After visiting Las Vegas over spring break, our entrepreneurial engineering stud

Question

After visiting Las Vegas over spring break, our entrepreneurial engineering student has abandoned his cargo hauling business and decided to take up gambling. Roulette is his game of choice because he figures that if he bets on red every spin then he has a 0.5 probability of winning. What he doesn't realize is that the probability of landing on red each spin is 18/38, not 0.5 His plan is to bet $1 each spin. If the wheel lands on red, he keeps the dollar he has bet and wins an additional S1. If it doesn't land on red, he loses the dollar he has bet. Suppose he starts with $3 (after all, he is a poor college student). He will keep playing until he either doubles his money (i.e., reaches S6 total) or goes broke, at which time he will leave the game. 1. Model the student's roulette experience as a terminating Markov chain. Use all positive funds balances as states, and add an additional state for going broke. Remember that absorbing states are typically given the highest state numbers and are grouped together. Provide state definitions and draw the transition diagram 2. How many spins can the student expect to play before he leaves the game 3. What is the probability that the student reaches S5 at some point in the game? 4. How many times can the student expect to have $3 as his funds balance'? 5. What is the probability that he leaves the game with $6? 6. Suppose now that the student finds a game that uses a European roulette wheel, in which case the probability of landing on red is 18/37. Now what is the probability that he leaves the game with $6 if he starts with S31

Explanation / Answer

as we know probability means number of possiblity of that task .

like he think there is probability of red is coming is 0.5 which means 1/2 .

so it means in 2 chances he win in 1 task and loss in one task ..

But actually probability is 18/38 which means

in 38 games 18 times red is coming and 20 times other color is coming .

so it means in 38 games he win $18 for 18 games and loss $20 for 20 games .

overall in 38 games = 18-20 =-2$ it means he loss $2 in 38 games .

1$ loss in 19 games .

for lossing 3$ he need to spin 19*3 =57 games .

2) he have 3$ in starting so for breakdown he need to loss these $3

and as per above solution he need to play 57 games for lossing $3.

3) red color come or win $1 probabilty =18/38

    red color not come or loss $1 probabilty =20/38

so for $5 he need to win first 2 games . or 3 red 1 other or 5 red or 3 not red and so on ..

RR + LRRR + LLRRRRR ... till + 18red 16 not red

18/38 * 18*38 + 18/38* 20/38* 18/38 * 18*38 + ..so on

18/38 * 18*38 ( 1+ 18/38* 20/38 + [18/38* 20/38]2 +.......)

so in bracket this is gp with 14 term and a=1 and r = 18/38* 20/38

we solve aand get the answer .

4) for $3 he need to balance one win one loss so overall 0 and 2 win 2 loss ..like 18 won and 18 loss

so as per above we get the probability .

5) same as $5 here we need $6 .

for there 3 red , 4 red one not red .. and so on .

6) now for red =18/37 and for not red =19/37

and as per ques number 5 we solve the answer .

For any query ,please comment .

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