Craps Math Question

Hi 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.

The inside numbers are of course 5,6,8 and 9.

Thank U

P.S. The average player rolls a little over 8 times, so just wondering what percentage of those numbers are inside numbers more or less.

If the average number of rolls is 8 (rounding down) like you said and you took say 100,000 session, then the answer would be 3. Out of the 8 rolls you got atleast 1 come out and (1) 7 out roll. That would leave you with 6 rolls. If you wanted to count the come out roll it would be 3.5 inside numbers before the seven out.

The probably of rolling the inside numbers is 50% each roll.
(6,8 are 5/36 or .13888889)
(5,9 are 4/36 (1/9) or .1111111)

So by adding them all together you get 18/36 (1/2) or .5

Hope this helped.

P.s. if anyone spots an error in my math please let me know.

I get 1.619839371.

Probability of rolling
0: 0.250000
1: 0.231579
2: 0.298086
3: 0.089252
4: 0.131083

Quote: Wizard

I get 1.619839371.

Probability of rolling
0: 0.250000
1: 0.231579
2: 0.298086
3: 0.089252
4: 0.131083

Is it really easier to roll 4 than 3?

Wizard,

When it comes to math I am a bit thick headed.

If I understand you correctly, the average is 1.619839371 inside numbers after the come out?

Thank U

Quote: RouletteProdigy

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.

Question: at what point does the counting begin?

i.e. do you want to know the average number of inside numbers rolled from the time a shooter is passed the dice to when he sevens out?

For example, in this case:
point 6, 3, 12, 6
natural 11
point 5, 4, 7 out
Do you want the three comeouts counted separately, or combined (in which case, the number is 3 if you include the comeout rolls, or 1 if you do not)?

Quote: ThatDonGuy

Question: at what point does the counting begin?

i.e. do you want to know the average number of inside numbers rolled from the time a shooter is passed the dice to when he sevens out?

For example, in this case:
point 6, 3, 12, 6
natural 11
point 5, 4, 7 out
Do you want the three comeouts counted separately, or combined (in which case, the number is 3 if you include the comeout rolls, or 1 if you do not)?

Hi Don,

I would like to know the average number of inside numbers rolled after the point is established to when the average player sevens out.

Thank U

Quote: Ayecarumba

Is it really easier to roll 4 than 3?

3 ways to roll a 4: 1/3 3/1 2/2

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

Quote: RouletteProdigy

I would like to know the average number of inside numbers rolled after the point is established to when the average player sevens out.

I got that part. What I don't quite understand is, is it from when the shooter first establishes a point after taking the dice?

That is, what happens if a point is made after, for example, three inside numbers are rolled? Does the count continue from three, or does it reset to zero?

Quote: ThatDonGuy

I got that part. What I don't quite understand is, is it from when the shooter first establishes a point after taking the dice?

That is, what happens if a point is made after, for example, three inside numbers are rolled? Does the count continue from three, or does it reset to zero?

This is an important clarification, as I think the figure "8" for "average number of rolls per shooter" includes all the "non-point establishing" rolls. Consider this sequence of rolls, "New Shooter Coming Out!:
7
12
11
2
3
4 - Mark the 4
10
7 - Out! Line away

It's 8 rolls, but only one after the point is established. After a point is established, I thought the average number of rolls was 3 and change?