Answer:
Step-by-step explanation:
a)
If you toss independently a fair coin and you count the number of tosses until the first tail appears.
This is a geometric distribution with [tex]P=\frac{1}{2}=0.5[/tex](win or lose)
The probability mass function is,
[tex]P_x(k)=(1-p)^{k-1}p\\\\(1-\frac{1}{2})^{k-1}(\frac{1}{2})\\\\=(\frac{1}{2})^{k-1}(\frac{1}{2})\\\\=(\frac{1}{2})^{k-1+1}=(\frac{1}{2})^k[/tex]
The payout is [tex]Y=2^k[/tex] and this payout has the same PMF as X
The expected value of the Y
[tex]E(Y)=\sum\imits^\infty_{k=1}yp_y(k)\\\\=\sum\imits^\infty_{k=1}2^k(\frac{1}{2})^k\\\\=\sum\imits^\infty_{k=1}\frac{2^k}{2^k}=\sum\imits^\infty_{k=1}1=\infty[/tex]
b)
If you are faced with the choice of playing for give fee or not playing at all and your objective is to make the choice that maximize your expected net gain. You would be willing to pay any value of fee. However, this is in strong disagreement with the behaviour of individuals. The discrepancy is due to a presumption that the amount one is willing to pay is determined by the expected gain.
Answer:
a) (2^nCr*(1/2)^r*(1/2)^2^n-r) * 2n
b) 2^nC2^n*(1/2)^r *2n
Step-by-step explanation:
Number of coin tossed = n = 2^n
Let p, the probability of tail appeared = 1//2
Let q, the probability of tail not appeared = 1/2
∴ P(x= r) = 2^nCr*p^r*q^n-r, where r = number of tosses until p appears
P(x=2^n) =2^ nC2^n *(1/2)^r*(1/2)^2^n-r
Hence, amount expected to play this game = (2^nCr*(1/2)^r*(1/2)^2^n-r) * 2n
b) Amount that willing to pay to play this game = 2^ nC2^n*(1/2)^r *2n
