August 3rd, 2011 at 9:11:25 AM
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What are the odds of throwing all 6 point numbers before a 7? I can't find a proof, I get a different result depending on what sequence the numbers are rolled. need help

August 3rd, 2011 at 10:19:28 AM
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More info is needed.

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?

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?

Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?
Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition.

August 3rd, 2011 at 10:34:35 AM
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He probably means winning the Fire Bet, which has been discussed before. I indicate the answer in my blackjack appendix 4.

It's not whether you win or lose; it's whether or not you had a good bet.

August 3rd, 2011 at 10:51:20 AM
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yes, any 7. its a prop bet. 2,3,11,12 and repeats are stand offs.

August 3rd, 2011 at 10:53:38 AM
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Quote:WizardHe probably means winning the Fire Bet, which has been discussed before. I indicate the answer in my blackjack appendix 4.

craps Wiz

August 3rd, 2011 at 11:25:30 AM
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I'm sure I'm over-simplifying it, but I kinda think the answer is:

( ( 3/36 ) * ( 4/36 ) * ( 5/36 ) * ( 5/36 ) * ( 4/36 ) * ( 3/36 ) ) / ( 6/36 ) = 0.0000099229Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?
Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition.

( ( 3/36 ) * ( 4/36 ) * ( 5/36 ) * ( 5/36 ) * ( 4/36 ) * ( 3/36 ) ) / ( 6/36 ) = 0.0000099229

August 3rd, 2011 at 11:32:16 AM
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Assuming that the first roll includes a non-come out roll?

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

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

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You want the truth! You can't handle the truth!

August 3rd, 2011 at 1:16:46 PM
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The odds of hitting X before 7:

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%

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%

August 3rd, 2011 at 2:33:54 PM
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I 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.

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.

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You want the truth! You can't handle the truth!

August 3rd, 2011 at 3:08:23 PM
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If the point numbers appear in this order 4,5,6,8,9,10 I get this answer:

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?

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?