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November 9th, 2011 at 7:52:09 AM permalink

NowTheSerpent

Member since: Sep 30, 2011
Threads: 11
Posts: 278

Quote: DJTeddyBear
I was afraid to admit that I have no idea what he's even referring to when he says "as everyone knows...."

I'm sure he's attempting a bit of humor. And stochastic means purely probabilistic, with no intelligence controlling the outcome.

Never maintain as merely a "humble" opinion that which you are not prepared to defend; If that which you hold can be rigorously supported, don't be so "humble" as to call it "your" opinion, for indeed its reality transcends and holds YOU.

November 9th, 2011 at 7:53:48 AM permalink

NowTheSerpent

Member since: Sep 30, 2011
Threads: 11
Posts: 278

Quote: Wizard
You're right, thanks for the correction. The column should be correct now.

I could count the number of girls on this forum with one hand, so don't expect one of them to answer. Personally, I think that is the mathematician from the show Numbers.

Doctor Charles Epps

November 9th, 2011 at 1:53:36 PM permalink

7craps

Member since: Jan 23, 2010
Threads: 10
Posts: 332

Quote: CrystalMath
I actually don't have it any more and it would be pretty time consuming to recreate. I used a Markov Chain instead of a matrix though. They essentially do the samem thing but the chain will use less space in Excel.

I figured you might not still have your work.
You use Excel. That is a lot of work for matrices.
There are free softs (Winmat at Peanut soft) Winmat
and online like
Matrix Algebra Tool that are even easier and faster.

Is not the matrix below correct except I do not know what to place in row 9 to make this 9x9?
(A 1 in 9,9 would be for 8 or more and that is a different question.)

0.50000 0.50000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
0.50000 0.00000 0.50000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
0.50000 0.00000 0.00000 0.50000 0.00000 0.00000 0.00000 0.00000 0.00000
0.50000 0.00000 0.00000 0.00000 0.50000 0.00000 0.00000 0.00000 0.00000
0.50000 0.00000 0.00000 0.00000 0.00000 0.50000 0.00000 0.00000 0.00000
0.50000 0.00000 0.00000 0.00000 0.00000 0.00000 0.50000 0.00000 0.00000
0.50000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.50000 0.00000
0.50000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.50000
0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
We would be concerned with 8,1 a tail after 7 heads correct?
Plugging in a few numbers does not arrive at your answer.

I can do the work, I just need to know how to start off.
Thanks

Life in the Key of F#...a.crap=(gambling) - (math) b.math=(crap) / (gambling) c.gambling=(crap) / (math)

November 9th, 2011 at 8:13:38 PM permalink

CrystalMath

Member since: May 10, 2011
Threads: 3
Posts: 476

I thought about this again since I just can't stay away from interesting questions. Anyhow, the answer I was trying to find was the probability of getting exactly one streak of 7. Because of that, we have to keep track of when the streak exceeds 7 and also we need to keep track of whether or not a streak of 7 has already been reached. In the matrix below, "streak 7,n" represents the likelihood of having reached a streak of 7 and then currently having a streak of size n. When you calculate T^100, using the following transition matrix, the probability of ending in each state will be on the top row. Then, we need to add the probabilities of streak 7; streak 7,0 through streak 7,6; and streak 7,8+. When you do, you will see that I made an error in my original answer. In my original answer, it appears as though I also added streak 7,7.


streak 0	0.5	0.5	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
streak 1	0.5	0	0.5	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
streak 2	0.5	0	0	0.5	0	0	0	0	0	0	0	0	0	0	0	0	0	0
streak 3	0.5	0	0	0	0.5	0	0	0	0	0	0	0	0	0	0	0	0	0
streak 4	0.5	0	0	0	0	0.5	0	0	0	0	0	0	0	0	0	0	0	0
streak 5	0.5	0	0	0	0	0	0.5	0	0	0	0	0	0	0	0	0	0	0
streak 6	0.5	0	0	0	0	0	0	0.5	0	0	0	0	0	0	0	0	0	0
streak 7	0	0	0	0	0	0	0	0	0.5	0.5	0	0	0	0	0	0	0	0
streak 8+	0.5	0	0	0	0	0	0	0	0.5	0	0	0	0	0	0	0	0	0
streak 7,0	0	0	0	0	0	0	0	0	0	0.5	0.5	0	0	0	0	0	0	0
streak 7,1	0	0	0	0	0	0	0	0	0	0.5	0	0.5	0	0	0	0	0	0
streak 7,2	0	0	0	0	0	0	0	0	0	0.5	0	0	0.5	0	0	0	0	0
streak 7,3	0	0	0	0	0	0	0	0	0	0.5	0	0	0	0.5	0	0	0	0
streak 7,4	0	0	0	0	0	0	0	0	0	0.5	0	0	0	0	0.5	0	0	0
streak 7,5	0	0	0	0	0	0	0	0	0	0.5	0	0	0	0	0	0.5	0	0
streak 7,6	0	0	0	0	0	0	0	0	0	0.5	0	0	0	0	0	0	0.5	0
streak 7,7	0	0	0	0	0	0	0	0	0	0.5	0	0	0	0	0	0	0	0.5
streak 7,8+	0	0	0	0	0	0	0	0	0	0.5	0	0	0	0	0	0	0	0.5

I heart Crystal Math.

November 9th, 2011 at 9:05:53 PM permalink

7craps

Member since: Jan 23, 2010
Threads: 10
Posts: 332

Quote: CrystalMath
I thought about this again since I just can't stay away from interesting questions. Anyhow, the answer I was trying to find was the probability of getting exactly one streak of 7.

OK.
Excellent work but I now think you are answering a different question.

Now this seems confusing.
Starting over...

From the Wizard:
"The probability of getting "one or more" streaks of exactly seven heads in a row out of 100 coin tosses is 17.29%"
I get the same answer from computer sims.

"The probability of getting one or more streaks of "at least" seven heads in a row out of 100 coin tosses is 31.7520%"
The Wizard has the correct solution of 31.7520% but had the matrix not square in his article. He did correct it.

June 16, you wrote, "Using a Markov chain, I calculate the probability of exactly one 7 coin streak to be 0.172920892."
So then you must have added all the values?
I must have misread that statement when I first read it.

But is sure looks like the theoretical answer that matches the answer the Wizard gave for "one or more". The Wizard has never shown his work on this so I must assume he also ran a computer simulation.

I will work with your matrix and see if I can duplicate your first answer by adding all the values. It should work.
Thank you for your efforts.
Excellent work as always.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Added: Excellent. I came up with 0.1729209 and I am sure if I added the extra places the results will match perfectly to your answer.

Worked exactly as you described and I now understand the process.

I did not use Excel to calculate T^100, I used Winmat (free Peanut soft) and it took just a few seconds to paste in the matrix once I had it set up in Excel.

I have seen others use inclusion-exclusion for these type of problems and they are really messy to work with.

I hope others can understand your work as well, if they want to of course.

Again, Thanks very much.

Added Nov 10.
Here is my T^100

streak 0	streak 1	streak 2	streak 3	streak 4	streak 5	streak 6	streak 7	streak 8+	streak 7,0	streak 7,1	streak 7,2	streak 7,3	streak 7,4	streak 7,5	streak 7,6	streak 7,7	streak 7,8+
0.414358508	0.207589542	0.104000322	0.052103139	0.026103161	0.013077427	0.006551663	0.003282319	0.003295345	0.085641492	0.042410458	0.020999678	0.010396861	0.005146839	0.002547573	0.001260837	0.000623931	0.000610905
0.412720599	0.206768966	0.103589221	0.051897182	0.025999978	0.013025734	0.006525765	0.003269344	0.003282319	0.087279401	0.043231034	0.021410779	0.010602818	0.005250022	0.002599266	0.001286735	0.000636906	0.000623931
0.409451255	0.205131057	0.102768644	0.051486081	0.025794021	0.012922551	0.006474071	0.003243446	0.003256318	0.090548745	0.044868943	0.022231356	0.011013919	0.005455979	0.002702449	0.001338429	0.000662804	0.000649932
0.40292549	0.201861713	0.101130735	0.050665505	0.02538292	0.012716594	0.006370889	0.003191753	0.003204419	0.09707451	0.048138287	0.023869265	0.011834495	0.00586708	0.002908406	0.001441611	0.000714497	0.000701831
0.389899757	0.195335948	0.097861391	0.049027596	0.024562344	0.012305493	0.006164931	0.00308857	0.003100827	0.110100243	0.054664052	0.027138609	0.013472404	0.006687656	0.003319507	0.001647569	0.00081768	0.000805423
0.363899778	0.182310214	0.091335626	0.045758252	0.022924435	0.011484917	0.00575383	0.002882612	0.002894052	0.136100222	0.067689786	0.033664374	0.016741748	0.008325565	0.004140083	0.00205867	0.001023638	0.001012198
0.312002596	0.156310236	0.078309893	0.039232487	0.019655091	0.009847007	0.004933254	0.002471512	0.00248132	0.187997404	0.093689764	0.046690107	0.023267513	0.011594909	0.005777993	0.002879246	0.001434738	0.00142493
0.208413375	0.104413054	0.052309914	0.026206753	0.013129326	0.006577663	0.003295345	0.001650935	0.001657487	0.291586625	0.145586946	0.072690086	0.036293247	0.018120674	0.009047337	0.004517155	0.002255315	0.002248763
0.416002918	0.208413375	0.104413054	0.052309914	0.026206753	0.013129326	0.006577663	0.003295345	0.003308423	0.083997082	0.041586625	0.020586946	0.010190086	0.005043247	0.002495674	0.001234837	0.000610905	0.000597827
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625
0	0	0	0	0	0	0	0	0	0.5	0.25	0.125	0.0625	0.03125	0.015625	0.0078125	0.00390625	0.00390625

0.1729208930 is my final answer

Life in the Key of F#...a.crap=(gambling) - (math) b.math=(crap) / (gambling) c.gambling=(crap) / (math)

November 10th, 2011 at 8:29:12 AM permalink

CrystalMath

Member since: May 10, 2011
Threads: 3
Posts: 476

After re-reading the entire post, I see that we were trying to confirm the Wizard's assertion that the probability of getting "one or more" streaks of length 7 is 0.1729. In my original answer, I said "exactly one" streak of length 7, but my answer was for "one or more."

So, it looks like we agree.

Thanks for the Winmat suggestion. I'll need to look into that. Also, I need to finally try to learn about inclusion/exclusion.

I heart Crystal Math.

November 10th, 2011 at 8:44:28 AM permalink

7craps

Member since: Jan 23, 2010
Threads: 10
Posts: 332

Quote: CrystalMath
Thanks for the Winmat suggestion. I'll need to look into that. Also, I need to finally try to learn about inclusion/exclusion.

Thank you. You are very welcome.

inclusion-exclusion is a great tool to have. It can get very messy when using many terms.

a good example is here:
askthewizard/278
4th question down
"What is the expected number of rolls of two dice for every total from 2 to 12 to occur at least once?"

Also BruceZ over at 2+2 really does a great job on explaining inclusion-exclusion
The Wizard explains it also in the article.

Life in the Key of F#...a.crap=(gambling) - (math) b.math=(crap) / (gambling) c.gambling=(crap) / (math)

November 10th, 2011 at 10:01:58 AM permalink

7craps

Member since: Jan 23, 2010
Threads: 10
Posts: 332

Quote: CrystalMath
Thanks for the Winmat suggestion. I'll need to look into that.

Winmat is a C++ program from Peanut software, its free, that does matrices and much more.
An excellent program as is the suite of Peanut math programs.

I suggest you try it out with the:
"I once hit six royals in single-line video poker within 5,000 hands. In my lifetime I have played about 25 million hands. What are the odds?"
matrix that you did.
six royals in single-line video poker within 5,000 hands-ask the wizard 277

You can just copy and paste the matrix into Winmat and calculate T^25,000,000 in a split second. (or 25,000,000-5000) I have not studied that article yet.

I have only used the Winmat program a few times but what an excellent tool to have IMO.

CrystalMath, you rock!
Thanks for all your excellent work!

Life in the Key of F#...a.crap=(gambling) - (math) b.math=(crap) / (gambling) c.gambling=(crap) / (math)

February 14th, 2012 at 11:17:03 AM permalink

mustangsally

Member since: Mar 29, 2011
Threads: 5
Posts: 170

Wow what a tread for exact precision!

Quote: 7craps
From the Wizard:
"The probability of getting "one or more" streaks of exactly seven heads in a row out of 100 coin tosses is 17.29%"
I get the same answer from computer sims.

"The probability of getting one or more streaks of "at least" seven heads in a row out of 100 coin tosses is 31.7520%"
The Wizard has the correct solution of 31.7520% but had the matrix not square in his article. He did correct it.

"The probability of getting "one or more" streaks of exactly seven heads"

We need to focus on the meaning of the word "exactly"

Here are 5 popular run statistics in current use.

The Wizard, CM and 7Craps in this thread all are using the 4th method, E(n,k)
the number of success runs of exact length k followed by failure until n trials.

And in a finite number of trials this can happen in 3 ways.
HHHHHHHT (8trials) at the beginning of the sequence only
THHHHHHH (8trials) at the end of the sequence only (the exception where the run does not end in a failure)
THHHHHHHT (9trials) from the beginning of the sequence to the (n-(r+1))th trial

Here is the problem with this counting method E(n,k).

It does not fit all statistical problems.
Really that simple.

HHHHHHH HT would not count as a run of exact length 7 using the E(n,k) method
but as a run of length 8.

But it does contain a run of consecutive successes of exactly length 7.
This is Feller's counting method. N(n,k)

This can easy lead to misunderstanding and misapplying of the data results.

Feller's method starts the count over once the length has occurred.
So, HHHHHHH HHHHHHHT would be a frequency of 2 for length 7 using N(n,k)
and frequency of 1 for length 14 using E(n,k)

Of course, it is easy to count using E(n,k) and then convert the data to N(n,k).
But, impossible to do the reverse.

OK Sally, why bring this up?
Because so many mis-use the results to seriously understate the probability of seeing a run of "exactly" length 7.
Why? Because there are 2 meanings of the word "exact" in the methods of counting runs.

Gamblers always want to know the probability of a win or lose streak. Some even are wise enough to consider the number of trials that they can happen in.

So, I add to this thread that the probability of at least one of an exact length 7 in 100 trials using N(n,k), I think this is more useful info to any gambler,
is 31.7520% - the same as the G(n,k) method

(A gambler would have a "fit" if HHHHHHH then HHHHHHH followed when he was betting on tails for a 7 step Marty. It happened twice! But the E(n,k) counting method would say it did not even happen 1 time. Ouch!)

And I am not splitting any hairs.
I have no split ends.
I love tight ends.

Sally
Happy Valentine's Day
I am on vacation!
Yeah!

I Heart Vi Hart

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