The Roulette Conundrum

[eric l]

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:

 

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

 

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]

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.

[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]

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]

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.

[Farsight]

I don't think it's a contradiction Simon. You're forgetting about the intervening zeros. To back up what you were saying earlier:

 

Suppose I'm betting when the next zero will come up, and I'm wondering whether it'll come up on the first spin or the second.

 

Like you said there's a 1/37th chance it'll come up on the first spin. Which means the chances of it NOT coming up on the first spin are 36/37th.

 

The chances of it coming up on the second spin are 1/37th, but the chances of it NOT coming up on the first spin AND coming up on the second spin are 36/37th * 1/37th.

 

Which means I'm better off betting on the first spin, regardless of what came up last time. If I'm foolish enough to bet it'll come up next on the 37th spin the odds are 36/37th^36 * 1/37. Which is probably your 1/99.

[Qfwfq]

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?

It is not a contradiction at all!

 

The probability of the next spin giving a zero is indeed 1/37 but this is a different thing than "the probabilty of spinning exactly 36 consecutive non-zero spins followed by a zero", which does have a lower probability, slightly less than 1/99.

 

The difference between the two is that your player 2 is really betting on a single spin, the fact that he waits for 36 in a row to be non-zero has no bearing on the probability of the following result. If you're willing to bet against a result that is already known to have occurred, let me know and I'll finally get rich! :)

 

The formula that Ron gave is true, based on the fact that the results of each run are independent of each other, and is a calculation of the probability that, in a set of n + 1 specified spins, only a specified one of them will be zero. That isn't what player 2 is achieving if he waits until 36 in a row have given no zero. If he only bets at that point, the house will never grant him a ratio greater than 37 (indeed they would grant the customary 36).

[Qfwfq]

Note:

 

The computation is valid even if the word "specified" is understood to mean as you go along, providing the spin is specified before the result, along with either zero or not zero. The player could agree with the house that, before one spin, he can say "zero this time" and the opposite before each of n other spins, at his choice. Neither order nor even consecutivity make any difference to the computation of probability, as long as the player specifies spins before the result is predictable. They could even agree to the player going from wheel to wheel or coming back another day, no difference.

[Simon]

Agree with all of the above.

 

Incidentally, I did have "none" in mind when I first created the problem. Ron was the first person to choose that answer without prior influence. Others needed some serious persuasion. Well done Ron! :confused:

 

There's another twist to the two player scenario.

 

We're agreed that if probability is played out, the casino will win from both players. The House pays only 36-to-1 on a 1/37 bet.

 

The question is which player is likely to lose more?

 

Both players will be betting £1 on zero.

 

Player 1 will only bet after a zero is spun.

Player 2 will only bet after a zero is spun + at least 36 other numbers.

 

The chance of getting Player 1's conditions is 1/37

The chance of getting Player 2's conditions is 1/99

 

Player 1 will bet £1 almost three times more often than Player 2.

 

So Player 1 is likely to lose nearly three times as much as Player 2.

 

On the other hand, if the House paid in the player's favour (e.g 38-to-1), Player 1 will win nearly three times more than Player 2.

 

This illustrates why there is no contradiction. Win or lose, Player 1 is likely to get more betting opportunities than Player 2. This is simply because an unconditional zero will occur nearly three times more than a zero + at least 36 other numbers.

 

By the same ratio, two consecutive zeros is nearly three times more likely than than two zeros separated by exactly 36 other numbers.

 

This is where we came in. :)

 

Simon