0
   

Swedish bingo

 
 
Tier
 
Reply Tue 5 Dec, 2017 02:53 pm
Hello,
Swedish bingo is a 5x5 numbers bingo where you dont get any free numbers and have to make full house that is all 25 numbers in order to win. There are number drawn from 1-75. B 1-15, I 16-30 and so on.
I have data from about 400 games played. Averagely it takes 40-45 numbers until someone wins and of course there can be number of combinations that can win simultaneously. And of course when back testing you can find combinations that would have won earlier.
Now I took 219, lets say half, of mu data set and began looking fir winning combinations that would have won in different games. I found 7 different combinations that would have won in 4 different games each, 44 different combinations that would have won in 3 different games each. Others were only two timers or one timers and I was not interested in them. Now my thought was that there are thousands of combinations in bingo game but there are much fewer that appear in multiple games over time and that was proven, 7 4timers and 44 3timers out of these 219 games played. Now my second logic was that if there are combinations that come up 4 times or 3 times out of 219 games then the same combinations will come up again in next 200 games but when I tested this I got no matches at all. Not one combination appeared again and not even once.
Now what am I missing here? Where is the error in my logic?
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Question • Score: 0 • Views: 916 • Replies: 2
No top replies

 
AngleWyrm-paused
 
  1  
Reply Tue 5 Dec, 2017 07:35 pm
When repetitive gambling scenarios come up there's often a conversation about how coins don't have memory and the next flip is also 1/2 likely to be a heads. This is a mistaken answer to a mistaken question. The observer continues to use the same starting point, but doesn't count the narrowing of the probability space.

This is how we say the next single flip is 1/2 chances per outcome of being heads, but also given that we are in the process of flipping a third coin, only 1/2^3 = 1/8th of the outcomes contain that answer: We're just already down two of those 1/2's leaving only one remaining.
0 Replies
 
ekename
 
  1  
Reply Wed 6 Dec, 2017 04:44 am
@Tier,
Quote:
Now what am I missing here? Where is the error in my logic?


You've tried your first and your "second logic", now try the logic that the numbers are random and that past patterns are irrelevant.



0 Replies
 
 

Related Topics

 
  1. Forums
  2. » Swedish bingo
Copyright © 2024 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 05/29/2024 at 06:58:16