No zero roulette house edge ?

I am sure this has been discussed but I could not find it .. - I want to play online at a no zero roulette. (36 numbers and NO 0`s)

My strategy would be to bet 1 unit on 35 numbers out of the total 36 existing numbers, excluding the one which previously landed.

Basically I would risk 35$ to win 1$ betting that there will not come twice in a row the same number. - OK, now my question is:

What are the exact odds (%) to come the same number twice in a row on a no zero roulette ? - I am gambling still on a 0% HE ?

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Quote: Ibeatyouraces

1 in 36 or 0.027777.....%

google it ; there are claims that it is zero house edge.

for example:

http://roulettepayouts.org/no-zero-roulette.htm

PS: Imaginary conversation at online casino:

Guy: Hey boss how can we have zero house edge on this?

Boss: Stupid! in our online casino, money can only come in. Money never goes out. How can we lose?

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Quote: flipdisk

If you believe in the Probability Theory then the chances are reduced after the first time it is shown. Isn't a Mathematical Law but still most gamblers use it on games of chance. Example: Flip a coin, you have 50/50 chance of getting head. The next time, the odds are the same as before and that is true but with that Theory, it is still true but its not. If you do it 100 times, will you get 100 heads? Chances are greatly low but the odds are still the same. I have a chart showing that with European Roulette somewhere.

"Probability Theory" is the opposite of that. The term you're looking for is "Gambler's Fallacy."

Quote: PlayHunter

What are the exact odds (%) to come the same number twice in a row on a no zero roulette ? - I am gambling still on a 0% HE ?

The odds to come the same number twice in a row is 1/36 = 2.77%. (It is the same probability that any number you guess will come up.)

Second question: Yes, your strategy will yield a house edge of 0%. (In fact any strategy has a house edge of 0%, because each single bet has a house edge of 0%, and each spin is statistical independent.).

One note: Because the no-zero roulette is fully symmetrical between you and the house, the total variance of betting 35 numbers with $1 is exactly the same as betting $1 on a single number. Thus you play at the highest variance possible for your unit bet of $1.

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Thank you MangoJ .. excellent ! -- BTW do we have a "Thanks!" button here ? (Wizard would be great if we can have this feature)

Quote: MangoJ

One note: Because the no-zero roulette is fully symmetrical between you and the house, the total variance of betting 35 numbers with $1 is exactly the same as betting $1 on a single number.
Thus you play at the highest variance possible for your unit bet of $1.

Which gives the best chance to double a bankroll or hit any win goal?
EV and Variance being the same.

Now we look to the Skew.
Very positive for the 1# bet and very negative for the 35# bet.

Example: $100 bankroll and $200 target goal (double a $100 bankroll)

Betting $1 on 1# = about 47.231% chance of success to win at least $100
about a 44.634% chance of winning more than $100

Betting $35 to win $1 = about 43.683% chance of winning $100

Thank you guido111 ! Very useful information as well !

Quote: guido111

Example: $100 bankroll and $200 target goal (double a $100 bankroll)

Betting $1 on 1# = about 47.231% chance of success to win at least $100
about a 44.634% chance of winning more than $100

Betting $35 to win $1 = about 43.683% chance of winning $100

Betting $100 on black: 50.000% chance of winning $100.

Which one should you choose ?

Quote: MangoJ

Betting $100 on black: 50.000% chance of winning $100.

Which one should you choose ?

yeah, i thinking about that.
$100 on a color, or odd/even or high/low.

Quote: MangoJ

Betting $100 on black: 50.000% chance of winning $100.

Which one should you choose ?

Betting $100 on black is a different question from the OP.
I choose to bet $100 on red... because Red wins 50% of the time.
And one can still lose 10 times in a row with a 0% house edge.


OP first asked about betting $35 to win $1.

He is thinking out loud about the highest winning probability to win $1 on one bet.
He has it.

Then OP talks about losing twice in a row.

Seems to be, and I may be 100% wrong on this, he wants to make this bet for only 2 spins.

OP, speak up, this is your thread.
What do you really want??
How much is your bankroll and how many spins do you expect to bet on and how much do you want to win?
This is the real world.

Then his chances of losing both spins would be (1/36)^2 before his first bet.
Happy to win a big $2?

He should be asking the probability of winning 5 times in a row, or even better,
maybe 36 times in a row to show a profit before the first loss.

Quote: 7craps

Then OP talks about losing twice in a row.

Seems to be, and I may be 100% wrong on this, he wants to make this bet for only 2 spins.

As I understand OPs question, he evaluates the house edge for this strategy (betting all but the last number on a no-zero roulette game).
Since he loses all his bets every time two consecutive numbers show up, it *is* a legit question to ask for the probability the same number will come up twice in a row (1/36). If one does not oversee the whole game in the first place (no HE on any bet, and hence EV=0 on any strategy), this is the correct way to determine the EV of the chosen strategy.

Players EV is then $1 * 35/36 - $35 * 1/36 = 0 (of course).