richbailey86
Joined: May 8, 2014
• Posts: 325
November 28th, 2014 at 4:53:45 PM permalink
How likely is it to NOT win at least 2 bets IN A ROW within 12 decisions.

Lets say I choose Pass line and only pass line

I play at this table for a period of 12 pass/don't pass decisions.

What is the likelihood that I will not win back to back at least one time

I can lose 10 out of 12 bets as long as the 2 wins are back to back
An idea whose time has come cannot be stopped by any army or any government. – Ron Paul
Mission146
Joined: May 15, 2012
• Posts: 15308
November 28th, 2014 at 5:02:07 PM permalink
Quote: richbailey86

How likely is it to NOT win at least 2 bets IN A ROW within 12 decisions.

Lets say I choose Pass line and only pass line

I play at this table for a period of 12 pass/don't pass decisions.

What is the likelihood that I will not win back to back at least one time

I can lose 10 out of 12 bets as long as the 2 wins are back to back

The probability of winning two in a row:

.4929 * .4929 = 0.24295041

Therefore, the probability of playing two Pass Line bets and losing one, the other, or both is .75704959.

If you look at twelve total PL bets, then you had the potential to win two in a row on eleven different occasions, because you couldn't have won two in a row STARTING with the twelfth bet.

1-(.75704959)^11 = 0.95318687703

Therefore, the probability that you will make eleven pairs of two bets and win at least two in a row once is 95.31869% and the probability that you will not win two in a row at all is 4.68131%.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
richbailey86
Joined: May 8, 2014
• Posts: 325
November 28th, 2014 at 5:05:51 PM permalink
thank you very much Mission, I appreciate you taking the time to answer that was great
An idea whose time has come cannot be stopped by any army or any government. – Ron Paul
Mission146
Joined: May 15, 2012
• Posts: 15308
November 28th, 2014 at 5:16:26 PM permalink
No problem!
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5417
November 28th, 2014 at 5:22:36 PM permalink
I did a brute force calculation, and got 9.8798%

Mission - are each of the 11 "pairs" necessarily independent, considering that each consecutive pair shares a result?
bahdbwoy
Joined: Aug 23, 2013
• Posts: 163
November 28th, 2014 at 6:27:32 PM permalink
7craps had .0987 of no run with a similiar question

http://wizardofvegas.com/forum/gambling/craps/16060-beauty-of-probability/4/#post301921
richbailey86
Joined: May 8, 2014
• Posts: 325
November 28th, 2014 at 6:39:49 PM permalink
Quote: bahdbwoy

7craps had .0987 of no run with a similiar question

http://wizardofvegas.com/forum/gambling/craps/16060-beauty-of-probability/4/#post301921

Thanks
An idea whose time has come cannot be stopped by any army or any government. – Ron Paul
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
November 30th, 2014 at 10:40:17 PM permalink
Is OP creating a betting system?
I love them
Quote: ThatDonGuy

Mission - are each of the 11 "pairs" necessarily independent, considering that each consecutive pair shares a result?

I do not even know what mission came up with. His method to me loooooks to calculate the expected number of runs, or lack of them.
not any known probability in our universe, but we all could be wrong, yes?
maybe a new pair of shoes might help ;)
Quote: ThatDonGuy

I did a brute force calculation, and got 9.8798%

really? you listed all 2^12 outcomes and counted?
that is using the force, I say. I love brute or is that BRUT...
sounds the same to me but smells great, in me opinion

exercise is good for many too.

a streak calculator would be an easy solution and just subtract that value from 1 as most calcs figure the prob of at least 1 success.

it only works for 50% probs like a fair coin flip but it should give us the lowest probability as the pass line bet success rate is close to 49.3%, slightly lower than 50%
that # = 377 (can be found too in Pascal's triangle and who does not like triangles?)

377/2^12 = 377/4,096 or arounds about 0.092041016
Ah, mission was way off Houston (i was just there too)

Here, I solved this as did a 10 year old using very very simple simple math ands a spread sheet (I used excel)

Merry Christmas and Happy Holidays and Happy New Years
in cell D14 lies the gift

the cell formulas to do this yourself or is that elves

if you do not get it
just try it out
A2 to C2 are easy, just 1,0,0
A3 to D3 are easy too.
just fill them down (A3 to D3) how far you wants.
It really is up to yous

Sally says
I Heart Vi Hart
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5417
December 1st, 2014 at 5:39:32 PM permalink
Quote: mustangsally

really? you listed all 2^12 outcomes and counted?

Yes - I couldn't think of another way to calculate it accurately.

Well, "I" didn't count them - my 4.7GHz PC did.

Oh, and remember, you can use [sup] and [/sup] for superscripts - e.g. 212 instead of 2^12
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
December 1st, 2014 at 6:59:21 PM permalink
Quote: ThatDonGuy

Yes - I couldn't think of another way to calculate it accurately.

Ah, Eye see
I figured wrong then, Me thought that you would do a simple Markov chain
like so

I use a udf
but P*P= P^2*P=P^3*P^2*P^2 would work innnnn Excel too, ugh
Quote: ThatDonGuy

Well, "I" didn't count them - my 4.7GHz PC did.

Oh, and remember, you can use [sup] and [/sup] for superscripts - e.g. 212 instead of 2^12

I was going to actually put 4,096 right from my win7 calc
but I was in a hurry as usual.
My calc uses 2^ and I like it.

I still LOVE triangles!
Sally

almost forgot, for those that use R
(I do a lot)
`> #####################################################> # runs.m(m,N,p)> #> # Probability of at least 1 run of length m or more> # in N trials with probability p of success.> # Uses Markov chain.> #####################################################> > require(expm)Loading required package: expmLoading required package: MatrixLoading required package: latticeAttaching package: ‘expm’The following object is masked from ‘package:Matrix’:    expm> > runs.m = function(m,N,p) {+   q = 1-p+   A = matrix(rep(0,(m+1)^2),nrow=m+1)+   A[1,1:m] = q+   A[2:(m+1),1:m] = diag(p,m,m)+   A[m+1,m+1] = 1+   v = c(1,rep(0,m))+   (A%^%N %*% v)[m+1]+ }> print(1- runs.m(2,12,244/495),12)[1] 0.0987981138517> `

that comes from the basic matrix program
`###################################################### runs.m(m,N,p)## Probability of at least 1 run of length m or more# in N trials with probability p of success.# Uses Markov chain.#####################################################require(expm)runs.m = function(m,N,p) {  q = 1-p  A = matrix(rep(0,(m+1)^2),nrow=m+1)  A[1,1:m] = q  A[2:(m+1),1:m] = diag(p,m,m)  A[m+1,m+1] = 1  v = c(1,rep(0,m))  (A%^%N %*% v)[m+1]}print(runs.m(2,12,244/495),12)`

U B Good
I Heart Vi Hart