Given a $5$ out of $36$ lottery ($5$ unique numbers out of pool of $36$ numbers ranging $[1,36]$), the probability that a draw (5 numbers appearing in each game) has at least one pair of consecutive numbers (like $22, 23$) is $0.49$. Then probability of NOT-having-at-least-one pair-of-consecutive-numbers-in-a-draw is $(1 - 0.49)$.
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 consecutive 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 of at least one pair of consecutive numbers is very high: $1-0.03 = 0.97$. So I would play some numbers having one pair of consecutive 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 consecutive numbers like 22 23) and 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?
$\dfrac1{39270}$ vs $\dfrac 1{376992}= \dbinom{36}5$ 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 another instance of Gambler's Fallacy, but I cannot understand where is the mistake?
The mistake is in $(1-p)$, you misinterpreted what $(1-p)$ means.
Let's call any pair-of-consecutive-numbers - $dif1$ (pairs 1 2 or 34 35 etc all give difference = 1).
"probability of having 5 draws (5 games) in a row each (game/draw) NOT-having-at-least-one pair-of-consecutive-numbers (not dif1) is very small: $(1−0.49)5=0.03$." - it is true.
So $p=0.03$.
But what exactly $(1-p)$ means?
"Everything besides" means following (alternative) scenarios of 5 draws (aka games) being NOT $[NOTdif1,NOTdif1,NOTdif1,NOTdif1,NOTdif1]$:
This is why this (1-p) is so huge!
$(1-p)$ does not define probability of next draw being dif1.
Must be very careful and strictly define p and especially $(1-p)$.
Probability of next draw being dif1 is same every draw, with no regard what happened before. I think there is no (correct) formula for probability of next dif1 depending on what happened before...
Moreover, in a single draw dif1 probability is p=0.47. Then (1-p) = 0.53 is NotDif1 probability.
The last reasoning seems strange when you look at computational statistics of a real lottery - even 20-30 long streaks of successive NotDif1 are very rare in the general picture. If they are rare, why not think that "me personally would not hit that very rare streak" - I cannot explain it so far. Maybe somebody can add to the topic to clear it in more details than usual dull arguement about lotteries "past information cannot be used to predict future draws."
Very detailed reasoning about lotteris trains combinatorics and probability understanding very much.