buzzpaff
buzzpaff
Joined: Mar 8, 2011
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October 3rd, 2011 at 8:59:24 AM permalink
Quote: gog

So you are 'educating' people who make their living in part from calculating advanced mathematics/statistics, and come up with a method to beat an entire industry that makes their money off these numbers at their own game while using only a thousandth of the sample size, by copy and pasting sophomore year textbooks? And you are willing to share this knowledge with us for a bargain price of $25? I can't believe anyone can be this deluded, so that leaves only one other possibility.



THIEF, no more, no less !
statman
statman
Joined: Sep 25, 2011
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October 3rd, 2011 at 10:40:45 AM permalink
Nothing but rudeness and contempt! Very disappointing! Well, in case you think I can't calculate an accurate answer to Shack's problem, here it is to 100 digits. If you want 1,000 just say the word.

The problem is "What is the probability of at least one number being unhit in 200 spins of a true roulette wheel?"

Answer: 0.1698457156512422824774366227571399727305239397937597577397915962140059107818329286640551237007763387
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
boymimbo
boymimbo
Joined: Nov 12, 2009
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October 3rd, 2011 at 12:56:13 PM permalink
I think Math was quite clear in stating what was wrong. When you do a simulation it is to verify that your theoretical result matches the experimental one and when it's off, it's off. It means that your theoretical method or your simulator is wrong and Guido111, Math, and Dorothy have all verified that.

So, I'm guessing that what you are saying is that you are trying to detect a biased wheel through the occurence of non-hits based on a series of spins, and that theoretically, more non-hits means more hits, because if other numbers are hitting more, then other numbers are therefore hitting less.

Balderdash. Let's say that a wheel is biased towards three numbers, say 00, 0, and 1 with a frequency of .04 and that the frequency of all other numbers is .88 / 35 = .025143. Does the non-occurence of numbers now appear more frequently? Well of course it does, as the frequency of hitting the other numbers are less. But can you DETECT the non-frequency before detecting the bias. I say the answer is no.

In my little empircal analysis of 120,000 spins which represents 600 trials of 200 spins each where 00, 0, and 1 appear .04 of the time and the remainder appear evenly distributed:

No occurences of a number once occured on 98 occurences (16.333%). No occurences of a two numbers occurred 12 times (2%). No occurences of 3 numbers appeared twice (0.33%) . In the occurence where no occurences of 3 numbers, the most frequent numbers was not 00, 0, and 1, but 28 (10 times), 12 (appeared 11 times), 8, and 1 (appeared 10 times).

Inotherwords, to summarize, EVEN if you could conclude (and you cannot) that the non-appearance of numbers is correlated with a biased wheel, you cannot make that determination after 200 spins, and even if you could, the numbers that appear the most frequency in those 200 spins are not the ones that are biased.
----- You want the truth! You can't handle the truth!
statman
statman
Joined: Sep 25, 2011
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October 3rd, 2011 at 4:11:12 PM permalink
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A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
mustangsally
mustangsally
Joined: Mar 29, 2011
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October 3rd, 2011 at 6:45:50 PM permalink
Quote: statman

Nothing but rudeness and contempt! Very disappointing! Well, in case you think I can't calculate an accurate answer to Shack's problem, here it is to 100 digits. If you want 1,000 just say the word.

The problem is "What is the probability of at least one number being unhit in 200 spins of a true roulette wheel?"

Answer: 0.1698457156512422824774366227571399727305239397937597577397915962140059107818329286640551237007763387


So you mean "What is the probability of at least "ANY" one number being unhit in 200 spins of a true roulette wheel?"

Since there is a difference between a "specific" number and "any" one number when using a formula, your answer looks to be different from the answer in the table from your Kindle document.
Does this mean you now used a different formula?

The inclusion-exclusion formula that the Wizard used in his article appears to me to be the correct one to use for "any" number.
I also verified his results in Excel.

Your posted formula earlier in this thread that MathE explained looks to be for one "specific" number.

The results are very close but they are different and now I can see why others are bashing your math.

One must compare apples to apples and it looks like you are now doing that.
Yes?
Sally
I Heart Vi Hart
statman
statman
Joined: Sep 25, 2011
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October 4th, 2011 at 1:05:03 AM permalink
mustangsally: I love cowgirls!

For a complete discussion of the Wizard's problem look in his "Roulette" section. I am not especially interested in this problem but the hyenas here have been using his results to call me a hack. I now realize that the Wiz posts answers only to the problems he can solve. I posed my problem to him and instead of answering it he suggested I post it to the forum. No one on the forum knows how to solve it either. MathE makes a lot of noise but I have not yet seen him do a single calculation.

The question I posed to the Wiz was "What is the probability of at least one number coming up n times in b spins of a roulette wheel?" When he did not provide an answer I went to work on it myself. The results passed certain tests and since I had never seen this particular problem treated anywhere I thought I had something original. Now it appears that some of my figures may be off in the third or fourth decimal place. That doesn't matter for a statistical test that requires only two decimal places, but the hyenas are demanding perfection. My approach was to calculate the probability for a single number and then apply the union-of-events rule to include all the numbers on the wheel. This requires complete independence of the events and that may be the source of error.

The Wizard's result was easy to duplicate. There is a combinatorial function T(m,n) that gives the number of ways of distributing m distinguishable objects to n distinguishable boxes in such a way that no box is left empty. T(200,38)/38^200 is the probability that all slots will be filled. Subtract that from 1 and you have the probability that at least one will be left empty. I don't know how to apply this approach to the more general problem. I hope you didn't buy the table from Amazon. You can get it from me for free. I have also submitted it to www.freebookspot.es. I'll be glad to hear from you any time.
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
statman
statman
Joined: Sep 25, 2011
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October 4th, 2011 at 1:32:03 AM permalink
MATERIAL REMOVED

Some of it eventually may reappear on the web site of the
Rancocas Valley Journal of Applied Mathematics
Please flag this page so that it may be deleted.
Many thanks to those who have been helpful.
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
statman
statman
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October 5th, 2011 at 3:01:19 AM permalink
Quote:

MathExtremist: I'm interested in what is statman's interpretation of that number (whether it's 0.81 or 0.83) and what it means to his understanding of whether the wheel is biased.


If the tables say that the probability of a certain number of hits on a number in a given number of spins is 0.81, it means the probability of that happening by chance. For 0.81, that number of hits is very likely to occur by chance and does not indicate a biased wheel. A value of 0.00001, such as we get for Murphy's 0, means that that number of hits is unlikely to occur by chance and indicates a biased wheel.
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
rdw4potus
rdw4potus
Joined: Mar 11, 2010
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October 5th, 2011 at 11:09:54 AM permalink
Quote: statman

No one on the forum knows how to solve it either.



Riiiiiight. Maybe they can't solve it. Or maybe they just think that you're a tool and the problem isn't worth their time. Seeing as how at least 3 members of this forum are widely respected professional gaming mathematicians, I'm guessing that the latter is the case...
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
statman
statman
Joined: Sep 25, 2011
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October 5th, 2011 at 12:12:09 PM permalink
MATERIAL REMOVED

Some of it eventually may reappear on the web site of the
Rancocas Valley Journal of Applied Mathematics
Please flag this page so that it may be deleted.
Many thanks to those who have been helpful.
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant

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