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.

  • 1
    "It means that when I see 5 draws in a row each NOT-having-at-least-one pair-of-consecutive-numbers, then I can immediately play such 5 numbers that they have at least one pair of consequtive numbers with probability 1−0.034=0.97 (97%)." This is where you lost me. – Rahul Nov 8 at 20:03
  • By the first paragraph I think a single "draw" is a selection of 5 distinct numbers out of $\{1, ..., 36\}$, and then you are talking about 5 independent such draws. I did not understand your second paragraph "then I can immediately play such 5 numbers that they have at least one pair of consequtive numbers with probability." Also the meaning of "I see" and "I play" are not clear to me. [edit: I see that Rahul above points out the same sentence that I found to be unclear, 20 seconds before my own comment. I leave my comment just as further evidence that people cannot understand the question] – Michael Nov 8 at 20:03
  • 1
    It is exactly the gambler's fallacy. Past history gives you no information about the future draws. – Ross Millikan Nov 8 at 20:06
  • @RossMillikan : While "past history gives you no information about future draws" is generally true in memoryless situations, I am surprised that you could understand the question enough to conclude that this advice applies. – Michael Nov 8 at 20:10
  • @Michael: I admit I do not understand the strategy being proposed, but it seems to wait for a number of draws without pairs next to each other, then conclude they are more likely in the next draw. – Ross Millikan Nov 8 at 20:12
up vote 1 down vote accepted

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?

It means everything besides $[NOTdif1,NOTdif1,NOTdif1,NOTdif1,NOTdif1]$ = (5 games) in a row each (game/draw) being NOTdif1.

"Everything besides" means following (alternative) scenarios of 5 draws (aka games) being NOT $[NOTdif1,NOTdif1,NOTdif1,NOTdif1,NOTdif1]$:

  1. probability of $[dif1,dif1,dif1,dif1,dif1]$ PLUS
  2. probability of $[NOTdif1,dif1,dif1,dif1,dif1]$ PLUS
  3. probability of $[dif1,NOTdif1,dif1,dif1,dif1]$ PLUS
  4. probability of $[NOTdif1,dif1,NOTdif1,dif1,dif1]$ PLUS
  5. probability of $[dif1,dif1,dif1,dif1,NOTdif1]$ PLUS
  6. probability of $[NOTdif1,NOTdif1,dif1,dif1,dif1]$ PLUS
  7. probability of $[dif1,NOTdif1,NOTdif1,dif1,dif1]$ PLUS
  8. probability of $[NOTdif1,dif1,NOTdif1,NOTdif1,dif1]$ PLUS
  9. ...
  10. etc.

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.

It means it is better ALWAYS PLAY NotDif1 ! Even after 100 times in succession play NotDif1 next! I believe it is pure math, but personally I wouldn't dare do that, thou I believe it is true...

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.

  • If you are betting even money on a 0.47 vs 0.53 trial (and the trials are independent), then yes you should bet the 0.53 outcome even if it has happened 24 times in a row. After 24 in row, you are more likely to have 25 in a row on the next draw than to have it the streak end on the next draw. – Ned Nov 9 at 0:18
  • The first sentence of this answer currently reads "The mistake is in $(1−p)$, you misinterpreted what $(1−p)$ means." While I will not downvote this answer, I think it is unusual that: (i) The answer mentions a value $p$ that is never defined or used in the question or in any followup comments; (ii) The asker seems to be referring to himself/herself as "you." While I fully support an asker answering his/her own question, I would expect that answer to start something like "I believe my mistake is..." – Michael Nov 9 at 19:02
  • PS: I still think it would be helpful for you to answer my question "how does it jump?" in the above comment, regarding clarification of your (incorrect) sentence "then at the next game the probability that there would appear at least one pair of consequtive numbers is very high: 1−0.034=0.97." Note also that if you play $\{1,2,3,4,5\}$ each and every time, you have just as much chance of winning (over, say, 100 games played) as if you play different sequences each time. I do not recommend playing the lottery. A friend of mine used to call it "a tax on stupid people." – Michael Nov 9 at 19:05

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.