I've read an article in local newspapers where they explain that probability of having ten reds in a row is the same as one red and nine blacks (actually that all combinations in ten spins have the same probability) trying to explain that betting on red when black came in last three spins doesn't make a difference to betting on black again.

Important note: let's have a roulette without zeros (so only same number of blacks and reds) for simplicity.

There was a list of all possibilities like this: RRRR, RRRB, RRBR, RBRR, BRRR, RRBB, RBRB, BRRB, BRBR, BBRR, RBBB, BRBB, BBRB, BBBR, BBBB. (I might have missed some but I'm sure you get the idea). From this probability of RRRR and RRRB is the same: p = 1 / 15 which is the same for all the above mentioned sets.

I got confused here a little bit because when I look at it there are only two one colored options (RRRR, BBBB) and a lot of its combinations so isn't probability of a row full of one color lower than probability of the row being interrupted: p(RRRR, BBBB) = 2 / 15, p(all others) = 13 / 15? So wouldn't that mean that betting on the other color is more likely to be "luckier" choice?

The explanation stated that roulette doesn't have a memory so the probability on each try is always 1/2 for red and black which I understand but why should one colored row have the same probability as not one colored rows, I don't get it.

Could please someome tell me if I'm mixing wrong terms here or perhaps misunderstood it?

Thank you.


There are 16 equally likely outcomes.

So $\rm \color{red}{RRR}B$ and $\rm BBBB$ are both equally likely.

Then are the odds of the fourth try being black better if the results of the first three tries are $\rm BBB$ or if they are $\rm \color{red}{RRR}$? What about $\rm \color{red}{R}B\color{red}{R}$ or $\rm B\color{red}{R}B$?

  • $\begingroup$ I don't think they depend but that just you are more likely to meet non-single-color streaks. But I guess I got the information from the article in a wrong way. $\endgroup$ – Vico Lemp Aug 21 '16 at 7:04

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