# Expected values probability help

A fair coin is tossed 3 times, and a player wins $26 if 3 tails occur, wins$13 if 2 tails occur, and loses $26 if no tails occur. If 1 tail occurs, no one wins. what is the expected value? i dont really understand what does the "if 1 tail occurs , no one wins. how do you set up that? i only set up up to ($26*3/8) + ($13* 2/8) -(26* 4/8) how do i set up the if 1 tail occurs? ## Answers and Replies Stephen Tashi Science Advisor i dont really understand what does the "if 1 tail occurs , no one wins. how do you set up that? Use$0 for the winnings.

i only set up up to ( $26*3/8) + ($13* 2/8) -(26* 4/8) how do i set up the if 1 tail occurs?[/

Your probabilities add up to more than 1, so they are wrong.

if i use 0 for the winnings that 1 tail occurs that just means i add 0 to the end of my equation?
($26*3/8) + ($13* 2/8) -(26* 4/8) + 0 (1/8)

isnt that still the same?
so what is wrong with my equation?

Stephen Tashi
if i use 0 for the winnings that 1 tail occurs that just means i add 0 to the end of my equation?

($26*3/8) + ($13* 2/8) -(26* 4/8) + 0 (1/8)
isnt that still the same?
Yes. The zero just shows a grader than you know what you are doing.

so what is wrong with my equation?

You have computed the probabilities incorrectly. For example, the probability of getting 2 heads in 3 tosses isn't 2/8. Of the 8 possible results of tossing 3 coins, there are 3 possible ways that one can get 2 heads and 1 tail.

You have computed the probabilities incorrectly. For example, the probability of getting 2 heads in 3 tosses isn't 2/8. Of the 8 possible results of tossing 3 coins, there are 3 possible ways that one can get 2 heads and 1 tail.

oh wow, i can't believe i made such a stupid mistake, thank you for helping me out