Answer:
In the long run, ou expect to lose $4 per game
Step-by-step explanation:
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.
Assuming X be the toss on which the first head appears.
then the geometric distribution of X is:
X [tex]\sim[/tex] geom(p = 1/2)
the probability function P can be computed as:
[tex]P (X = n) = p(1-p)^{n-1}[/tex]
where
n = 1,2,3 ...
If I agree to pay you $n^2 if heads comes up first on the nth toss.
this implies that , you need to be paid [tex]\sum \limits ^{n}_{i=1} n^2 P(X=n)[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2[/tex] ∵ X [tex]\sim[/tex] geom(p = 1/2)
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}[/tex]
[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =6[/tex]
Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6
= $4
∴
In the long run, you expect to lose $4 per game
