Let's take not odds/even example, but a more complicated one, but same logic can be applied to both.
Given some lottery, say $5$ out of $36$ lottery ($5$ unique numbers out of pool of $36$ numbers ranging $[1,2,…,36]$), I know the probability that a draw (5 numbers appearing in each one game) has at least one pair of consecutive numbers (like $22, 23$ -a pair of numbers whose difference $23-22 = 1$). Say this probability is $0.49$ (stackoverflow confirms it). Then probability of NOT-having-at-least-one pair-of-consecutive-numbers-in-a-draw is $(1 - 0.47)$.
Now, probability of having 5 draws (5 games) in a row each (game/draw) NOT-having-at-least-one pair-of-consecutive-numbers is very small: $(1 - 0.49)^5 = 0.03$. Say I don't play and just observe. While observing, when I see 5 strictly consequtive draws each NOT-having-at-least-one pair-of-consecutive-numbers, then I know that if I immediately play (next game), then at the next game the probability that there would appear at least one pair of consequtive numbers is very high: $1-0.03 = 0.97$ (97%). So I would play some numbers having one pair of consequtive numbers.
Guessing having a pair-of-consecutive-numbers with 97% probability I still have to pick one concrete pair out of 35 pairs-of-consecutive-numbers (say 21 22 or 1 2 or etc.).
But the trick is - if I found a place where previous draws give me also two additional indicators: "play every second number (12 14)" with 90% and "play every third number (17 20)" with 87%, etc, then I make union of those three pairs (including consequtive numbers like 22 23) and
suddenly get exactly 5 numbers and play them (if union gives not exactly 5 - I wait next such "time-frame" to play). My chances to guess correctly are $(1/35)*(1/34)*(1/33) = 1/39270$ (because in this scenario I need to guess correctly 1 out of 35 pairs of consecutive numbers, then 1 out of 34 pairs of every second number (like 12 14) and then 1 out of 33 pairs of every third number (like 12 14).
If I play like this (once in a week/month), do I have better chances?
1/39270 vs 1/376992 (binomial(36,5)) is a huge (10 times!) leverage. Yes, I need to adjust 1/39270 chances by the fact that I had 97%, 90% and 87% (not 100%) for pair groups, but it won't be 10 times I think.
Where is the fallacy in my reasoning? I don't believe this is not another trick of Gambler's Fallacy, but I cannot understand where is the mistake? Math shows all is fine like that and there seems no mistake.