Question

Two gamblers, call them Gambler A and Gambler B flip a coin repeatedly. The coin is unfair and comes up heads 2/3 of the time. Gambler A wins one dollar from Gambler B, when a head is tossed. Conversely Gambler B wins one dollar from Gambler A when a tail is tossed. The coin tosses are independent. The game ends when one of the gamblers runs out of money. There are 5 dollars in the pot. Determine the probability that Gambler A wins the game if he starts with I dollars. Here I

Answer 1

Answer:

≈ 0.52

Step-by-step explanation:

P( head ) = 2/3 , P( tail ) = 1/3

when a head is tossed ; Gambler A wins $1

when a tail is tossed : Gambler B wins $1

Determine the P( Gambler A wins the game ) if he starts with I dollars

Assuming I = $1

n = 5

p ( head ) = P( winning ) = 0.66

p( losing ) = 0.33

applying the conditional probability in Markov which is ;

Pₓ = pPₓ₊₁ + (1 - p) Pₓ₋₁

P( 1) = 0.66P₂ + 0.33P₀

resolving the above using with Markov probability

P( 1 ) = 0.51613

hence the probability of Gambler A winning the game if he starts with $1

≈ 0.52