There are 10 balls in a bucket with each ball marked 1 through 10 and are chosen at random from a bucket each day. The bucket is fair and all 10 balls are reset each day in the bucket. With a wager of 1:10 for guessing the ball number correctly, you win 10 dollars, or 9 dollars profit, for every $1 you bet. You can bet on only 1 of 10 numbers per day.
Now, let's say you also have the history for this particular game from inception for all 10 balls. A ball hasn't been selected from the bucket for over 50 consecutive days only 15 times in this game's entire history. Out of the 15 times this has happened, the ball was selected 14 of 15 times within the next 10 days. 1 of 15 was drawn on the 64th day. Currently, ball number 2 hasn't been selected from the bucket for 50 consecutive days. You are allowed to use the Martingale system or a similar system where you increase your wager each day you lose.
Does this mean winning at this point has a 93.3% (14/15) chance within the next 10 days? With the example above, would it be a good idea to make a bet on ball number 2 since it hasn't come out in 50 consecutive days?
No. If "the bucket is fair", then all the information about the game's history and the recent draws is entirely irrelevant. The probability ball two will be drawn next is $1/10$, because that's what "the bucket is fair" means.
The expected profit from any bet, no matter what ball you bet on, is zero:
$$ \frac{1}{10}(+9) + \frac{9}{10}(-1) = 0 $$
(However, some patterns of persistent gambling will eventually lead to losses even with an entirely "fair" game like this: See the Gambler's Ruin.)