Question

Late at night, you are again greeted by the stranger in a dark silk cape. This time he has a coin with probability 0.2 of coming up heads. He suggests a game in which you flip the coin repeatedly, earning $1 for each heads and losing $1 for each tails. The game will continue until you go broke or have a fortune of $3.

a) If you begin the game with $2 in your pocket, what is the probability that after 2 flips, you will have $2?

b) If you begin the game with $1 in your pocket, what is the probability you will ever reach $3?

Answer 1

a) I can have $2 after 2 flips only if one of the flips result in a head and the other results in a tail,

P[after 2 flips. you will have $2]

= P[head in the first throw, tail in the second throw] + P[tail in the first throw, head in the second throw]

= 0.2*0.8 + 0.8*0.2 = 0.32 (since, probability of getting a head in the coin toss = 0.2)

b) If I begin the game with $1, then at the next flip, he may lose $1 with probability 0.8, which stops the game or he gains $1, and has $2, at the second flip he may lose $1 and the same situation arises as it did when he had $1 at the beginning, or he could gain $1 which would stop the game.

so, probability that he ever reaches $3=

P[he gets 2 heads at two flips consecutively] + P[head->tail->head->head in four flips, will result in a 3]+ .....and so on

= 0.2*0.2 + (0.2*0.8*0.2)*0.2+ (0.2*0.8*0.2*0.8*0.2)*0.2+....

= (0.2)²+ (0.16)*(0.2)²+ (0.16)²*(0.2)²+....

= 0.04+ (0.16)*0.04+ (0.16)²*0.04+....

= 0.04 [1+ 0.16 + (0.16)²+.....] , which is a sum of infinite g.p. series

= 0.04*(1/1-0.16) = 0.0476