Question

in a game the number 1 to 20 are written on tickets and placed in a bag. a player draws out a number at random. he or she wins $3 if the number is even $6 if the number is a square number and $9 if the number is both even and square

How much should be charged to play the game so that it is a fair game?

Answer 1

here below is the probability distribution of winning amoount from this game:

P(X=3)=P(even number which are not square)=8/20 (As 8 even number whch are not square between 1 and 20: these are 2,6,8,10,12,14,18,20)

P(X=6)=P(square but not even)=2/20 ( 2 number whcih are square but not even ; 1,9)

P(X=9) =P(square and even)=2/20 (2 numbers whcich are square and even: 4,16)

hence expected value of game =xP(x)=3*(8/20)+6*(2/20)+9*(2/20)=2.7

hence for game to be fair cost of game should be equal to  expected value of game which is $2.7.