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Roulette and Maths

AuthorRoulette and Maths
First of all this is NOT a roulette suggestion thread. This is a thread related to mathematical problem which I got from the roulette.

For a moment let us ignore 0,00 part of the roulette. And assuming I can only bid RED or BLACK. My spin history says there are 5 consecutive BLACKS. So will the probability of getting a RED in the sixth turn be higher or getting 6 consecutive BLACKS be more.

My first though is that we can treat this as one-dimensional random walk problem and black + 1 red is higher. But One issue is that the number of trials is not fixed. Will my hypothesis be still true?
Probability of Red after one Black = 0.5
Probability of Red after two Blacks = 0.5
Probability of Red after n Blacks = 0.5

If you are only talking about the probability of the last bet then it doesn't matter. But if you calculate overall probability of n blacks and a red of then it might be different.
You are walking into the trap that makes casinos rich ;) : Red and Black are no mathematical concepts, they are psychology. Roulette is only numbers: you get a random pick out of 36 (normally 37, 38) numbers. Cha
Chances of getting one number is always 1/36.
chances of getting red next is always 50%. Chances of getting black next is also 50%.
Yeah, the spins are independent of each other, and thus independent probability.

Getting 6 blacks in a row has the same probability as black red black red black red.
It's simple do some random math calculations with respect to previous spins and bet on whatever number u get with those calculations.
U can never predict as to what number will come.
In the end it's all about luck
That's single probability you guys are talking about. But consider that I have 8 turns. Probability of getting 8 Reds is 1/2^8. While probability of getting 7 reds and 1 black is 8C7* (1/2)^8. Which is significantly higher.
@Dariel don't worry. I don't gamble I just want to figure out the mathematics behind it.
I am assuming every who comments here knows random walk problem in statistics.
Well since there is a third option the 0/green both red/black have a slightly lower chance lets call it 49%. SO in the long term you will slowly lose money. Realistically the way to profit would be to wait till your up and just ditch and never play again

Also again previous rolls are irrelevant the chance to win black or red is the exact same every single spin (even if say the last 100 spins where black that will never influnce the chance to get black again)
The chance of 8 red and 7 red and 1 Black are the same. These events are statistically independant i.e one event does not influence the next. https://en.wikipedia.org/wiki/Independence_(probability_theory)
Go through this. Independence makes tbe two probabilities equal. But given a fixed number of games it doesn't make all combinations equally likely.
After three reds the probabilty of getting a black is 0.5.
But after 4 turns probabilty of getting 3 reds and one black is higher than probabilty of getting 4 reds. That's my point.
Thats not correct though

If you have 4 turns:

Case 1: Red (50% chance), Red (50% chance), Red (50% chance), Red (50% chance)
0,5 x 0,5 x 0,5 x 0,5 = 0,0625 - 6,25% of this combination happening

Case 2: Red (50% chance), Red (50% chance), Red (50% chance), Black (50% chance)
0,5 x 0,5 x 0,5 x 0,5 = 0,0625 - 6,25% of this combination happening

The only thing different here is the fact outcome 4 is called black instead of red, mathematically that does not make any difference though.

The only way in a situation like this where the chance would change is if the previous event (event 3 = red) would have an influence on the next event. But the fact that you hit red does not increase the chance of hitting black the next time.

Lets say every time you hit a red number that number no longer can be hit, in that case the next event you would be more likely to get black. However in roulette every single roll is independent. After each outcome the whole system resets and the exact same possibilities are back. Again, you could literally get red 1 million times in a row, the chance for black would still be 50%.
//Case 2: Red (50% chance), Red (50% chance), Red (50% chance), Black (50% chance)
0,5 x 0,5 x 0,5 x 0,5 = 0,0625 - 6,25% of this combination happening///
I see. Actually BRRR = RBRR = RRBR = RRRB.
So this makes the difference if assume random walk but. Each of this combination is as equal as RRRR. My confusion is over. Thanks.
That means there are actually no safe bet in roulette as you are targetting a specific combination only. Thanks I have a better understanding of probability theory now. Guy studying Statistical mechanics should have known better. But anyways better late than never :P.
That means there are actually no safe bet in roulette as you are targetting a specific combination only.

I could've told you this straight away. :P
Well you said this is not a roulette suggestion thread and only focused maths.
You can never predict roulette, not on any probability or anything ;)
@optimus prime.
I was focussing on math only. I had trouble understanding the concept of equal a priori probability on stat mech and then most probable distribution. I was seeing roulette and found that this was a mathematically equivalent problem. Roulette would spice things up in lordswm forum that's why.. I was going through the comments and it hit me then that there are no safe bets in roulette.
closed by cosmicdust97 (2018-04-09 10:34:59)
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