Question :A gambler is invited to play a simple game in which two coins are
tossed. He will receive nothing if both coins show tails, or \$12,if one
shows heads and one shows tails, or else $60 if both coins show heads.
What is the gambler expected return in this game?
My attempt : I thought the available outcomes are HH, HT and TT so if I have expected return divided by the number of possibilities $($$12+\$60+\$0)/3 which is 24, I will get the expected return. However, the answer to this question is \$21.
How is \$21 expected return?
Best Answer
The HT outcome is twice as likely (as the HH and TT ones) to occur, since it can occur as HT or TH, which makes for four events of equal probability. Thus $$E=\frac{0+2×12+60}4=21$$