ANSWER
The expected value is $0
EXPLANATION
The wins or loses of this game are:
• $1 if HH
,• $5 if TT
,• -$3 if TH or HT
H is for heads and T is for tails.
First we have to see the probabilities of these events happen:
[tex]P(HH)=P(TT)=\frac{1}{4}[/tex]This is because the posible outcomes are 4: HH, TT, HT, TH. And in only one of these the coins are both heads or both tails.
The probability that they are different is:
[tex]P(HT\text{ or }TH)=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}[/tex]The expected value is:
[tex]E=1\cdot P(HH)+5\cdot P(TT)-3\cdot P(HT\text{ or }TH)[/tex][tex]E=1\cdot\frac{1}{4}+5\cdot\frac{1}{4}-3\cdot\frac{1}{2}=\frac{1}{4}+\frac{5}{4}-\frac{3}{2}=\frac{1+5-6}{4}=\frac{6-6}{4}=0[/tex]The expected value is 0.
It is a fair game, because the expected value of the game is 0. This means that the cost of playing (which would be the cost of losing) is the same as the cost of winning.
