craps math question

After a random 100 points being established(or whatever number makes the math easier) What the % of points that will be passed? Is it somewhere around 40%?

Quote: vert1276

After a random 100 points being established(or whatever number makes the math easier) What the % of points that will be passed? Is it somewhere around 40%?

it is fair to say that folks here get weary from saying the past does not matter in craps. After xyz the odds are the as before xyz.

Quote: odiousgambit

it is fair to say that folks here get weary from saying the past does not matter in craps. After xyz the odds are the as before xyz.

I took the question to mean, out of 100 points established at random, how many are winners? So you'd have to figure how many 4s/5s/6s/8s/9s/10s out of 100, and then how many would win. You'd get more established points of 6 and 8, and those would be easier to convert, I'd say 40% is a good guess.

Quote: odiousgambit

it is fair to say that folks here get weary from saying the past does not matter in craps. After xyz the odds are the as before xyz.

No, I FULLY understand the past doesnt matter. I just want to know the math on whats my % on passing a random point?

For example I can just look at odds on a put bet and see my odds of passing on a ........

4,10 is 33%
5,9 is 40%
6,8 is 45%

But I cant just average those number together to see what the % is of passing a random point. Because the 6,8 point will be established more than the 4,10 point. I suppose I could do the math. I just wanted to know if someone new the % off the top of their head is all.

Quote: cclub79

I took the question to mean, out of 100 points established at random, how many are winners? So you'd have to figure how many 4s/5s/6s/8s/9s/10s out of 100, and then how many would win. You'd get more established points of 6 and 8, and those would be easier to convert, I'd say 40% is a good guess.

Ya, you got what I was saying. This really all steems from me just wonder about wincraps LOL. I just want to be able to look at the summary after 100 points are pass and see if im experiencing negative or positive variance. Of course this still wont tell me much because if I had a high number of 4,10 in relation to 6,8 points being established it really doesn't matter because that will throw off the numbers.

Quote: vert1276

What the % of points that will pass? Is it somewhere around 40%?

Yes. It is precisely 40.066 percent. at all times, in all circumstances, on all tables, even those that point North or have a Boxman that smiles. Its precisely 40.066 percent no matter how many times you want to "stop the clock" and add up all the points that have been established and count which ones the shooter did indeed make that point.

Will any one session be 40.066 percent? Heck no... how can you have .066 percent of a roll? And if you arbitrarily start counting at one particular roll, the very next one may meet your criterion so that would be 100 percent if you stop counting right then and there.

But the probability is that if you happen to stand there guzzling free booze and enjoying the "scenery" whenever you stop and look at your notes, its gonna be somewhere around 40.066 percent. If you take larger and larger numbers you will get closer and closer to that 40.066 percent.

If you look at the figures you will undoubtedly see some sort of pattern, be sure to tell the dice what that pattern is and they may thank you for reminding them. However, do not speak rudely to the dice since that might make them abandon their pattern just to spite you.

But the figure is 40.066 percent. It don't change.

Quote: vert1276

Ya, you got what I was saying. This really all steems from me just wonder about wincraps LOL. I just want to be able to look at the summary after 100 points are pass and see if im experiencing negative or positive variance. Of course this still wont tell me much because if I had a high number of 4,10 in relation to 6,8 points being established it really doesn't matter because that will throw off the numbers.

The other posts did a good job of telling you what you need to know. How about some formulas so you can figure it out by yourself if you so desire.

The question you ask is a binomial distribution. Since there is only success or failure as a single outcome.

100 trials = n
0.406061 = p (exact is 201/495) probability of a point winning.
q = 1-p

answer for the mean or average per 100 trials is (n*p)
standard deviation = sqrt(n*p*q) or (n*p*q)^.5
I will let you do the math.

Here is a distribution table for 100 points established.
Interesting how close the probabilities of 40 and 41 point successes are.

Point wins	Exact prob	or less
11	0.000000010%	0.000000010%
12	0.000000030%	0.000000030%
13	0.000000120%	0.000000150%
14	0.000000510%	0.000000660%
15	0.000002000%	0.000002660%
16	0.000007260%	0.000009920%
17	0.000024520%	0.000034430%
18	0.000077290%	0.000111730%
19	0.000228050%	0.000339780%
20	0.000631450%	0.000971230%
21	0.001644590%	0.002615820%
22	0.004037490%	0.006653310%
23	0.009361100%	0.016014420%
24	0.020533140%	0.036547550%
25	0.042675400%	0.079222950%
26	0.084161650%	0.163384590%
27	0.157699710%	0.321084310%
28	0.281089390%	0.602173700%
29	0.477120070%	1.079293760%
30	0.771993250%	1.851287010%
31	1.191786810%	3.043073820%
32	1.756897460%	4.799971280%
33	2.475084980%	7.275056260%
34	3.334530750%	10.609587020%
35	4.298919820%	14.908506840%
36	5.306631470%	20.215138310%
37	6.275464910%	26.490603220%
38	7.112979960%	33.603583180%
39	7.730852630%	41.334435820%
40	8.060202730%	49.394638550%
41	8.064214780%	57.458853320%
42	7.744859330%	65.203712660%
43	7.142032120%	72.345744780%
44	6.325468620%	78.671213400%
45	5.381668540%	84.052881940%
46	4.399167830%	88.452049770%
47	3.455533910%	91.907583680%
48	2.608546110%	94.516129790%
49	1.892581150%	96.408710940%
50	1.319785670%	97.728496610%
51	0.884610240%	98.613106850%
52	0.569893140%	99.182999990%
53	0.352864140%	99.535864120%
54	0.209971500%	99.745835620%
55	0.120061620%	99.865897250%
56	0.065959510%	99.931856760%
57	0.034810000%	99.966666750%
58	0.017643840%	99.984310600%
59	0.008586950%	99.992897550%
60	0.004011630%	99.996909180%
61	0.001798450%	99.998707630%
62	0.000773430%	99.999481060%
63	0.000318940%	99.999800000%
64	0.000126060%	99.999926070%
65	0.000047730%	99.999973800%
66	0.000017310%	99.999991110%
67	0.000006000%	99.999997110%
68	0.000001990%	99.999999100%
69	0.000000630%	99.999999730%
70	0.000000190%	99.999999920%

between 36 and 45 point wins (about 1 sd) or 69.144%
How about a graph

Quote: vert1276

Ya, you got what I was saying. This really all steems from me just wonder about wincraps LOL. I just want to be able to look at the summary after 100 points are pass and see if im experiencing negative or positive variance.

I missed the Wincraps statement.
In the games log in wincraps there is a histogram and more data than you know what to do with. I think you have to track the point wins in an auto bet file first using a chip stack #1 thru 19 and it will then display the actual data for you.

I can look into that later if you would like.

update: Games Log will only show stats after a game/session ends and a new one begins. This will cause all data in the summary page and elsewhere to be reset unless you save it to a text file. Not a good thing if you still want to see the last session data.

One actually can have WinCraps, in an auto bet file, do the math for you and show the current expected value and range (standard deviation) in a few chip stacks.

I could set that up later and post a link to the auto-bet file so you can download it to your computer if you would like, if you know how to run auto-bet files. Just ask.

Or just figure out the math from my above post and you will know the mean and standard deviation for any value of trials, be it 100 or 30 or 10,000!
Enjoy

Unless I've missed something it would average out to be 40.606060...%

Going to hit a point # 1/3 of the time, but that's irrelevant. The question is when a point is established.

If a point is established the probability of it being a:

4 or 10 is 0.125
5 or 9 is 0.166666...
6 or 8 is 0.208333...

Add them up you get 1

Probablity of a 4 or 10 winning is 0.333...
Probablity of a 5 or 9 winning is 0.4
Probablity of a 6 or 8 winning is 0.454545...

Multiply them out:

4 = 0.041666667
5 = 0.066666667
6 = 0.09469697
8 = 0.09469697
9 = 0.066666667
10 = 0.041666667

total = 0.406060606