The Roulette Conundrum

[Erasmus00]

I think the best way to bet would be to determine the average number of spins of the wheel between two hits of 0.

As Ron previously mentioned, the probability of rolling n consecutive spins is given by



With this in hand we are free to take the average number of spins between zero rolls, given by



This sum gives us the average number of spins (the probabiltiy of that number of spins, multiplied by the number).

With some mathematical trickery (if anyone wants to see the mathematics, let me know and I'll type it up) we find that this sum has a closed form: and gives exactly 36 spins. (This should not be surprising to anyone, but it is nice to know that math works out).

We can also take the root mean squared spin number. This simply involves replacing the leading n with n^2 in the sum above. This too has a closed form, and gives around 50. Hence our deviation is rather high (in fact, almost equal to the mean).

Sorry for not typing up the math, I just ran out of steam. If anyone wants to see it, let me know and I'll type it up.

Edit to clarify: Since the variance is essentially the mean, we expect this to follow a poisson distribution. (This is common for the case of random walks with each step unlikely) Hence, we can instantly assert the following: our mean value is 36, as is our mode.
-Will

[Simon]

You have certainly calculated the Mean: it is indeed 36.

I maintain that you have not calculated the Mode.

The probabiltiy of 36 consecutive non-zero spins is just above 1/99. Whereas the probabilty of a single non-zero spin between two zeros is just above 1/38. The probability of none - i.e two consecutive zeros - is 1/37.

If you were betting on which of these would occur the most often in a large unbiased sample - say 50,000 spins - I'm afraid you would almost certainly lose!

If probabilty was played out, you would find more than twice as many events of consecutive zeros than you would find events of zeros separated by 36 other numbers.

I realise that all this is even more counter-intuitive than the Monty Hall problem!

Simon

[Erasmus00]

Ahh, I reread the original post and I'm afraid I misunderstood the bet. I thought it was the standard bet it won't hit 0 as long as you can untill you lose. You are looking at hitting EXACTLY the right number. Missing that put me at cross purposes. Sorry.
-Will

[Turtle]

Originally Posted by Erasmus00

I think the question is about an imaginary, perfect roulette wheel. Obviously, in practice, any single roulette wheel will have a bias. Alternatively, we could rotate through a new roulette wheel every spin, to try and average out any bias effects. Or, given a real situation, we could try to calculate where the ball will land, given some initial conditions of the throw, which is, I believe, what the stanford guys did.
-Will

My point exactly...or close enough. In my view, using the roulette wheel as an example is in the least disengenuous and in the most an outright deception. If the point of the original post is just to go through some series of probability calculations of an idealised state, then all that ought to be given is the numbers and no connection made to any reality. Otherwise there is an implication that the calculation results actually represent reality regardless of how many times one lists caveats to the contrary.

All of probability - no matter how simple or complex - rests on a single assumption and all reality - no matter how simple or complex - denies that assumption; that is, all things being equal. Nothing ever was, is, or will be all equal. The devil is in the details, so I sometimes play his advocate.

[Simon]

Oh well Erasmus, you got there in the end.

Let's see if others beleive it.

I'm curious though. You said you missed the parameter of hitting exactly the right number of spins. But earlier in the thread you were already using the appropriate term "Mode".

I can't figure out, in this context, how Mode can mean anything except a specific number of spins that occurs the most often. Is it possible to have two numbers that are a subcategory of the same Mode? Of course you weren't actually saying that.

Nevertheless, you were thinking in terms of allowing any number of spins up to the most likely Mode, which you said was 36. In what sense did you think this was true? Were you saying that zero would occur "anywhere-before-the 37th-spin" more often than "anywhere-before-the-Nth" (where N is any other number)? In that case, we seem to be into overlapping scenarios where two numbers of spins belong to the same Mode. If the Mode referred to the most likely maximum number of spins, why stop at 36. Why not cover yourself and pick the highest one possible?

I'm just trying to get my head around what you thought you meant.

Simon

[Simon]

Oh and Turtle, I beleive I covered all the angles when I said: "given an unbiased roulette wheel"

This meant that the question should be treated purely as a problem in probabilty - nothing more, nothing less.

Giving tangible examples is more engaging - even for mathemeticians - whether reality happens to conform to not.

If an audience prefers all hypothetical scenarios - dice, cards etc - to be kept out of probability, then I'll need to rethink.

Personally it helps me.

Simon

[Turtle]

Originally Posted by Simon

Oh and Turtle, I beleive I covered all the angles when I said: "given an unbiased roulette wheel"

This meant that the question should be treated purely as a problem in probabilty - nothing more, nothing less.

Giving tangible examples is more engaging - even for mathemeticians - whether reality happens to conform to not.

If an audience prefers all hypothetical scenarios - dice, cards etc - to be kept out of probability, then I'll need to rethink.

Personally it helps me.

Simon

An euphemism for 'covered my ass' inasmuch as probability theory originated with Laplace trying to quantify dice and cards for his gambling friends. If an audience doesn't know what shows are available, what is the probability they will chose one they don't know about? My intent is nothing else than to make you(all) rethink in light of new discovery. Personally, the status quo offends me.

[eric l]

Originally Posted by Simon

Here is a probabilty puzzle, the answer to which may be disputed.

Given an unbiased roulette wheel with a single zero:

"How many non-zero spins are most likely to occur consecutively?"

To avoid any ambiguity, imagine you had to bet on precisely how many 1-36 numbers will come up between two zero spins. Assume that if you guessed correctly you would win a large money jackpot - and you're allowed to bet on any figure you like.

Which figure would you choose?
Or are they all equally probable?

Simon

Reading the question put in bold I can not agree with the proposed answer "NONE".

The chances that your first spin is a non-zero spin is 36/37 (on your roulette). There is one chance out of 37 that you can not go on for n consecutive non-zero spins.

The chances that you have two consecutive non zero spins is (36/37)*(36/37) or 0,9467; and so on.

[Simon]

Erik, the way I would explain it is this.

"How many non-zero spins are most likely to occur consecutively?"

In other words: how many 1-to-36-spins will most likely occur before a zero spin?

The probability of none is:
1/37 = 0.27027

The probability of one is:
(36/37)*(1/37) = 0.0262965

The probability of two is:
(36/37)*(36/37)*(1/37) = 0.0255858

And so the chance decreases.

It is always based on the same formula that was stated earlier:



In each case, the chance of "n" non-zero spins and one zero spin must be taken into account and multiplied to get the correct probabilty.

Convinced yet?

Simon

[eric l]

Originally Posted by Simon

The probability of none is:
1/37 = 0.27027

Convinced yet?

Simon

Not at all !

For starters, in my book 1/37 = 0.027. That is the chance for a zero spin.

This means that the chance for a non-zero spin is (1 - 0.027).

The chance for two consecutive non zero spins is (1 - 0.027) * (1 - 0.027).

This of course will go on decreasing as you increase the number of spins, until you reach a value below 0.50, at which the chance that you have at least one zero spin in the series becomes higher than the cance that you have no zero spins.