Another Math Question...

1. 75

   DzyMsLizzyposted 5 years ago

   Okay, y'all know by now that my choo-choo often jumps the tracks into a no-man's -land of queries that have every one else wondering whether I'm cuckoo.  Well, yes, probably. But I hold firm to the saying, "it's good to be crazy; it keeps you from going insane!" LOL
   
   Now, on to my weird puzzle that I lack the math skills to figure out...kind of a probability & statistics theme. Here is the background information:
   
   As  I'm sure you know, the game of BINGO uses cards divided into 25 squares, 24 of which each contain a random number with a spread of 15 ascending possible numbers assigned by columns. There are 75 numbers available for use by column, as follows:
   B
   1-15
   I
   16-30
   N
   17-45
   G
   46-60
   O
   61-75
   
   So, what are the odds (for or against) of the same number occupying the same position on any group of 4 cards?
   
   You're welcome: Happy New Year!

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2. 61

   paradigmsearchposted 5 years ago

   My best guess...
   
   There is one chance in three the number will show up at all.
   
   There is one chance in five it will occupy the same position.
   
   Thus, 1/3 times 1/5 equals 1/15 or odds of 1 in 15.
   
   Calculating odds can be a tricky business, in fact I've already realized a flaw in my logic. My method is good for an approximation, but a more accurate answer requires a more complicated method involving the use of "not" as opposed to the use of "is".

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   1. 75

      DzyMsLizzyposted 5 years agoin reply to this

      Cool, Para!  Thanks!  I do sometimes find that happening, but not often enough that it struck me as common.
      Math and I are sworn enemies; I am deathly allergic to the topic.  LOL

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   2. 75

      wildernessposted 5 years agoin reply to this

      I think there is far more to it than that.  For instance, out of each group of 15 numbers, there WILL be 5 chosen.  There will be 5 numbers between 1 and 15, for instance; it is not possible to have all numbers higher than 15, even though random choice dictates that it is very possible.
      
      In addition, once all the numbers of a card are chosen there are now 24 numbers, any one of which could be on all four cards, and that will affect probabilities as well.  So will the center "N" value - isn't that a "free number" or some such?
      
      I think the question is far more complex than it seems because the numbers are NOT totally random but chosen with certain rules in mind.

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      1. 75

         DzyMsLizzyposted 5 years agoin reply to this

         Thanks, wilderness; you are correct, the center spot is a 'free' space containing no numbers.

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