How is Average # of rolls calculated in Craps ?

Hi all,
in craps, after a number has come out, how to calculate the average number of rolls before a 7 or the number is rolled out? i like to know the formula.

thanks.

point is 4 or 10
the game ends when the point number or a 7 is rolled.

9/36 = that probability (p)
for the expected number of rolls = 1/p

so 1 / 9/36 = 1 * 36/9 = 4

the distribution looks like this
from (1-p)^N * p (on roll N)

roll	by roll N	on roll N
1	0.25	0.25
2	0.4375	0.1875
3	0.578125	0.140625
4	0.68359375	0.10546875
5	0.762695313	0.079101563
6	0.822021484	0.059326172
7	0.866516113	0.044494629
8	0.899887085	0.033370972
9	0.924915314	0.025028229
10	0.943686485	0.018771172
11	0.957764864	0.014078379
12	0.968323648	0.010558784
13	0.976242736	0.007919088
14	0.982182052	0.005939316
15	0.986636539	0.004454487
16	0.989977404	0.003340865
17	0.992483053	0.002505649
18	0.99436229	0.001879237
19	0.995771717	0.001409428
20	0.996828788	0.001057071
21	0.997621591	0.000792803
22	0.998216193	0.000594602
23	0.998662145	0.000445952
24	0.998996609	0.000334464
25	0.999247457	0.000250848
26	0.999435592	0.000188136
27	0.999576694	0.000141102
28	0.999682521	0.000105826
29	0.999761891	7.93698E-05
30	0.999821418	5.95274E-05

5 or 9 point
p=10/36
average = 36/10 = 3.6
what is that .6?

the distribution

roll	by roll N	on roll N
1	0.277777778	0.277777778
2	0.478395062	0.200617284
3	0.623285322	0.144890261
4	0.727928288	0.104642966
5	0.803503764	0.075575475
6	0.858086052	0.054582288
7	0.897506593	0.039420541
8	0.925976984	0.028470391
9	0.946538933	0.020561949
10	0.961389229	0.014850296
11	0.972114443	0.010725214
12	0.979860431	0.007745988
13	0.985454756	0.005594325
14	0.989495101	0.004040346
15	0.992413129	0.002918027
16	0.994520593	0.002107464
17	0.996042651	0.001522057
18	0.997141914	0.001099264
19	0.997935827	0.000793913
20	0.998509208	0.000573381
21	0.998923317	0.000414109
22	0.999222396	0.000299079
23	0.999438397	0.000216001
24	0.999594398	0.000156001
25	0.999707065	0.000112667
26	0.999788436	8.13708E-05
27	0.999847204	5.87678E-05
28	0.999889647	4.24434E-05
29	0.999920301	3.06536E-05
30	0.999942439	2.21387E-05

6 or 8 point
p=11/36
average = 36/11 = 3 and 3/11
what is that?

the distribution

roll	by roll N	on roll N
1	0.305555556	0.305555556
2	0.517746914	0.212191358
3	0.665102023	0.14735511
4	0.767431961	0.102329937
5	0.838494417	0.071062456
6	0.887843345	0.049348928
7	0.922113434	0.034270089
8	0.945912107	0.023798673
9	0.962438963	0.016526856
10	0.973915947	0.011476983
11	0.981886074	0.007970127
12	0.987420885	0.005534811
13	0.991264503	0.003843619
14	0.993933683	0.00266918
15	0.99578728	0.001853597
16	0.9970745	0.00128722
17	0.997968403	0.000893903
18	0.998589169	0.000620766
19	0.999020256	0.000431087
20	0.999319622	0.000299366
21	0.999527515	0.000207893
22	0.999671886	0.00014437
23	0.999772143	0.000100257
24	0.999841766	6.9623E-05
25	0.999890115	4.83493E-05
26	0.999923691	3.35759E-05
27	0.999947008	2.33166E-05
28	0.9999632	1.61921E-05
29	0.999974444	1.12445E-05
30	0.999982253	7.80869E-06

Sally

Like Sally said, it depends on the point.

If it is 4 or 10, one of two things happens on each roll: either you roll your point or a 7 (probability 3/36 + 6/36 = 1/4) and stop, or you roll something else (probability 1 - 1/4 = 3/4) and keep rolling.
You will require N rolls with probability (1/4 + (N-1) x 3/4), since your last roll (point or 7) has probability 1/4, and the other N-1 rolls have probability 3/4.
The "average" is the sum of (1 x 1/4) + (2 x 3/4 x 1/4) + (3 x (3/4)² x 1/4) + (4 x (3/4)³ x 1/4) + ...
= 1/4 x (1 + 2 x 3/4 + 3 x (3/4)² + 4 x (3/4)³ + ...)
= 1/4 x ( (1 + 3/4 + (3/4)² + (3/4)³ + ...)²
= 1/4 x (1 / (1 - 1/4))²
= 1/4 x 4² = 4

In general terms, let P be the probability of rolling the point or 7; obviously (1-P) is the probability of rolling something else.
The average = (1 x P) + (2 x (1-P) x P) + (3 x (1-P)² x P) + (4 x (1-P)³ x P) + ...
= P x (1 + 2 (1-P) + 3 (1-P)² + 4 (1-P)³ + ...)
= P x (1 / (1-(1-P))²
= P x (1 / P)²
= 1 / P
For a point of 4 or 10, P = 1/4, so the average number of rolls is 4
For a point of 5 or 9, P = 4/36 + 6/36 = 5/18, so the average number of rolls is 18/5 = 3.6
For a point of 6 or 8, P = 5/36 + 6/36 = 11/36, so the average number of rolls is 36/11 = 3.272727

Quote: ilikevp

after a number has come out, how to calculate the average number of rolls before a 7 or the number is rolled out? i like to know the formula.

for "before any number has come out", not knowing what point number it will be
we would have to weight each average

4 = 36/9 * 3/24 = 108/216 = 0.5
10 is the same as 4
only 4 more to do
and add them up = 3.563636364
or 196/55

How about the average number of rolls until a point decision counting the come out roll.
we already have 196/55
and add to that 36/24 for the average number of rolls it takes to establish a point

that comes to 5.063636364
or 557/110

is this information useful?
Sally

Quote: mustangsally

point is 4 or 10
the game ends when the point number or a 7 is rolled.

9/36 = that probability (p)
for the expected number of rolls = 1/p
...
Sally

Why is "expected number of rolls = 1/p" ? Actually that is what i have been wandered.


Thanks.

Quote: ThatDonGuy

Like Sally said, it depends on the point.

...
In general terms, let P be the probability of rolling the point or 7; obviously (1-P) is the probability of rolling something else.
The average = (1 x P) + (2 x (1-P) x P) + (3 x (1-P)² x P) + (4 x (1-P)³ x P) + ...
= P x (1 + 2 (1-P) + 3 (1-P)² + 4 (1-P)³ + ...)
= P x (1 / (1-(1-P))²
= P x (1 / P)²
= 1 / P

Why is "The 'average' is the sum of (1 x P) + (2 x (1-P) x P) + (3 x (1-P)² x P) ..." ?
and, does this series have a former name?

Thanks

Quote: ilikevp

Why is "expected number of rolls = 1/p" ? Actually that is what i have been wandered.

Thanks.

from
http://www.mathsisfun.com/data/probability-events-independent.html


Probability of an event happening =
Number of ways it can happen DIVIDED BY
Total number of outcomes

we can easily invert this to
Total number of outcomes DIVIDED BY
Number of ways it can happen
a very simple way to get an average
it is how independent events are

example
coin flip for first Head wait time.
p=1/2
average number of flips to see a Head = 1/ 1/2 or 1 * 2/1 = 2

another way for the average that Don first showed
it could happen on the first flip by 1*p
if not the first flip it could happen on the second flip. 1-p * p * 2
the 3rd flp = (1-p)^2 (2 failures in a row) * p * 3
it could happen on the 123,456th flip
it could happen way into the future. this is an infinite geometric series.
we can sum them all to see what number the series converges to
the first 17 flips
0.5 = 1*p
0.5 = 1-p * p * 2
0.375 = (1-p)^2 * p * 3
0.25 and so on with (1-p)^(N-1) * p * N
0.15625
0.09375
0.0546875
0.03125
0.017578125
0.009765625
0.005371094
0.002929688
0.001586914
0.000854492
0.000457764
0.000244141
0.0001297
this sums to 1.999855042
the first 30 flips = 1.99999997

sure looks like we are going towards 2 here
same a 1/p

the infinite geometric series (or wait time for a success) describes the probability distribution
more on that here
http://www.mathsisfun.com/algebra/sequences-sums-geometric.html

there still must be a reason you want to know this about rolls to a point decision
Sally

i was kinda interested in the math of craps, and happened to find that a sum of infinite series of p(roll out a push) equals the average # of rolls. i think it's not coincident but can't explain it, that's why i asked the question in the first place.

btw,
the 3rd flp = (1-p)^2 (2 failures in a row) * p * 3
where does '* 3' come from ?



thanks