Question

Three fair coins are tossed. If all three coins show heads, then the player wins $15. If all three coins show tails, then the player wins $10. If it costs $5 to play the game, what is the player's expected net gain or loss at the end of two games?

Answer 1

`-$3.75`

Explanation:

expectation is defined as `sum_i^n P(x)x`. In this case we know that the possible outcomes for a round is

• x_1 =3 heads, $15
• x_2 = 3 tails, $10
• x_3 =anything else, 0$

so `P(x=x_1)=1/8` because there is only 1 way to accomplish this.
`P(x=x_2)=1/8` because this is the only way for this one too.
this means that `P(x=x_3)=6/8`

for two games we could have
`P(x=x_1, x= x_1)=1/64, 30`
`P(x=x_1, x= x_2)=1/64, 25`
`P(x=x_1, x= x_3)=6/64, 15`
`P(x=x_2, x= x_1)=1/64, 25`
`P(x=x_2, x= x_2)=1/64, 20`
`P(x=x_2, x= x_3)=6/64, 10`
`P(x=x_3, x= x_1)=6/64, 15`
`P(x=x_3, x= x_2)=6/64, 10`
`P(x=x_3, x= x_3)=36/64, 0`

Leading to
#1/64*30 + 1/64*25 + 6/64*15 + 1/64* 25 + 1/64*20 + 6/64* 10 + 6/64* 25 + 6/64* 20#

simplifying

`1/64(30 + 25 + 25 +20) + 6/64(15 + 10 + 15 + 10)`
`1/64(100) + 6/64(50) = 400/64 = 6.25`

Another way to think about this is knowing that the expectation for a two games can be stated as `E(game1+game2) = E(game1) +E(game2)` thus `15*1/8 + 10 *1/8 +0*6/8 = 25/8` for one game.
for two games it should be `50/8 = 6.25` which it is. Now we can subtract the amount to play per game leaving

`6.25 - 10 = 3.75`. Playing the two games should result in a loss of `$3.75`