Suposse that you bet to one number in the roulette, the probability of hit is 1/37.
Use the binomial distribution for calculate the probabily of hit at least one time in k spins is:
$$1 - \binom{k}{0} * (1/37)^0 * (1 - 1/37)^k $$
So
$$1 - (36/37)^k $$
I want win how minimum 10:
¿Is correct?
If I bet 26 times in a row to the same number, the probability of winning would be 0.510, if I bet 100 0.934, then I only have to win once to recover what I lost and more.
On
The question is not clear, but no series of losing bets can have a positive expectation. It is quite possible to have a chance higher than $0.50$ to win, but that will be overbalanced by a small chance to lose a large amount.
If you bet $26$ times (whether the same number or not) you are correct that the chance to lose them all is about $0.490479$ and your chance to win at least one (and show a profit on the series) is about $0.509521$ If you lose them all you lose $26$ times your bet. Your chance to win exactly once is about $0.3542347$ and you win $10$. If you add up all the probable outcomes, your expectation is $-\frac {26}{37}\approx -0.7027$
No, this does not work. The probability your number is the winning one is $\frac{1}{37}\ \ ^1$ every try and the win is only 36 times your bet. You can understand that that green zero is your lose.$^2$
$^1$ this is the probablity with only one green "zero", of course if you are playing in an old fashioned casino with roulette allowed to carry the double zero the probability of winning is $\frac{1}{38}$
$^2$I don't want to discourage you, but trust me, if there would have been a way of winning at the roulette no man would still be working lol