The Roulette Conundrum

[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

[eric l]

The probability to have something else than a zero is 36/38 (on a common roulette, with a 0 and a 00). That is for one spin.

For n spins, the probability is (36/38)^n (36/38 to the power n).

If this value reaches 50 % (or 0.5), the probability that you have no zero spin is equal to the probability that you have had at least one zero spin.

With two zeroes this will be after 12 or 13 spins, with only one zero this would be after 25 or 26 spins. You can check this easily with a formula in Excel.

Of course, even with an unlimitted number of spins, you can never be sure that at least once you had a zero. An unbiased roulette "forgets" what numbers have come out before !

[Simon]

I wrote:

Given an unbiased roulette wheel with a single zero

 

In Britain we only have single zero roulette wheels. :)

 

Eric, just so I've I understand your answer: you've said that assuming we have an unbiased single zero roulette wheel (i.e one with 37 numbers rather than 38), the most probable Mode number of consecutive non-zero spins should be 25 or 26. Whereas if it's a double-zero wheel, your answer is 12 or 13 spins.)

 

Have I understood you correctly?

 

If so, I must disagree in both cases.

 

Simon

[ronthepon]

Allright, here's my stab at the question.

 

It will be one if the first event gives a negative (non-zero) and the second gives a positive.

 

Chances are = (probability of negative in first)(probability of positive in second)

 

= (37/38)(1/38)

 

Right? now taking two as our hopes,

 

Chances = (probability of negative in first)(probability of negative in second)(probability of positive in third)

 

= (37/38)(37/38)(1/38)

 

Similarly for three,

 

= (37/38)(37/38)(37/38)(1/38)

 

and for the n'th,

 

= ((37/38)^n)(1/38)

 

Seeing that 37/38 is less than one, I'd put my money on one. If possible, zero, because it gives a probability of 1/38, the highest one.

[eric l]

Well, Simon, you make me curious for your answer, and for the way you come to it.

[Simon]

Ok Erik, here is my answer: NONE.

 

I agree with Ronthepon.

 

By my calculation, you are more likely to spin a zero straight away than achieve ANY chosen number of consecutive non-zero spins. (This answer applies to both to a single-zero and double-zero roulette wheel).

 

In second place, my answer is ONE - you are more likely to achieve just a single 1-36 spin than any other number of non-zero spins. Again, I agree with Ronthepon.

 

In other words, I'm saying that if you compare any two figures of consecutive 1-36 spins, you should always bet on the smaller one.

 

NONE is smallest figure of all.

 

Many will find this unbeleivable! :D

 

Simon

[Erasmus00]

Ok Erik, here is my answer: NONE.

 

I agree with Ronthepon.

 

By my calculation, you are more likely to spin a zero straight away than achieve ANY chosen number of consecutive non-zero spins. (This answer applies to both to a single-zero and double-zero roulette wheel).

 

In second place, my answer is ONE - you are more likely to achieve just a single 1-36 spin than any other number of non-zero spins. Again, I agree with Ronthepon.

 

In other words, I'm saying that if you compare any two figures of consecutive 1-36 spins, you should always bet on the smaller one.

 

NONE is smallest figure of all.

 

Many will find this unbeleivable! :)

 

Simon

 

I'm going to have to disagree with this analysis. First, on the grounds of common sense: Anyone who has ever played roulette will tell you that landing on any one specific number is pretty rare. In fact, all things being equal and averaged over a long period, about 1/37 of the rolls will hit 0. If these were equally distributed, 36 would be the perfect bet, but, obviously, these are not going to be equally distributed. If this analysis were correct, it would be very common to get two zeroes in a row (or, by symmetry, two of ANY number in a row). I think thus far, I most agree with Eric.

 

So whats wrong with your analysis? Well, you are looking at the odds of rolling EXACTLY one roll then hitting 0 compared to the odds of rolling exactly two rolls then hitting 0, etc. What we should be looking at is the average number of rolls between hitting 0 twice.

 

So whats wrong with the following analysis: Compare the probability of rolling x consecutive non zero rolls and comparing this to the 0 roll as follows?

[math] \left(\frac{36}{37}\right)^x > \frac{1}{37} [/math]

 

This method suggests (incorrectly I think) we should roll 130 times or so.

 

-Will

[Turtle]

Seriously, have you all missed out on chaos theory?

A quick Google of "Stanford Chaos Theory guys roulette" gave this:

http://www.cs.brown.edu/research/ai/dynamics/tutorial/Documents/CrackingWallStreet.html

It's a long read, but it's a seriously complex subject. But then chaos favors the prepared imagination.;) :D

[Erasmus00]

Seriously, have you all missed out on chaos theory?

A quick Google of "Stanford Chaos Theory guys roulette" gave this:

http://www.cs.brown.edu/research/ai/dynamics/tutorial/Documents/CrackingWallStreet.html

It's a long read, but it's a seriously complex subject. But then chaos favors the prepared imagination.;) :D

 

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

[Simon]

Erasmus wrote:

All things being equal and averaged over a long period, about 1/37 of the rolls will hit 0.

 

That is correct.

 

What we should be looking at is the average number of rolls between hitting 0 twice.

 

The actual average is, perhaps unsurprisingly: 36 (or 19 for a double-zero wheel).

 

However that is not what is being asked. The question is explicitly requiring the most probable Mode number of consecutive 1-36 spins between two zeros. In other words, if you took a large enough sample, which single number of consecutive non-zero spins would actually occur the most often.

 

I stand by the same answer: NONE. If it had to be higher than that, I would bet on ONE.

 

I acknowledge that this conclusion will be contraversial. :D

 

Simon

[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

 

[math] \left(\frac{36}{37}\right)^n\frac{1}{37} [/math]

 

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

 

[math] \sum_{n}n\left(\frac{36}{37}\right)^n\frac{1}{37} [/math]

 

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]

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.:cup:

[Simon]

Oh well Erasmus, you got there in the end. :cup:

 

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. :cup:

 

Simon

[Turtle]

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. :hihi:

 

Simon

 

An euphemism for 'covered my ***' 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.:naughty: