Question
You flip a coin,
If it is heads, use 1 die
If it is tails, use 2 dice
If the total of the die/dice is X≤5, you lose
If the total of the die/dice is X=6, it is a push
If the total of the die/dice is X≥7, you win
Find the probability of winning, expected and house advantage.
This is how far I got (I may have made a mess of things, and I don't know if I'm correct to assume "expected" means "expected value"):
For expected value,
E[X|Tails]= (1/6)(1+2+3+4+5+6)= 3.5
E[X|Heads]= (2+12)(1/36)+(3+11)(2/36)+(4+10)(3/36)+(5+9)(4/36)+(6+8)(5/36)+7(6/36) = 7
Using the linearity of expectation,
E[X]= (3.5)(1/2)+(7)(1/2) = 5.25
For probability of winning,
P(X≥7|tails)= 0
P(X≥7|heads)= 21/36
P(X≥7)= (21/36)*(1/2) = 21/72
For house advantage,
Given a win pays a to 1, where a>0, and the bet size is b>0, then,
P(X=ab)=p, p>0, where p= 21/72 (from above)
P(X=-b)=q, q>0, where q= (1- 21/72 - (1/6+5/36)*1/2) = 40/72 (from above)
P(X=0)=r, r≥0, where r=(1/2)(1/6 + 5/36) = 11/72 (from above)
and
H0(B,X) = −ap+q = -a(21/72)+40/72
and
H(B,X) = (−ap+q)/p+q = (-a(21)+40)/ 61
As you say, your chance of a win is $\frac {21}{72}$. Your chance of a push is $\frac 12 \cdot \frac 16 + \frac 12 \cdot \frac 5{36}=\frac {11}{72}$ where the first term comes from heads+$6$ and the second comes from tails plus a sum of $6$. Your chance of a loss is then $\frac {40}{72}$ and your expectation on a bet of $1$ is $\frac {21-40}{72}=-\frac {19}{72}$