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Posers and Puzzles

Posers and Puzzles

  1. 02 Apr '05 19:47 / 1 edit
    Bill has n+1 coins and Ben has n coins.
    Both players 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 obtains more heads than Ben?
  2. Standard member The Plumber
    Leak-Proof
    02 Apr '05 20:20
    Originally posted by THUDandBLUNDER
    Bill has n+1 coins and Ben has n coins.
    Both players 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 obtains more heads than Ben?
    50%
  3. Standard member Bowmann
    Non-Subscriber
    02 Apr '05 22:51
    1/2.
  4. 03 Apr '05 04:55
    Originally posted by Bowmann
    1/2.
    But Bill has more coins than Ben.
  5. Standard member ark13
    Enola Straight
    03 Apr '05 05:03
    Yeah, but it's 50% that he'll get MORE heads than him. If they've got an equal number, it doesn't count. I haven't worked out the probability, but I know that much.
  6. Standard member Bowmann
    Non-Subscriber
    03 Apr '05 15:18
    Originally posted by THUDandBLUNDER
    But Bill has more coins than Ben.
    Are you saying I'm wrong, or that I didn't explain the solution?
  7. Standard member Palynka
    Upward Spiral
    03 Apr '05 15:33
    Originally posted by ark13
    Yeah, but it's 50% that he'll get MORE heads than him. If they've got an equal number, it doesn't count. I haven't worked out the probability, but I know that much.
    It's easy. In n coins the expected value of heads is the same, so it's 50% on the last coin.
  8. Standard member Bowmann
    Non-Subscriber
    03 Apr '05 15:34
    Originally posted by Palynka
    It's easy. In n coins the expected value of heads is the same, so it's 50% on the last coin.
    Hmm...?!
  9. Standard member Palynka
    Upward Spiral
    03 Apr '05 15:42 / 1 edit
    Sorry. I meant the expected value of the NUMBER of heads in n tosses, obviously.
  10. 03 Apr '05 18:13 / 1 edit
    Originally posted by Bowmann
    Are you saying I'm wrong, or that I didn't explain the solution?
    The second.

    What about if they both had n coins.

    What would the probability be then?
  11. Standard member The Plumber
    Leak-Proof
    03 Apr '05 19:12 / 1 edit
    Originally posted by THUDandBLUNDER
    What about if they both had n coins.

    What would the probability be then?
    That is a function of n. For n=1, the probabilty is 1/4, and I believe as n approaches infinity, the probability approaches 1/2 (someone better at math than I will have to confirm that).

    Not sure what that has to do with your original problem, since the answer is always 1/2 there, regardless of n.
  12. 04 Apr '05 01:35
    Not sure what that has to do with your original problem, since the answer is always 1/2 there, regardless of n.[/b]
    I don't think the problem is as easy as everyone seems to think it is. I think the answer is 1/2 only if n=0, which is why the above answers are only conditionally correct. Here's what I think it is:

    if they both had n coins, then the probability of Bill getting more heads would be n/(2n+2). This is very easy to show.

    So my approach is the following: consider they both throw n coins, and the probability of Bill having more heads is given above. Then consider that Bill has one more coin to throw which of course can be heads with 1/2 probability. Then the overall probability that Bill gets more heads should simply be a weighted average of these probabilities:

    (n(n/2n+2) + 1/2)/(n+1).

    This only equals 1/2 for n=0, which makes sense. for all other possibilities, it depends on n.

    THUD, is this right?
  13. 04 Apr '05 01:38
    oops, i think i forgot a parentheses. i meant:

    {n[n/(2n+2)] + 1/2}/(n+1)
  14. 04 Apr '05 02:12
    Originally posted by davegage


    if they both had n coins, then the probability of Bill getting more heads would be n/(2n+2). This is very easy to show.

    THUD, is this right?
    Actually, Bowmann was right for n+1 coins but he gave no reasoning.

    When they both have n coins the calculation is a bit more complicated.
  15. 04 Apr '05 02:57
    yeah, after i look at it, Bowmann is right: the answer is 1/2 regardless of n.

    the n/(2n+2) is right for n coins each, but i don't think i can take the weighted average like i did.

    the proof that the answer to the original question is 1/2 is straightforward now that i think about it:

    bill has n+1 coins, ben has n:

    consider that ben has gotten i heads (i = 0,1,2,3,...,n). Then Bill has gotten j heads (j = 0,1,2,...,n+1). Then there are clearly (n+1)(n+2) possibilities (you have to include the zero). But the number of times that Bill has more heads than Ben is only given by SUMMATION(from i=1 to i=n+1) of i. this is equal to (1/2)(n+1)(n+2) (this is easy to show by simple number theory). then the probability we are looking for is (1/2)(n+1)(n+2)/((n+1)(n+2)) = 1/2.


    that was probably clear as mud, but it works out...little hard to generalize in clear terms.

    good problem.