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

≈ 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

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

the diameter is 16 cm. are you a teacher?

step-by-step explanation: