wudged
wudged
Joined: Aug 7, 2013
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December 13th, 2013 at 12:30:59 PM permalink
Assuming the dice are fair/unbiased and otherwise equal to each other, changing the dice has no effect at all. Each face on each die still has the same 1/6 probability of occurring.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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December 14th, 2013 at 11:16:48 AM permalink
If anyone is interested in the formula for the expected number of attempts (e.g. rolls) for a particular event to occur N consecutive times before another event takes place, the formula is:
(p + q)n - pn / (q pn), where p is the probability that the wanted event takes place and q is the probability that the unwanted event takes place.

For example, earlier in the thread, it was asked how many rolls would it take to roll 6 or 8 3 times before rolling a 7.
In this case, p = 5/18 and q = 1/6, so the result is ((5/18 + 1/6)3 - (5/18)3) / (1/6 x (5/18)3) = 18.576 rolls.

If p + q = 1, then this is the previously mentioned Feller's formula.



Let Fk be the expected number of attempts needed if you already have k consecutive successful attempts.
Since the target is n consecutive successful attempts, Fn = 0
Fk = 1 + p Fk+1 + q F0 + (1 - p - q) Fk
Solve for F[0]

Let r = 1 / (p + q)

Lemma: Fn-k = ((pr)k - 1) / (pr - 1) x (r + qr F0)

Proof by induction:
If k = 0: Fn-k = 0 = Fn
Assume it is true for k
Fn-k = ((pr)k - 1) / (pr - 1) * (r + qr F0)
= (1 + pr + (pr)2 + ... + (pr)k-1) * (r + qr F0)
Fn-(k+1) = r + qr F0 + pr Fn-k
= r + qr F0 + pr * ((1 + pr + (pr)2 + ... + (pr)k-1) * (r + qr F0))
= r + qr F0
+ pr2 * ((1 + pr + (pr)2 + ... + (pr)k-1)
+ pqr2 F0 * ((1 + pr + (pr)2 + ... + (pr)k-1)
= r + pr2 * ((1 + pr + (pr)2 + ... + (pr)k-1)
+ F0 * qr
+ F0 * pqr2 ((1 + pr + (pr)2 + ... + (pr)k-1)
= r + pr2 + p2r3 + ... + pKrk+1
+ F0 * qr (1 + pr + (pr)2 + ... + (pr)k)
= r (1 + pr + (pr)2 + ... + (pr)k)
+ F0 * qr (1 + pr + (pr)2 + ... + (pr)k)
= (r + qr F0) * (1 + pr + (pr)2 + ... + (pr)k)
= (r + qr F0) * ((pr)k+1 - 1) / (pr - 1)
Therefore, if it is true for k, then it is true for k + 1, and since it is true for 0, it follows that it is true for all integers >= 0

F0 = Fn-n = ((pr)n - 1) / (pr - 1) * (r + qr F0)
F0 = r * ((pr)n - 1) / (pr - 1) + F0 * qr * ((pr)n - 1) / (pr - 1)
F0 * (1 - qr * ((pr)n - 1) / (pr - 1)) = r * ((pr)n - 1) / (pr - 1)
F0 * ((pr - 1) - qr * ((pr)n - 1)) = r * (pr)n - 1)
F0 = r * (pr)n - 1) / ((pr - 1) - qr * ((pr)n - 1))

Substitute 1 / (p + q) for r, and eventually you get:
F0 = ((p + q)n - pn) / (q pn)
bahdbwoy
bahdbwoy
Joined: Aug 23, 2013
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December 15th, 2013 at 12:39:03 PM permalink
Quote: mustangsally


I like the .711 probability of hitting 2 in a row by the 7th bet.
Maybe the don't pass player likes it too.

Sally



why do you like it?

if i said what is the probability that you would not have 2 DP winners in a row (exclusing push) is that p= 1-(.4929^2) ?

if so then is the probability of not getting a streak of 2 in 7 attempts p^7 = .142518 ?

thank you
7craps
7craps
Joined: Jan 23, 2010
  • Threads: 18
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December 22nd, 2013 at 8:03:31 PM permalink
Quote: bahdbwoy

if i said what is the probability that you would not have 2 DP winners in a row (exclusing push) is that p= 1-(.4929^2) ?

here is what I get.
yes, and that would be for just the very first 2 bets made (actually resolved)
=1 - (949/1925)^2 = 0.7569638
Quote: bahdbwoy

if so then is the probability of not getting a streak of 2 in 7 attempts p^7 = .142518 ?

not quite.
0.276245216

we have to calculate this at each step (each bet made) for the Run of 2 then subtract from 1.
look at the below table
almost twice as high of a probability of *no run of 2* in 7 resolved bets
# betsat least 1 run of length 2no run 2
101
20.2430361950.756963805
30.3662587020.633741298
40.4894812090.510518791
50.5827561870.417243813
60.6608473790.339152621
70.7237547840.276245216
80.7751686130.224831387
90.8169598490.183040151
100.8509994770.149000523
110.8787037640.121296236
120.9012584310.098741569
130.9196186490.080381351
140.9345650750.065434925
150.9467322590.053267741
160.956637050.04336295
170.9647001060.035299894
180.971263890.02873611
190.9766071820.023392818
200.9809569230.019043077
210.9844978590.015502141
220.987380380.01261962
230.9897269160.010273084
240.9916371290.008362871
250.993192150.00680785
260.9944580250.005541975
270.9954885190.004511481
280.9963273990.003672601
290.9970102950.002989705
300.9975662110.002433789

I can break the table down like so (for column 2)
p = 949/1925
q = 1-p
run = 2

for bet #2 = p^run (0.243036195)
for bet #3 = Pbet#2+ (q*p^run) = 0.123222507 = 0.366258702

Now we have to look back to see if the Run of 2 did not happen earlier
and add that in
for bet #4 = Pbet#3+(1-Pbet#1)*q*p^run
for bet #5 = Pbet#4+(1-Pbet#2)*q*p^run
and so on down to complete the column in the table

runs or streaks are sneaky critters
but can be easily figured out using the right math
(and verified with simulation)
p=(949/1925)
n=7

grouped data items: 1000000
group middle freq freq/100
--------------------------------------------
-0.5 <= x < 0.50 0.00 276578 27.66%
0.50 <= x < 1.50 1.00 723422 72.34%

--------------------------------------------
0.00 276578
1.00 723422
--------------------------------------------
cumulative
--------------------------------------------
-0.5 <= x < 0.50 0.00 276578 27.66%
0.50 <= x < 1.50 1.00 1000000 100.00%

--------------------------------------------

Good Luck
winsome johnny (not Win some johnny)

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