Math一問~~~~~

actually, the expected value is normally base on how much the player bet expect some special case(fair game) like this.
Let the player bet $N. If the player win(probability is 1/2), he/she will win $N, if he/she loses(probability is 1/2), he/she will lose $N(-$N)
so the expected value=[1/2*$N+1/2(-$N)]=$0
So in this case, the expected value is $0 and it called a fair game.

do you have any problem with my explanation?

a ) A fair coin is tossed . A player has to guess whether it is a head or a tail . The player is offered the odds of 1 to 2 if his guess is correct , otherwise the player loses .

you know the meaning of odds?

In this question, it means that if my guess is correct, then I receive two dollars, since I have given 1 dollar to bet. My net profit is 1.

if my guess is wrong, I lose 1 dollar

The probability of winning the game is 1/2, the probability of losing the game is 1/2

So, the expected value

=1/2*1+1/2(-1)

=0