The Roulette Conundrum

[Simon]

Yes Erik, I did leave out the extra 0. (Sorry about that)

I meant to type: 1/37 = 0.027027

That is the exact chance of spinning a zero straight away, without hitting any 1-36 numbers. That is why I say "none" has the highest probability.

Admittedly, this was the question I put in bold:

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

Potentially this might ignore whether zero comes up at all. But in my view, the word "consecutively" in this context does mean "before the first zero comes up" Just in case that wasn't everyone's interpretation, I wrote immediately afterwards:

To avoid any ambiguity, imagine you had to bet on precisely how many 1-36 numbers will come up between two zero spins.

This leaves no doubt. I'm asking how many consecutive spins are most likely to occur before the first zero comes up. I maintain that NONE is more likely than any other number.

Now, if you chose the other interpretation which might be "How many consecutive non-zero spins are most likely to occur, regardless of what follows?" - then we're into potential confusion. If a sequence of 12 non-zero numbers is followed by another non-zero number, in what sense is it a sequence of 12 but not 13? Obviously it would be a sequence of both.

What you're doing is focussing on the probabilty of getting N non-zero spins, while ignoring what comes after N. But ok, let's do that.

The chance of at least one non-zero spin, followed by anything is of course (36/37)*(37/37) = 0.9729729.

The chance of at least 26 zero spins, followed by anything is 0.490477, which is the benchmark taking you below 50%.

From this, you have deduced that 26 is the most likely number of non-zero spins to occur.

I'm compelled to ask: in what sense does 0.490477 (the chance of at least 26 non-zero spins) have a greater probability than 0.9729729 (the chance of at least 1 non-zero spin)?

Using your approach, the answer has to be that one non-zero spin has a greater probabilty than any other number of consecutive non-zero spins.

Also - again following your method - you could argue that the chance of at least 0 non-zero spins, followed by anything, is 100%. So NONE wins here as well.

Nevertheless, all these calculations inevitably overlap with each other.

In order for any number of consecutive non-zero spins to be distinctive, it must stop at that number.

6 consecutive non-zero spins has to mean a sequence of 6 non-zero spins followed by a zero spin - otherwise it will also belong to any higher number of non-zero spins.

Do you agree?

[eric l]

Simon, I've been thinking it over (and over and over) and I see your point.
What has happened is that you look at it from the player's viewpoint, while I looked at it from the viewpoint of the bank. And I was biased by the paradox that the chances to have consecutive non-zero spins are lower than the chances that you have at least one zero spin with well below 36 or 37spins.

The player is normally not interested whether the spin is zero or a non-zero. Any number that is not the number het put his money on, makes him loose.
But to the bank, zero-spins are all important : the bank can collect all the money on the table without having to pay anything to anyone. To the bank non-zero spins are zero-operations (probabilitywise). The dollar (or pound, euro or yen) that Abe put on black will go to Bill who put his dollar on red (or vice versa if the number coming up was black); likewise the dollar put by Charley on odds will go to Dave who put his dollar on evens (or vice versa). If a zero turns up, it is considered to be neither black nor white, neither odd nor even, so the bank collects from all four. I could extend this to playing on numbers, or "a cheval" on two or four numbers... it would only make this posting longer (and more boring).
The paradox that after only 26 spins, the chance that you have had at least one zero-spin is higher than the chance for consecutive non-zero spins, is one of the reasons the bank is usually the winner at the end of the day (or night). The other reason is that there are always players who think they can outsmart the roulette, but by their "clever" system actually improve the bank's chances rather than their own.

[Farsight]

Apologies, I haven't read the thread, but I do know something about casinos. I used to work in a casino.

If you want to win money in a casino play blackjack. There's a simple set of rules to use that tilt the odds in your favour.

http://en.wikipedia.org/wiki/Blackjack

You can also win at roulette. Just through chance. But it's such fun that people don't stop when they're ahead. They only stop when the night is done or they're cleaned out. If you have the self-control to stop when you're ahead, you end up getting banned. No kidding.

[ronthepon]

Originally Posted by Popular

Apologies, I haven't read the thread, but I do know something about casinos. I used to work in a casino.

If you want to win money in a casino play blackjack. There's a simple set of rules to use that tilt the odds in your favour.

http://en.wikipedia.org/wiki/Blackjack

You can also win at roulette. Just through chance. But it's such fun that people don't stop when they're ahead. They only stop when the night is done or they're cleaned out. If you have the self-control to stop when you're ahead, you end up getting banned. No kidding.

Actually, this was about probability.

[TheBigDog]

This is always a weird study. The fact will always remain that no sample of size x will have any greater probability than any other sample of size x. The proximity of an event that happens outside of the sample having an effect on the sample, while tempting, is meaningless.

Bill

[Farsight]

Suppose I'm only allowed to bet on zero. I've just won. Now, should I miss a go? Should I miss ten goes and go to the bar? No. There isn't any most likely number of goes before the next zero comes up. It's just a 1/37th probability every time, whether it came up last time or not. Surely? Is that what you're saying Simon?

[ronthepon]

No. It's like betting money on: "There'll be ten spins, and if the eleventh spin show's the first zero, then I win."

[Simon]

Popular wrote:

Suppose I'm only allowed to bet on zero. I've just won. Now, should I miss a go? Should I miss ten goes and go to the bar? No. There isn't any most likely number of goes before the next zero comes up. It's just a 1/37th probability every time, whether it came up last time or not. Surely? Is that what you're saying Simon?

Yes, the chance of getting a zero on any individual spin is always 1/37, regardless of what has come up before.

Nevertheless, I'm saying that you're more likely to get a zero immediately than get any chosen number of non-zero spins first.

Just to re-state what should be very obvious: the scenario of my original post is an excercise in probabilty, not related to any known wagering rules.

No betting strategy for any casino is being recommended!

If the scenario was played as a game, it would be like a special one-off promotion. The croupier would say:

“We’ve looked at the last ten million spins.
We found the first zero.
We then found the second zero.
We counted how many 1-36 spins occurred between the first and second zero..
We then found the third zero.
We counted how many 1-36 spins occurred between the second and third zero.
And so on.

Whichever number of 1-36 spins appeared between two zeros the most often is the winner.

For £1 you can buy a ticket with any figure of your choice from 0 to infinity. This will be the number of 1-36 spins that you think has occurred the most often between two zeros.

Those who are correct will share the jackpot.”

Assuming an unbiased roulette wheel (chaos theory notwithstanding), I think ten million spins is a large enough sample for probability to be played out.

Those who chose NONE as the most frequent number of consecutive 1-36 spins between two zeros would be most likely to win the prize.

That is the answer that may seem unbeleivable.

Simon

[Simon]

Popular you raised a challenging question - one that I missed.

Suppose there are two players betting on zeros at different times.

Player 1 is hoping for succesive zeros. After each zero comes up, he bets on the next spin.

Player 2 is hoping for zeros separated by 36 other numbers. After each zero, he waits for 36 non-zero spins. Only then does he bet.

It seems clear that for each bet, the chance of a zero is the same for both players - 1/37.

Nevetheless, the following propositon is also true.

The probability of spinning a zero immediately is 1/37
The probabilty of spinning exactly 36 consecutive non-zero spins followed by a zero is about 1/99.

This appears to be a contradiction.

But is it?

Simon

[LaurieAG]

Originally Posted by Simon

The probability of spinning a zero immediately is 1/37
The probabilty of spinning exactly 36 consecutive non-zero spins followed by a zero is about 1/99.

This appears to be a contradiction.

But is it?

Simon

Simon, it's interesting when you look at the probability of one persons bet winning over the other persons during any test period.

Once the average number of spins before spinning zero goes above 36 the person betting on 36 non-zero spins will have more unsuccessful bets than the person betting on zero after zero spins. Once the average goes below 36 the bias goes to the zero after zero option.

There are other betting related penalty biases that occur when there is a run of zero throws as opposed to a run of any other numbers. A run of zeros will result in a run of successful bets (run-1) for one person while the other person will make just as many uncertain bets (run -1) blanket covering a period 36 days hence. This bias does not impact on the number of bets because no losing bets are made during runs of non zero numbers followed/preceded by other non zero numbers by either person.

At the extremes, the person betting on zero after zero has a better probability of winning when the average spins are much less than 36, loses when the average spins are 36 or greater and gains greatly when a sequence of zero's are spun. The person betting on zero after 36 spins only wins when the average number of spins before zero is close to 36 and can lose very big when a run of zero's are spun.

It does seem to put the zero after zero bet into a favourable position though.