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Posers and Puzzles

Posers and Puzzles

  1. 23 Aug '06 04:04 / 1 edit
    Bill has n+1 coins and Ben has n coins. They both throw all of their coins simultaneously and observe the number that come up heads. Assuming all the coins are fair, what is the probability that Bill gets more heads than Ben?
  2. 23 Aug '06 14:10
    combining your title and me not seeing the 's' on heads i didnt know where you was going with this..
    the answer is 100%? (hoping this is a brainteaser and not a maths one)
  3. 23 Aug '06 14:35
    Damnit I've forgotten stats. Is it n+1/n?
  4. 23 Aug '06 15:04
    Originally posted by ThudanBlunder
    Bill has n+1 coins and Ben has n coins. They both throw all of their coins simultaneously and observe the number that come up heads. Assuming all the coins are fair, what is the probability that Bill gets more heads than Ben?
    50%
  5. 23 Aug '06 15:06 / 1 edit
    lol, that's more than 100%.

    edit: replied to the (n+1)/n
  6. Standard member PBE6
    Bananarama
    23 Aug '06 15:08
    Originally posted by ThudanBlunder
    Bill has n+1 coins and Ben has n coins. They both throw all of their coins simultaneously and observe the number that come up heads. Assuming all the coins are fair, what is the probability that Bill gets more heads than Ben?
    The probability of Bill getting "m" heads with his (n+1) coins is:

    (n+1)choose(m)/2^(n+1)

    The probability of Ben getting (m-1) or less heads with his "n" coins is:

    SUM{(n)choose(i)/2^n}, where i=0...(m-1)

    The probability of both these events happening simultaneously is simply their product. Now, to count all the possible outcomes we use the double sum:

    P = SUM{(n+1)choose(j)/2^(n+1) * SUM{(n)choose(i)/2^n}}, where j=1...n+1, and i=0...j-1

    For example, for n=10 at some point we would calculate the probability that Bill got 3 heads (j=3), and Bill got 0, 1, and 2 heads (i=0,1,2), at some other point we would calculate the probability that Bill got 8 heads (j=8), and Ben got 0, 1, 2, 3, 4, 5, 6, and 7 heads (i=0,1,2,3,4,5,6,7), etc...

    I don't know if there's a good way to simplify the above expression, or if there's an easier "aha!" solution, but this one will give you the correct answer if you grind it out.
  7. 23 Aug '06 16:01 / 3 edits
    Originally posted by PBE6
    I don't know if there's a good way to simplify the above expression, or if there's an easier "aha!" solution, but this one will give you the correct answer if you grind it out.
    I'm not sure if this is worth an "aha!":

    Probability of Bill getting more heads = probability of Bill getting more tails = 50% (as they can't both happen).
  8. 23 Aug '06 17:38 / 2 edits
    Originally posted by ThudanBlunder
    I'm not sure if this is worth an "aha!":

    Probability of Bill getting more heads = probability of Bill getting more tails = 50% (as they can't both happen).
    Yes I had:

    If A has k heads and n+1-k tails and B has j heads and n-j tails then

    P( k > j ) = P( n+1-k > n-j ) (*)

    by symmetry of heads/tails. Finally, for integers k and j

    n+1-k > n-j if and only if ( k < j or k=j )

    so the two events in (*) partition the sample space so each have probability 1/2.

    Yours is definitely "aha" though!
  9. Standard member Bowmann
    Non-Subscriber
    23 Aug '06 22:17
    Always recycle
  10. Standard member Freddie2008
    9 Edits
    23 Aug '06 22:22 / 1 edit
    [i]what is the probability that Bill gets more head than Ben?[/b]
    [/i]I imagine this will have a lot to do with looks/size.
  11. 08 Oct '06 15:04
    it depends on the value of n
  12. 16 Oct '06 02:15
    I'd say it was still a 50% probability because there'd now be an odd number of coins between them and the most probable outcome would be that the +1 would be the deciding coin.
  13. 16 Oct '06 03:59
    0.5