Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What distribution should i use?

  1. Sep 22, 2009 #1
    Player1 makes the bet.
    the 2nd player has a 32% chance to accept the bet.
    the 3rd player has a 56% chance to accept the bet.
    the 4th player has a 20% chance to accept the bet.

    there can only be one winner of said bet, so player one is interested in knowing the EXPECTED number of players that will call his bet (how much competition will he have)

    (we can make the assumption that the % chance of a player accepting the bet does not change depending on the # of players accepted before him..they aren't that smart)

    anyone have an idea how I can approach this?

    haven't taken stats in a while
     
  2. jcsd
  3. Sep 23, 2009 #2
    The question doesn't seem to be clear . I assume there are only 2 outcomes to a bet - win or lose . Now you say that no more than 1 player can win the bet , but how can that be :
    If player 1 makes the bet , and then player 2 and player 3 accepts the bet , but player 4 declines .

    If player 1 loses the bet then that means player 2 and 3 have won . But you said that no more than 1 player can win the bet. How come ???
     
  4. Sep 23, 2009 #3

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    The average number of players accepting the bet is 32% + 56% + 20%, of course.
     
  5. Sep 23, 2009 #4
    lol so obvious i didnt notice.
     
  6. Sep 23, 2009 #5

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    ...and that's what we're here for.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook