Question about probabilities on roulette

[MARAGATO]

Hi everyone

I have a doubt about math of roulette, and think this is the place to ask someone to help me.


lets talk a bet in same collor/odd four times in sequence. Lets say you bet the most basic sequence ever, 1, 2, 4 and 8 chips.

you lose 15 when you dont hit, and win 1 when hit.

the probabilities to a bet in 18 numbers in 37 do not hit in 4 is calculated by (18/37)^4, 0,056012264%.

so in a infinite cicles of attempts, in average:


you win 1 chip 94,39877364%, or 94 chips each 100 cicles
you lose 15 chips 0,056012264%, 84 chips each 100 cicles

was that a correct favorable theorical perspective?

[ati]

Hi, welcome to the forum.
I'm not a math expert so I could be wrong but here's what I think.

Your calculation seems correct, but it is for no zero roulette.
For single zero, the probability of losing 4 EC bets in a row is (19/37)^4 which equals to 0.0695, so in 100 cycles you would lose about 6.95*15=104 chips , and win 93 chips.

Keep in mind that these are only calculations and averages. Real results will be different.

[MARAGATO]

Ty

So if a make a little increasei the bets to adapt to percentage, lets say:

1
2,4
4,7
1

so with new numbers (19/37), stay:

you lose 18.1  when you dont hit, and win 1,4 on average when hit.


you win 1,4 chips 93,0464352%, or 130 chips each 100 cicles
you lose 18,1 chips 0,69535648%, 126 chips each 100 cicles

in thesis, if use those bets in n cicles, will could have a theorical edge of 3,5%?

I understand you will lose because casino has infinite money to pay and you dont, so when the lose, that happends 0,7%, happens, lets say, 50 times in last 100 attempts, you will broke, and you cant do anything about it. I just wanna know  its mathematically correct.

[Steve]

Maybe the attached spreadsheet will help. It will tell you wins, losses, edge, and profit. You can extract probability from the calculations too. All you do is edit the values in the green boxes.

[BellagioOwner]

the probabilities to a bet in 18 numbers in 37 do not hit in 4 is calculated by (18/37)^4, 0,056012264%

The probability to bet 18 in 37 (european)and NOT hit is not (18/37)^4. This is the probability TO HIT. The probability to NOT hit is whatever outcome makes you lose. So actually (19/37)^4

Times you will lose:  (19/37)^4 = 0.0695
Times you will win: 1- (19/37)^4 = 0.9304
With that in mind... Let's calculate based on what you lose and win

1*[1- (19/37)^4] - 15*(19/37)^4 = 1*0.9304 - 15*0.0695 = 0.9304 - 1.0430 = -0.1126

On Zero Roulette you would just finish (theoretically in perfect balance situation) at even at your money. +- 0. That's why they take commission on winnings to keep them profitable.

Betting on 18  in 36 (No Zero Roulette) you would lose (18/36)^4 and win the rest of them [1-(18/36)^4]

So... 1* [1-(18/36)^4] - 15*(18/36)^4 = 1*0.9375 - 15*0.0625 = 0.9375 - 0.9375 = 0


PS: Don't get me started why people are still playing on American Roulette. I have no fricking  clue!!

PPS: Welcome to the forum.