May 8th, 2017 at 11:03:32 AM
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Quote:iamnomad3 ways to roll a 4: 1/3 3/1 2/2

2 ways to roll a 3: 1/2 2/1

This was in reply to the table the Wizard posted of inside number probabilities. According to his table, rolling 4 inside numbers is more probable than rolling 3 inside numbers.

America is all about speed. Hot, nasty, bad-ass speed. - Eleanor Roosevelt, 1936

May 8th, 2017 at 11:31:13 AM
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Assuming that the question is, "From the moment a shooter is handed the dice until he sevens out, and ignoring all comeout rolls, including those after successful points, what is the average number of times that an inside number will be rolled?"

Since comeout rolls are ignored, the only numbers that matter are 5, 6, 7, 8, and 9.

In any given roll, there is a probability 1/2 of rolling an inside number and 1/6 of rolling a 7, so if we are limited to just those numbers, there is a probability 3/4 of rolling an inside number and 1/4 of a seven.

The average number of inside numbers rolled is:

(0 x 1/4) + (1 x 3/4 x 1/4) + (2 x (3/4)^{2} x 1/4) + (3 x (3/4)^{3} x 1/4) +

= 3/4 x 1/4 x (1 + 2 x 3/4 + 3 x (3/4)^{2} + ...)

= 3/16 x (1 + 3/4 + (3/4)^{2} + ...)^{2}

= 3/16 x (1 / (1/4))^{2}

= 3/16 x 16

= 3

Since comeout rolls are ignored, the only numbers that matter are 5, 6, 7, 8, and 9.

In any given roll, there is a probability 1/2 of rolling an inside number and 1/6 of rolling a 7, so if we are limited to just those numbers, there is a probability 3/4 of rolling an inside number and 1/4 of a seven.

The average number of inside numbers rolled is:

(0 x 1/4) + (1 x 3/4 x 1/4) + (2 x (3/4)

= 3/4 x 1/4 x (1 + 2 x 3/4 + 3 x (3/4)

= 3/16 x (1 + 3/4 + (3/4)

= 3/16 x (1 / (1/4))

= 3/16 x 16

= 3

May 8th, 2017 at 1:16:18 PM
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Quote:ThatDonGuyAssuming that the question is, "From the moment a shooter is handed the dice until he sevens out, and ignoring all comeout rolls, including those after successful points, what is the average number of times that an inside number will be rolled?"

Since comeout rolls are ignored, the only numbers that matter are 5, 6, 7, 8, and 9.

In any given roll, there is a probability 1/2 of rolling an inside number and 1/6 of rolling a 7, so if we are limited to just those numbers, there is a probability 3/4 of rolling an inside number and 1/4 of a seven.

The average number of inside numbers rolled is:

(0 x 1/4) + (1 x 3/4 x 1/4) + (2 x (3/4)^{2}x 1/4) + (3 x (3/4)^{3}x 1/4) +

= 3/4 x 1/4 x (1 + 2 x 3/4 + 3 x (3/4)^{2}+ ...)

= 3/16 x (1 + 3/4 + (3/4)^{2}+ ...)^{2}

= 3/16 x (1 / (1/4))^{2}

= 3/16 x 16

= 3

Don,

You answered my question. I appreciate it.

If you feel like answering one more question. How many inside numbers occur from roll number 6 to a seven out on average. You would not count roll number 5. Roll number 5 includes the come out roll. ThankU

Last edited by: RouletteProdigy on May 8, 2017

May 8th, 2017 at 2:42:35 PM
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Quote:AyecarumbaThis was in reply to the table the Wizard posted of inside number probabilities. According to his table, rolling 4 inside numbers is more probable than rolling 3 inside numbers.

AHHHH...gotcha. Wondered why you were asking. I should've figured it out from rest of thread.

May 8th, 2017 at 6:58:13 PM
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I get slightly different numbers than the wizard.

0 .250 000

1 .231 579

2 .208 835

3 .178 503

4 .131 083

Average # hits = 1.709 091

The problem solved is how many of the inside numbers will a shooter roll during any series of rolls which ends on either a 7 or all four of the inside numbers have been rolled.

For instance there are six ways to throw a 7 and eighteen to roll 5 6 8 or 9. So the chances of not throwing any are 6/(4+5+6+5+4) = 6/24 = 1/4.

For one roll is SUM( Pr (5 before 7) * Pr (7 before 689) + Pr (6 before 7) * pr (8 before 589) etc.)

It is obvious that the probability of getting 3 is larger than getting 4. Whilst on 3 you stay of 3 if you roll a 7 or get to 4 if you roll the last number. Thus, as 7 is more likely, the chances of staying on 3 is bigger than getting to 4.

0 .250 000

1 .231 579

2 .208 835

3 .178 503

4 .131 083

Average # hits = 1.709 091

The problem solved is how many of the inside numbers will a shooter roll during any series of rolls which ends on either a 7 or all four of the inside numbers have been rolled.

For instance there are six ways to throw a 7 and eighteen to roll 5 6 8 or 9. So the chances of not throwing any are 6/(4+5+6+5+4) = 6/24 = 1/4.

For one roll is SUM( Pr (5 before 7) * Pr (7 before 689) + Pr (6 before 7) * pr (8 before 589) etc.)

It is obvious that the probability of getting 3 is larger than getting 4. Whilst on 3 you stay of 3 if you roll a 7 or get to 4 if you roll the last number. Thus, as 7 is more likely, the chances of staying on 3 is bigger than getting to 4.

May 8th, 2017 at 7:07:01 PM
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Quote:RouletteProdigyIf you feel like answering one more question. How many inside numbers occur from roll number 6 to a seven out on average. You would not count roll number 5. Roll number 5 includes the come out roll.

Assuming once again that all come out rolls are ignored, the answer is still 3. The dice don't know if the first roll that is counted is the second roll or the sixth - or the 13th, for that matter. They just roll.

May 8th, 2017 at 7:27:19 PM
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My earlier answer is how many different numbers will be rolled before the 7.

The question above is how many of (5689) might one roll before throwing a 7. If we ignore comeout rolls, where a 7 is a winner for the shooter rather than a 7-out, then other rolls are merely trying to make the point or 7-out. Thus they can be considered as a series of rolls that ultimately end with a 7-out.

There are 18 ways to roll (568 or 9) and 6 ways to roll (7) - hence if you rolled the dice often enough there would be 3 times as many inside numbers as 7's. Thus on average you would roll 3 inside numbers before the 7.

Another way of looking at it is in (say) a million rolls where you throw (5678 or 9) (i.e. ignore the ones 2 3 11 12) there are (about) 250k 7's and 750k inside numbers. So consider dividing the million rolls at each point where there is a 7; this creates 250k series where the last roll of each is a 7. The average length of a series is 4 (1m/250k) and the last roll of each series is always a 7. Thus the average series has 3 inside numbers (4-1) before the 7.

The question above is how many of (5689) might one roll before throwing a 7. If we ignore comeout rolls, where a 7 is a winner for the shooter rather than a 7-out, then other rolls are merely trying to make the point or 7-out. Thus they can be considered as a series of rolls that ultimately end with a 7-out.

There are 18 ways to roll (568 or 9) and 6 ways to roll (7) - hence if you rolled the dice often enough there would be 3 times as many inside numbers as 7's. Thus on average you would roll 3 inside numbers before the 7.

Another way of looking at it is in (say) a million rolls where you throw (5678 or 9) (i.e. ignore the ones 2 3 11 12) there are (about) 250k 7's and 750k inside numbers. So consider dividing the million rolls at each point where there is a 7; this creates 250k series where the last roll of each is a 7. The average length of a series is 4 (1m/250k) and the last roll of each series is always a 7. Thus the average series has 3 inside numbers (4-1) before the 7.

May 9th, 2017 at 9:12:12 AM
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Quote:charliepatrick...Another way of looking at it is in (say) a million rolls where you throw (5678 or 9) (i.e. ignore the ones 2 3 11 12) there are (about) 250k 7's and 750k inside numbers. So consider dividing the million rolls at each point where there is a 7; this creates 250k series where the last roll of each is a 7. The average length of a series is 4 (1m/250k) and the last roll of each series is always a 7. Thus the average series has 3 inside numbers (4-1) before the 7.

This may be misleading because in reality, there are many other sequences of rolls that contain Horn/outside numbers with 0 or more inside numbers before a 7. Shouldn't this make the actual number of "post point inside numbers before a 7" much lower than 3?

America is all about speed. Hot, nasty, bad-ass speed. - Eleanor Roosevelt, 1936

May 9th, 2017 at 9:32:48 AM
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It was early but I failed to say that all 2 3 4 10 11 12's are irrelevant to this question. So the only numbers that we're interested in are 5 6 7 8 9. One of 5(4) 6(5) 8(5) 9(4) = (18 perms) will happen three times more often than 7(6) = (6 perms).

May 9th, 2017 at 10:56:48 AM
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Interesting question.Quote:RouletteProdigyHi All,

I was wondering if any of you could assist me here.

How many inside numbers does the average craps player roll before he 7 out's.

Could you please include the figure that includes the come out roll and excludes it.

I do not know what an average craps player is (next to possibly being a lifetime loser, imo)

so,

I setup Wincraps real fast

counting all rolls

I get an average of 4.26 rolls

(1 million shooters)

part of the distribution

0 0.057336057

1 0.231608232

2 0.147214147

3 0.122207122

4 0.095114095

5 0.074770075

6 0.058461058

7 0.045865046

8 0.036345036

9 0.028325028

10 0.022298022

11 0.017440017

12 0.013759014

13 0.01044601

counting all rolls EXCEPT 5689 on the come out roll.

I get an average of 3 rolls

part of the distribution

0 0.25000825

1 0.187191187

2 0.140664141

3 0.105567106

4 0.07958508

5 0.058822059

6 0.044466044

7 0.033534034

8 0.025041025

9 0.018961019

10 0.014040014

11 0.01043301

12 0.007785008

13 0.005991006

hope this helps out

this should come close to calculated values

of course

Sally

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