4. Two people play a game. They take turns flipping a fair\' coin, the first to

Question

4. Two people play a game. They take turns flipping a fair' coin, the first to flip H wins. A round constitutes just one flip (only one person gets to "go".) (a) What is the probability that the player who goes second wins? (b) What is the probability that the game lasts at least 4 rounds? (c) Given they have played 4 rounds, what is the probability that the game lasts at least6 rounds? d) Suppose that the coin is biased with the probability of Heads equal to p. Suppose the game lasts exactly 4 rounds. What value of p maximizes the chances of the game lasting exactly 4 rounds? For k rounds?

Explanation / Answer

a)P(player goes second wins)=P(first flip T and then heads)+P(first three flip T and then heads)+P(first five flip T and then heads)+P(first seven flip T and then heads)+....

=(1/2)*(1/2)+(1/2)3*(1/2)+(1/2)5*(1/2)+(1/2)7*(1/2)+(1/2)9*(1/2)+...

=(1/2)*(1/2)/(1-1/22) =(1/4)/(3/4)=1/3

b)P(game lasts at least 4 runds) =P(in first 3 flips all tails)=(1/2)3 =1/8

c)

P(at least 6 rounds given at least 4 rounds) =P(X>=6|X>=4)=(1/2)5/(1/2)3 =1/22 =1/4

d)

P(X=4)=P(first 3 round tail and in fourth round it is heads)=P(A)=(1-p)3*p

differentaiting this with respect to p

(d/dp)*P(A) =(1-p)3-3p(1-p)2

putting above equal to 0 ;

1-p=3p

p=1/4 for maximizing the chance of exactly 4 heads

for k rounds :

P(A) =(1-p)k*p

differentaiting this with respect to p

(d/dp)*P(A) =(1-p)k-kp(1-p)k-1

putting above equal to 0 ;

1-p=kp

p=1/(1+k)

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