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#