Question

Roulette is one of the most common games played in gambling casinos in Las Vegas and elsewhere.

An American roulette wheel has slots marked with the numbers from 1 to 36 as well as 0 and 00 (the latter is called "double zero"). Half of the slots marked 1 to 36 are colored red and the other half are black. (The 0 and 00 are colored green.) With each spin of the wheel, the ball lands in one of these 38 slots.

One of the many possible roulette bets is to bet on the color of the slot that the ball will land on (red or black). If a player bets on red, he wins if the outcome is one of the 18 red outcomes, and he loses if the outcome is one of the 18 black outcomes or is 0 or 00. So, when betting on red, there are 18 outcomes in which the player wins and 20 outcomes in which the player loses. Therefore, when betting on red, the probability of winning is 18/38 and the probability of losing is 20/38.

When betting on red, the payout for a win is "1 to 1". This means that the player gets their original bet back PLUS and additional amount equaling their bet. In other words, they double their money. (Note: If the player loses they lose whatever amount of money they bet.)

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Scenario 1: Mike goes to the casino with $200. His plan is to bet $100 on red 7 consecutive times or until he either has increased his total to $500 or has lost all of his money. What is the probability he will go bankrupt (i.e. end up with $0)? (Give your answer correct to four decimal places.)


(Hint: Set this problem up as an absorbing Markov Chain with 6 states where the states keep track of his current amount of money. The amount of money he has will always be $0, $100, $200, $300, $400, or $500. The states where he has $0 or $500 are absorbing states since he quits playing whenever one of these states is reached.

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Scenario 2: Mike goes to the casino with $200 and will still be betting on red. As in Scenario 1, he will bet $100 each time he places a bet. However, instead of limiting himself to a maximum of 7 bets he decides to play indefinitely until he has reached $500 or goes bankrupt. If he uses this betting method, what is the probability he will eventually go bankrupt? (Give your answer correct to four decimal places.)

Answer 1

a) Set up the transition matrix X as follows where A = 20/38 and B = 18/38.

1 0 0 0 0 0

A 0 B 0 0 0

0 A 0 B 0 0

0 0 A 0 B 0

0 0 0 A 0 B

0 0 0 0 0 1

You need the probability that Mike will be bankrupt, but since bankruptcy is an absorbing state just multiply the transition matrix 8 times (X^8) and use the probability in the index (3,1) of the resulting matrix. This will give you the probability that given you start with $200, you will have $0 by the end of the 7th turn. The answer is 0.557 or 55.7% chance of going bankrupt.

b) You can use the same matrix as above, except you need to determine what the probability converges to as you keep betting. Since you can't multipy the matrix X an infinite amount of times, just use a computer or calculator to see when it seems that as you increase the power of X, the probability stays constant. For me, doing X^50 was enough, and gave an answer of 0.6618 or 66.18% chance of going bankrupt.