statman
statman
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September 27th, 2011 at 8:11:14 AM permalink
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heather
heather
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September 27th, 2011 at 9:00:39 AM permalink
Quote: statman

Another poster claimed that in the major casinos all the old imperfect
wheels have been replaced with new faultless ones. I referred to an
article entitled "Roulette Bias Exposed" by Jeff Murphy in Indian
Gaming
of July 2007, p. 40, in which he describes a wheel with a
chi-square of 94.59 at Seven Feathers Hotel & Casino Resort in
Canyonville, Oregon. Considering that the chi-square of a normal wheel
is around 30, this is a highly biased wheel Link



I said that I thought that they were unlikely to be found in legal US casinos. I said that I thought a study of casinos in Caracas might be interesting. I said that a study of out-of-the-way gambling locales in the Caribbean might be interesting. I even noted that the Wizard has innuendoed at very old wheels still being in use in Europe. I have never suggested that biased Roulette wheels do not currently exist.

On the first page of your first thread, you cited Allan Wilson. Good choice; I like him. So I talked about how even back in the 1950s when he had unquestionably biased wheels to play on, he and his group got backed off and the wheels repaired or replaced. Which is exactly what I would expect to happen even if you could find a biased Roulette wheel in a legal US casino; you would not be allowed to exploit it (as I've noted in two of your other threads). They were allowed to play on the biased wheel at Harold's Club only because they attached the highest high roller who played there to their crew, and that only bought them a little more time -- the owner ended up switching the wheel out with the biggest whale he'd ever netted sitting at the table.

And (although I haven't mentioned it yet), I also think that <10K spins will always be too small a sample. Unless I've missed something, your tables aren't even based on actual Roulette spins; they were created using Excel's buggy RNG. So a wheel whose results don't match results created by a nonwheel must be biased? I'm having trouble swallowing that one.
MathExtremist
MathExtremist
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September 27th, 2011 at 10:47:09 AM permalink
Quote: heather

Unless I've missed something, your tables aren't even based on actual Roulette spins; they were created using Excel's buggy RNG.


And a buggy formula. It's all well and good to repeat research into roulette bias detection, but let's not start the conversation by saying "I've made a new discovery in statistics" or "I've solved a 150-year-old problem." Using Excel's built-in RNG and plotting a histogram of the outcomes doesn't qualify as a new discovery in statistics.

Biased roulette wheels exist. Basically every mechanical wheel is biased to some degree. However, it's very hard to detect that bias, certainly without many trials and excellent recordkeeping (which includes far more than just the numeric outcome). Many casinos today use bias-detection software which will discover any bias sooner than a player can, and wheel mechanisms are much more precise than they used to be. So by all means, try to detect a bias if you wish. But you don't need any RNG trials to do that. The probability of a number X occurring N times in M spins of a fair roulette wheel is directly calculable.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
statman
statman
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September 27th, 2011 at 1:24:12 PM permalink
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gog
gog
Joined: Jan 7, 2011
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September 27th, 2011 at 1:57:20 PM permalink
Quote: statman

You say " The probability of a number X occurring N times in M spins of a fair roulette wheel is directly calculable." I agree, but can you do it?

I'll bet you can't. I don't mean the probability of a particular number, I mean the probability of any number.



Wrong guy to challenge buddy. His name should have been your first hint
DorothyGale
DorothyGale
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September 27th, 2011 at 3:28:24 PM permalink
Quote: statman

The probability of a number X occurring N times in M spins of a fair roulette wheel.


p(X occuring exactly N times in M spins) = (combin(M,N) * 37^(M-N))/38^M

--Ms. D.
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
statman
statman
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September 27th, 2011 at 3:29:43 PM permalink
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statman
statman
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September 27th, 2011 at 3:40:46 PM permalink
DorothyGale:

That's the weight of the binomial distribution and it gives the probability of some particular number such as 17 or 34 coming up exactly N times in M spins. In my paper I demonstrate why the binomial distribution doesn't apply to the game of roulette because it is not a simple win-lose game: it is a game with multiple outcomes. What is needed is the probability that some number or other of all of the 38 on the wheel will come up N times. It's like the probability of winning a lottery twice. The probability of a particular person doing so is very small but the probability of someone or other doing so is much greater.

I treat that in my other book on Amazon that Buzzpaff is so crazy about.
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
MathExtremist
MathExtremist
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September 27th, 2011 at 4:22:11 PM permalink
Quote: statman

Dear MathExtremist,

You are referring to the program I use in Corel Quattro Pro X5 to create the simulation histograms. Where is the bug? I am calling your hand.

My probability distribution is calculated using a formula.

You say " The probability of a number X occurring N times in M spins of a fair roulette wheel is directly calculable." I agree, but can you do it?

I'll bet you can't. I don't mean the probability of a particular number, I mean the probability of any number.


Seriously? The probability of "any number" occurring N times in M spins is 1 regardless of M and N. There is no such thing as a spin which does not yield "any number" -- if the ball bounces out, it doesn't count as a spin. If you don't mean what you actually said ("any number"), then you must be interested in the probability of some particular number appearing (or number from a particular set of numbers). If that's the case, the binomial distribution absolutely applies -- either the number appears with probability p, where p = n/38 and where n = the size of the set of target numbers, or the number does not appear with probability 1-p.

Edit: I think I deciphered what you're trying to say, which is that you don't care *which* number appears N times, but just that one or more of them does. In that case, the answer is found in the multinomial distribution. You want all partitions of M into 38 buckets where at least one of the coefficients is N. But I see you made it a bet. What do I win?

The bug I referred to was in your Excel formula. You were generating numbers in the range [0..38]. You said you observed too-small frequencies for a double-zero wheel, but that's because you weren't simulating one.

But buggier still is the idea that you need to generate random numbers at all to do roulette bias detection. The roulette wheel is the random variable. Introducing another is neither necessary nor productive.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
statman
statman
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September 28th, 2011 at 4:55:39 AM permalink
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