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!

Simon

[Erasmus00]

Originally Posted by 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!

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?


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/...allStreet.html
It's a long read, but it's a seriously complex subject. But then chaos favors the prepared imagination.

[Erasmus00]

Originally Posted by 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/...allStreet.html
It's a long read, but it's a seriously complex subject. But then chaos favors the prepared imagination.

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.

Simon