Best craps bonuses

foolshope
foolshope
foolshope
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Joined: Jul 15, 2011
December 21st, 2011 at 7:36:14 AM permalink
I'm curious, is there a way to predict how may Points/SOs will happen over X amount of throws? I know the numbers say about 59% will be SO and 41% points, but through that gauntlet, what are the odds of seeing say 5 points in a row? My experience has shown me that I rarely see that (though I did point 4x in a row the other day) but often I'll see 3, 4 even 10 SOs.

I'm sure there's a forumla - anyone have it so I can play around with it or is this something that can be done in this $20 simulator?

~N
guido111
guido111
guido111
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December 21st, 2011 at 5:47:37 PM permalink
Quote: foolshope

I'm curious, is there a way to predict how may Points/SOs will happen over X amount of throws?

As to the game of Craps,
"Predict" is not the correct word to use. The math will only show the probability of an event or a distribution.
It takes on average 557/110 or 5.060606 rolls to see a point decision. (This includes naturals on the come out roll)
the distribution from simulation
rollsrelativecumulative
218.839%18.839%
319.812%38.652%
416.140%54.791%
512.363%67.155%
69.080%76.234%
76.634%82.868%
84.761%87.630%
93.456%91.086%
102.485%93.571%
111.832%95.402%
121.290%96.692%
130.921%97.613%
140.685%98.298%
150.486%98.784%
160.335%99.119%
170.248%99.367%
180.176%99.544%
190.132%99.675%
200.090%99.765%
210.060%99.825%
220.048%99.873%
230.033%99.905%
240.026%99.931%
250.016%99.947%
260.016%99.963%
270.011%99.974%
280.006%99.980%
290.006%99.985%
300.005%99.990%


Quote: foolshope

I know the numbers say about 59% will be SO and 41% points, but through that gauntlet, what are the odds of seeing say 5 points in a row?


for the next 5 points a formula would be (201/495)^5.

If you want the probability of 5 in a row or more for any number of trials then check out the thread here Ask the Wizard correction
and you can also use the streak calculator here
Steak Probability Calculator

First pic below is the curves for Craps points win streak lengths.

Example: 2+ means a run of at least 2, 5+ a run of at least 5 (so 5 or more)

So, at least 1 run of 5 (5+) the 4th curve from the left shows about a 50% probability (median) at about 100 points (500 dice rolls)
The average number of points it takes to see a streak of 5 wins is 150.828 (sd of 146.9) and each point takes an average of 5.0606 rolls for an average number of rolls of 763.3
The distribution is not normal but very close to a geometric distribution.

I was also very lazy with my chart labels. Scale is log.
Craps Point win streak curves
x-axis (1, 10, 100 etc) are # of points

Next pic breaks the curves down further into the probabilities of multiple streaks per the number of points.
BruceZ over at 2+2 forum just recently created an Excel Macro worksheet calculating this with a recursive formula.
More on that to come after the holidays. Read More here
Function ProbRunMulti(n As Long, r As Long, p As Double, s As Long) As Variant

' This Formula Is Created By Probability Guru BruceZ; From twoplustwo.com
' Probability of 1,2,3,...s or more streaks of r or more consecutive heads in n tosses
' of a coin having probability p of heads.  Returns s values.
'
' P(j, i) = P(j, i - 1) + [P(j-1, i-r-1) - P(j, i-r-1)] * (1-p)p^r, for i > j*(r+1) - 1
'
' P(j, i) = p^(jr) * (1-p)^(j-1), for i = j*(r+1) - 1
'
' P(j, i) = 0, for i < j*(r+1) - 1
'
' P(0, i-r-1) = 1
'
' where j is the number of streaks, and i is the flip.
' Note that [P(j-1, i-r-1) - P(j, i-r-1)]is the probability of exactly j-1 streaks.
' j is taken modulo 2 in the code implementation.
' P is implmented as a linear array prob of size 0:n.
ReDim output(1 To s) As Double  ' output values for 1,2, ..., s or more streaks
    ReDim prob(1, n) As Double      ' array of probabiites for each flip.
    Dim c            As Double      ' (1-p) * p^r
    Dim curr         As Long        ' 0 or 1 to index first dimension of prob array
    Dim prev         As Long        ' 0 or 1 to index first dimension of prob array
    Dim first        As Long        ' offset of first flip where j streaks is possible
    Dim i            As Long        ' index of flips
    Dim j            As Long        ' index of streaks
    
    c = (1 - p) * p ^ r

' 1 or more streaks

    If r <= n Then prob(1, r) = p ^ r
    For i = r + 1 To n
        prob(1, i) = prob(1, i - 1) + (1 - prob(1, i - r - 1)) * c
    Next i
    output(1) = prob(1, n)
    
' 2, 3, ... s or more streaks

    For j = 2 To s
        curr = j Mod 2
        prev = (j + 1) Mod 2
        first = j * (r + 1) - 1
        If first <= n Then prob(curr, first) = p ^ (j * r) * (1 - p) ^ (j - 1)
        For i = first + 1 To n
            prob(curr, i) = prob(curr, i - 1) + (prob(prev, i - r - 1) - prob(curr, i - r - 1)) * c
        Next i
        
        output(j) = prob(curr, n)
        
        For i = 0 To n
            prob(prev, i) = 0       ' clear for reuse
        Next i
    Next j
    
    ProbRunMulti = Application.Transpose(output)
    
End Function

Example: a streak of 5 or more.
at 10 points (50 dice rolls) you can see about a 4.3824% chance of at least 1 run of length 5 or more and very close to 0% for 2 such streaks.
at 80 points (400 dice rolls) you can see about a 40.6251% chance of at least 1 run of length 5 or more and a 8.5128% probability of 2 such runs and a 1.0557% chance (1 in 100) of 3 such runs in 80 points.
snap shot of my Excel worksheet.

Quote: foolshope

My experience has shown me that I rarely see that (though I did point 4x in a row the other day) but often I'll see 3, 4 even 10 SOs.

I'm sure there's a forumla - anyone have it so I can play around with it or is this something that can be done in this $20 simulator?
~N

The longer one plays Craps, the more probable it is that longer such streaks will be witnessed. The streaks for no-points will be much longer and more of them per N trials since the probability of no point hitting is 294/495 or 59.3939% as you mentioned above.
Maybe I will show the same data for no-points later on. Holiday travel time.

The math behind streaks can be a bit challenging using a recursive formula or a Markov Chain approach.
The $20 for WinCraps would be well spent IMO.
Hope this helps.
Enjoy!
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