probability and fair coin toss?

Question

you must pay $1.75 to enter a game in which a fair coin is tossed 3 times and you are paid the same number of dollars as the number of tosses that come up as Heads. What is the expected value of this game? Is it fair?

Answer 1

I don't remember anything about probabilities, but I would have to say that if you think about it purely mathematically then the average amount won would equal $1.50, which is less than the $1.75 and so I would pass. On the other hand, I'm a gambling man and I would think that if I flip a coin three times I may just get 2 heads, so I would do it for the rush ;)

Answer 2

There are 8 different outcomes

HHH which pays $3
HHT which pays $2
HTH which pays $2
HTT which pays $1
THH which pays $2
THT which pays $1
TTH which pays $1
TTT which pays $0

4 of the 8 outcomes give you a profit.

The average payout is $(3+2+2+1+2+1+1+0)/8 = $1.50

Answer 3

I would imagine that if you kept playing, you would win an average of $1.50 a game. If you have to play $1.75 a game, you would be losing an average $0.25 per game you play.

f(x) = -$1.75x + $1.50x = -$0.25x

If you only played once, you could expect to win 1 or 2 times. This means you could win $0.25, or lose $0.75.

Doesn't seem very fair to me.