19/25
x 14/20
x 10/16
x 6/12
x 3/9 =1/22.556391 or 21.57 to 1
Am I missing something or is it just a matter of taking the average of all the combinations?
The Grand Victoria Casino in Elgin, Illinois offers a promotion called "Craps for Cash." A shooter wins a $4,000 cash bonus for making all six points on the same hand. All that's required is a bet on the pass line. How does this affect the house edge on this particular game? – John B. from Riverside, Illinois
The probability of making all six points is 1 in 6156. So the value of this to the shooter is $4000/6156 = 64.98 cents. As long as the shooter bets $45 or less on the pass line, and nothing else except the odds, he will have an advantage. April 12, 2007
If a man wants to bet that he can shoot all 6 points before a 7 what single price could he accept to know whether hes getting the best or the worst of the proposition?
There must be a mathematical proof!
Quote: boymimboI kind of thought of that approach but don't agree with it. The odds of throwing a 7 is always 1/6. The odds of throwing n before 7 is expressed as n/(n+6) where n is the number of dice probabilties.
If you think about a pyramid (D4), what is the probability of hitting a 1, 2, and 3 before a 4? The odds of hitting x before 4 is:
1: 1/2
2: 1/2
3: 1/2
Is the odds therefore of hitting 1, 2 and 3 before a 4: 1/8.
No. The experiment starts when a 1, 2, 3 is rolled.
What are the odds, that a different one of the 3 numbers is thrown before a 4. Well it's 2/3rds as there are only three other possibilities as the original number thrown is discarded.
So the odds of 1-2, 1-3, or 2-3 are 1/3 x 2/3 + 1/3 x 2/3 + 1/3 x 2/3 = 6/9
What are the odds then of getting the 3rd digit before the 4? Well it's 1 half as there is one way of getting the 3rd number and 1 way of getting the 4.
6/9 x 1/2 = 6/18 = 1/3.
So, the odds of hitting the 1,2, and 3 before the 4 is 1/3, not 1/8th.
Yeah you're right, although I disagree that the experiment starts after the 1st roll. I would say it starts immediately (since it's possible to 7 before hitting any numbers).
So for your example of (D4), it would be P(1st)*P(2nd)*P(3rd) = (3/4)*(2/3)*(1/2) = 1/4
Applying this to the problem at hand gets very complicated, because each number's probability differs, and isn't evenly distributed.
Gee, all SamsTown in Shreveport, Louisiana offers is a personalized Dice Sharks jacket if you make all six numbers before sevening out. I think they embroider your name on the jacket. I hear its a pretty nifty jacket too, but I'd prefer the four grand!!Quote: matildaThe Grand Victoria Casino in Elgin, Illinois offers $4,000 cash for making all six points on the same hand. All that's required is a bet on the pass line.
a) what are the odds of making all 6 points without 7-out and passing the dice (the fire bet), noting you are allowed to throw a 7 on the come out
b) what are the odds of rolling a 4,5,6,8,9,10 before a 7. Forget the rules of craps.
I personally think he wanted b.
In which case, I need one more piece of info: You make the prop bet, what happens if you roll a 7 before any of the other #s? no action, or lose?
Quote: dwheatleyI think it's still not clear whether the OP wanted to know:
a) what are the odds of making all 6 points without 7-out and passing the dice (the fire bet), noting you are allowed to throw a 7 on the come out
b) what are the odds of rolling a 4,5,6,8,9,10 before a 7. Forget the rules of craps.
I personally think he wanted b.
In which case, I need one more piece of info: You make the prop bet, what happens if you roll a 7 before any of the other #s? no action, or lose?
I actually envisioned a 3rd scenario - as if he was just betting across, and wanted to bink each bet before 7-out. That's why I felt the experiment starts right away, because they're typically not "on" until a point is established, at which time a 7 wipes you out (and you still have bets on all 6 numbers to hit).
I think it's clear that the OP's question is unclear.
Quote: dwheatleyI think it's still not clear whether the OP wanted to know:
a) what are the odds of making all 6 points without 7-out and passing the dice (the fire bet), noting you are allowed to throw a 7 on the come out
b) what are the odds of rolling a 4,5,6,8,9,10 before a 7. Forget the rules of craps.
I personally think he wanted b.
In which case, I need one more piece of info: You make the prop bet, what happens if you roll a 7 before any of the other #s? no action, or lose?
Thankyou, yes I did mean b. Nothing to do with pass lines, come outs, house edge or anything. A 7 before any of the point numbers is a lose!
Quote: steve2929Thankyou, yes I did mean b. Nothing to do with pass lines, come outs, house edge or anything. A 7 before any of the point numbers is a lose!
This is a variation of the "All Bet" that can be found at a few craps tables, where rolling all the numbers B4 a 7 wins.
This also sounds like a variation of the coupon collector problem. With Different probabilities for each coupon and we start over collecting when a 7 rolls.
There are a few scripts and a math example that can calculate the exact odds.
HERE at 2+2 forum
Or a simulation can also come very close.
I know the answer but will let you solve it first your way or using an idea from the above link.
Good Luck
