Roulette Probability (Black or Red)

[Steven]

Would be grateful if someone could help settle an argument:

The probability of hitting Red 10 times in a row on one table is 48.64%^9, very small chance indeed.

If you bet 5 times on black and therefore have lost, then move to a second table and again bet 5 times on black, is the mathetmatical probability of this occurring the same as 10 blacks appearing in a row on one single table?

The debate stems from the fact that in a two table scenario, we know the history of its last 5 spins. But my argument is that we are calculating the probability of one person receiving 10 blacks in a row rather than one table or any number of tables producing ten blacks in a row.

I suppose it may be the same as tossing a coin. The chances of tossing 10 heads in a row using one coin is the same as tossing 10 heads in a row swapping the coin after 5 throws.

Hope you can help.

 

[Automatic Monkey]

Assuming an unbiased wheel, yes

On the other hand, no roulette wheel is truly unbiased (although it may be a thousand years before the bias becomes mathematically significant.) So if I was a roulette player and saw a wheel give black 10 times in a row, this would not prove it significantly biased towards black but the likelihood of it having such a bias is greater than a wheel that has not shown this deviant result. Just imagine watching a coin flipper flip heads 10 times in a row... bet you a quarter he's cheating.

 

[stainless steel rat]

probability...

>>Would be grateful if someone could help settle an argument:

>>The probability of hitting Red 10 times in a row on one table is 48.64%^9, very >>small chance indeed.

I don't quite see the number you are giving. The probability for hitting Red is 47.4% (18/38, 18 reds, 18 blacks, 0 and 00). The probability for hitting two reds in a row is 47.4% ^ 2 (47.4 squared). To hit 10 in a row it should be 47.4% ^ 10, since there are ten consecutive trials and you multiply the probability for each together, giving 10 rather than 9 for the power.

>>If you bet 5 times on black and therefore have lost, then move to a second table >>and again bet 5 times on black, is the mathetmatical probability of this >>occurring the same as 10 blacks appearing in a row on one single table?

Depends on what you are asking. Is the probability for the _entire_ string of 10 the same? yes it is. After you see 5 in a row, is the probability for seeing 5 more in a row the same? No. It is just .47^5, since the first five have already happened.

>>The debate stems from the fact that in a two table scenario, we know the history >>of its last 5 spins. But my argument is that we are calculating the probability >>of one person receiving 10 blacks in a row rather than one table or any number >>of tables producing ten blacks in a row.

You have to be careful. I have a book where the author makes a total ass of himself, talking about a Martingale betting scheme. He correctly says "the bets can get high quickly." But then he incorrectly says that you can cheat this by watching the wheel, picking the color you want to bet on, and wait until it loses twice in a row. Now start your Martingale bet at the table min, pretending that you have already lost twice with a null-bet out. he says that the probability of hitting 10 more losers is way lower than the probability of hitting ten in a row without having seen the two prior losers. Dead wrong and utter garbage. If you hit 100 reds in a row, the probability of hitting 10 more in a row is exactly as computed above. Not .47 ^ 110.

It doesn't matter whether you play ten spins at 1 table, or 1 spin each at 10 tables, the probability of 10 reds in a row for _you_ is exactly the same, assuming there is no physical hardware bias to the wheel of course.

>>I suppose it may be the same as tossing a coin. The chances of tossing 10 heads >>in a row using one coin is the same as tossing 10 heads in a row swapping the >>coin after 5 throws.

you got it...

>>Hope you can help.

 

[stainless steel rat]

one addition

If you are talking about a "European wheel" (with just 37 numbers total including the green 0) then your percentage is correct, but you still need the power 10, not 9.

 

[Norm Wattenberger]

A few points *LINK* *PIC*

1. Biased wheels exist. But my understanding is that they are few and far between - at least in the US. This is not an effective method of Roulette play.

2. Red/Black would not be an effective method of determining, or playing, a biased wheel. Red and Black are interspersed and not useful indicators.

3. I believe Roulette can be beaten - without biased wheels. But, it is more difficult than BJ. And the analysis requires far more than casual observation. Read Laurence Scott for info as he is the expert.

 

[Cyrano]

Shoulda just said...

"What are the chances of me losing on Red (Black) 10 times in a row?"

 

[Steven]

Thanks

Guys, thanks for the input. You have confirmed what i have thought and settled the debate. The chances of getting 10 reds in a row is the same if you do it on one single table or ten.