Question

in a game a coin is flipped twice, if heads lands up either time, the player wins $4.60 otherwise the player looses $6.00. What is the expected value of this game?

Answer 1

4 equally likely situations here
HH: win 4.60
HT: win 4.60
TH: win 4.60
TT: lose 6
Expected value = 3/4*(4.60) + 1/4*(-6)

Answer 2

Thanks!

Answer 3

you can also do this as follow: play the game 4 times, you win \(3\times 4.60\) and lost \(-6\) for a total of \(3\times \$4.60-\$6\)
averaged over the 4 games, it is
\[\frac{3+4.60-6}{4}\]