Roulette Losing Streaks - What Is Possible

[Alicia Mitchell]

A lot of people ask me what kind of losing streaks are possible in
roulette, and the simple answer is really up to fate and the 'gods' if
you like. While not common in roulette, immense losing streaks have
been recorded and I've even seen some personally, and that's what I'd
like to talk about today.
To start, whenever we're speaking in roulette terms, we always think
of odds. The odds of a certain number coming up or the odds of a
certain number not coming up a certain amount of times in a row. But
the thing is that odds are just that; odds. They're not a guarantee of
anything and are just a guide to what will most likely happen, and
most likely does not mean definitely.
One of the largest losing streaks I'd ever heard of was at one of the
roulette tables inside the Venetian Casino in Las Vegas. Apparently
the '2nd 12' of the board had not come up for about 17 spins, which is
quite unlikely and according to the law of averages and odds should
only happen once every 1,500 spins or so. A punter at the casino
noticed this and began to make large $100 bets on this part of the
board, certain that it would come up in the next few spins.
After losing a further three spins in a row betting $100 each time
(which if he won would have evened out his losses) he continued to bet
in the hope he would recoup some kind of loss. 36 further spins later
and a total of $3900 down, the man gave up. Even after he did, the
ball failed to land on the '2nd 12' of the board another 7 times, for
a total of 63 consecutive spins of it not landing on that section of
the board.
But for me personally, the largest streak of what I'd call bad odds
(or bad luck) was in an Australian casino where the ball landed on the
'0' slot six times in a row. Now this might not sound as unlucky as
the last guy, but keep in mind he had a massive 33% chance of it
landing on that section of the board. For the ball to land on '0' even
once you only have a 2.7% chance, but six times consecutively?
0.0000000389%.
So just because it's unlikely, doesn't mean it won't happen. Keep that
in mind next time you're at the table.

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[The Midnight Skulker]

Post by Alicia Mitchell
One of the largest losing streaks I'd ever heard of was at one of the
roulette tables inside the Venetian Casino in Las Vegas. Apparently
the '2nd 12' of the board had not come up for about 17 spins, which is
quite unlikely and according to the law of averages and odds should
only happen once every 1,500 spins or so.

Sorry, but assuming a single-zero wheel I calculate once in about 785
sets of 17 spins. I'll leave it to Steve Fry to calculate how many
spins one can expect to make before seeing a streak of 17 non-second-
dozen results.

Post by Alicia Mitchell
A punter at the casino noticed this and began to make large $100
bets on this part of the board, certain that it would come up in
the next few spins.

A classic case of The Gambler's Fallacy.

Post by Alicia Mitchell
So just because it's unlikely, doesn't mean it won't happen.

Or as Redleathers was fond of saying, an unlikely event is certain to
happen, just not very often.


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[SteveF]

Post by The Midnight Skulker

Post by Alicia Mitchell
One of the largest losing streaks I'd ever heard of was at one of the
roulette tables inside the Venetian Casino in Las Vegas. Apparently
the '2nd 12' of the board had not come up for about 17 spins, which is
quite unlikely and according to the law of averages and odds should
only happen once every 1,500 spins or so.

Sorry, but assuming a single-zero wheel I calculate once in about 785
sets of 17 spins. I'll leave it to Steve Fry to calculate how many
spins one can expect to make before seeing a streak of 17 non-second-
dozen results.

I didn't see the original post. Was it in this newsgroup?
At any rate, this Roulette example displays the classic error in
computing statistics with streaks. A couple of years ago there was
heated discussion on the probability of streaks where I tried to
provide some education, and it prompted me to add an entire question

Post by The Midnight Skulker
Simple probabilities are easy to understand: if you know
that the probability of a seven showing when a pair of dice
are thrown is 1/6, then you conclude that on average every
six rolls will have a seven showing once. That is another
way of saying that the expected number of rolls to see a
streak of one seven is 6 rolls. However when streaks --
consecutive identical outcomes -- are involved things are
not so simple because the streak is more than one event.
The layperson may say that the probability of getting a
streak of two sevens in a row is 1/36, so the expected number
of rolls is 36 to get a streak. But that would be wrong
because it is ignoring the effects of other rolls in the
sequence that makes up the streaks and no-streaks of the
whole set of rolls. In fact, one needs 42 rolls to get a
streak of two sevens on average.
The equation for the expected number of events to get a
streak of "m" successes where the probability is "p" that
-m
p - 1
T = ------------
1 - p

In the quoted Roulette example, the issue is a streak of 17 spins without
the middle 12 numbers (I think that is called the middle street) showing.
In single-zero Roulette, there are 37 numbers, so the probability of no
middle street is 25/37, which is .6757 = "p".

For a streak of 17 spins ("m"=17) with no middle-12 numbers showing:
ignoring the nature of streaks one might compute the probability of that
happening as p^m, and the expected number of spins as the reciprocal,
p^(-m). For this example that is .6757^(-17) = 784.

However, the correct expected number of spins is 2415, using the equation
above. So the estimated 1500 spins quoted above was even too low (for
double-zero Roulette the expected number of spins is 2003.)

The concept of "expected number" of spins may be confusing to some. The
number is an average (mean) of a distribution, and it is not a normal (or
Gaussian) distribution, but one with a head on the left an a long tail on
the right. This probability distribution function is a function of the
probability of achieving the desired streak versus the number of spins
required to get the streak.

-- Steve F.

[The Midnight Skulker]

I didn't see the original post.  Was it in this newsgroup?

Yes, but the author is not a regular poster so, also considering the
subject, you may have ignored or filtered it. In fact I almost didn't
read it myself.

ignoring the nature of streaks one might compute the probability of that
happening as p^m, and the expected number of spins as the reciprocal,
p^(-m).  For this example that is .6757^(-17) = 784.

Well, I got 785 ->sets<- of 17 spins. IOW if one divides 13345 spins
into 17-spin sets, there will be, on average, one of those sets that
contains none of the numbers from the second dozen.

However, the correct expected number of spins is 2415 ... .

As we determined the last time we discussed a problem of this type,
the difference between your answer and mine is that my calculation
assumes that a set of 17 spins is completed even after one of the
middle dozen numbers appears whereas your answer assumes a set is
abandoned (and a new one started) when one of the "forbidden" numbers
shows.


I'm not sure what the author of the original post is trying to do.
She -- at least she is ostensibly female -- has posted a few articles
about roulette here and elsewhere. The advice therein, while rather
mundane, isn't all that bad, and there doesn't seem to be any attempt
to steer readers to a gambling site. Nevertheless, for some reason I
felt obligated to point out that Ms. Mitchell is numerically
challenged lest she attempt to sell us something in the future.


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9 * 3 aka Van Lewis
aka ***@gmail.com
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